Great problem, great explanation. However, I still have a problem. You prove the solution right by using simulation, but you don't explain WHY the options GB and BG are differente. Neither does wikipedia... Does anyone have a good explanation as to WHY this is so? I know it's the right solution, but why???

Good article. It's too easy to be fooled.

Check out this talk, which explains it quite nicely: http://www.youtube.com/watch?v=3pRM4v0O29o

Using the outcomes/all combinations is somehow strange - it would be easiest and less prone to these errors to use this approach:

1) first hypothetical toss after their forced ending:

50:50 that the result will be H or T => this means that either with probability 0.5 will H win and with probability 0.5 will the game continue with score 2:2

2) second hyptothetical toss - score 2:2:

50:50 that the result will be H or T => probability 0.5 for H to win, probability 0.5 for T to win

To sum up:

H either wins in the first toss with probability 0.5 or in the second toss with probability 0.5 (first toss wins T) * 0.5 (second toss wins H) = 0.25, total 0.75

T wins only in the following scenario - he wins first move AND second move - 0.5*0.5, total 0.25

---------------------------

Also, this approach is wrong:

HH HT TH TT

as there never will be the toss HT.

The second approach:

H TH TT

is correct, BUT we need to consider that the first option happens with greater probability, it should be depicted as:

H T(H T)

or by assigning probabilities based on the length of the outcomes:

H(0.5) TH(0.25) HT(0.25)

All what the first approach does is that it flattens the probabilities to the same value for each outcome by adding theoretical next toss... I don't know how about others, but it seems pretty unobvious to me compared to these. I don't see the reason why we should add theoretical unexisting moves, when we have other methods to show the same results in more clear way...

@ Bumhunter "but you don't explain WHY the options GB and BG are differente."

Because we do not know if the said girl is the older or younger child.

@axequalsb

They're different because you're enumerating all of the results of two coins being flipped:

1. coin A=heads, coin B=heads
2. coin A=tails, coin B=heads
3. coin A=heads, coin B=tails
4. coin A=tails, coin B=tails

Because having one heads, and one tails (HT TH) constitute two choices, that means having one heads and one tails is twice as likely as two heads (HH).

When we know at least one of the coins is heads, we eliminate an entire possibility:

1. coin A=heads, coin B=heads
2. coin A=tails, coin B=heads
3. coin A=heads, coin B=tails

Because having one heads, and one tails (HT TH) constitute two choices, that means having one heads and one tails is _STILL_ twice as likely as two heads (HH).

So any one choice is 1/3, but because "having one be tails" (or boy) exists in *two* of the choices, the resulting probability is 66%

@axequalsb: somebody in the previous post made it clear, at least to me: the distribution of the 2-child families is 25% BB, 25% BG, 25% GB and 25% GG - that's why BG and GB must be accounted as different groups.

It is simpler to think like this: a guy can have 2 girls, Mary and Jane, and a boy, Joe. But not 2 boys, because this was ruled out.

So the possibilities are
Joe and Mary
Joe and Jane
Mary and Jane

The other doesn't matter so I won't permute.

The key is that even though he said he got a girl, he could be refering to Mary or Jane, both are girls but they are not the same girl.

How does GG not include girls?

Argh, I mean "the order doesn't matter", of course.

It still amazes me how many people still get this wrong. I agree that it's not that intuitive first time round, but it's a topic covered in high school Mathematics courses.

I assume that most people who read and comment on a blog such as this are programmers or at least have some interest in programming and I'd like to think we programmers are above average intelligence, so what hope is there for the general population?! :-)

Thanks to Jeff for bringing it up, and well explained.

I answered it correctly in the previous post. but then i had read something similar recently. so can't take the credit :)

As for BG and GB being different: combination-wise they both are the same, but permutation-wise they are different. The person has only revealed that one of his children is a girl - he didn't specify whether its elder or the younger one. So GB for girl being first child and BG for girl being the second child.

hope someone can explain it in more mathematically.

Ric, I don't think this has anything to do with intelligence or even education..it is more of one of those "a-ha" puzzles, once you get it it is obvious, but it is natural to be confused. My gf is a PhD student in maths, fcking genius doing all sort of crazy shit in gazillion dimensios, and yet she got confused as well.

For those who still think it's 50% read 5 posts up, giving the kids names is the ultimate explanation.

This example is rubbish.

The question is simply "What is the probability that my other child is a boy". The answer is 50%. The sex of the first child has no bearing on the sex of the second.

Some people mentioned that the reason BG and GB are distinct is because you aren't told whether the first girl is the younger or older child. This reasoning holds for the GG case too.

Jeff suggests that there are 3 possibilities and therefore the chance is 2/3, but he's wrong. If you're differentiating between BG and GB then there are 4 equally likely possibilities, given that we're told one child is a girl:

1. Older boy, younger girl, and I told you about the younger girl
2. Older girl, younger boy, and I told you about the older girl
3. Older girl, younger girl, and I told you about the younger girl
4. Older girl, younger girl, and I told you about the older girl

So, the chance that my other child is a boy is 2/4 or 50%

In a nutshell, if you're differentiating between Bg and bG (where uppercase indicates the older child) then you also have to differentiate between Gg and gG.

I just wonder where all the dissent is. All of this was explained thoroughly in the comments many times, but people would still reply with "Nope, you're wrong, it's 50% and it's so obvious" with all sorts of broken justifications... but when Jeff says it, apparently all these same people lose their conviction. The arguments didn't change.

Note that there was one good and correct argument for 50%, which depends on how the known girl was selected, so they were not ALL broken...

Oh look, there's one brave enough to try with a broken explanation. Well done Gareth.

If sequence matters shouldn't this

> BB, GB, BG, GG

be like this

> BB, BB, GB, BG, GG, GG

and this:

> GB, BG, GG

be like this

> GB, BG, GG, GG

giving a 50% chance for both boy and girl?

Because GG (or BB) have a sequence too: G1-G2 or G2-G1

So are you actually saying that once you've had one girl child the odds of having a boy increase????

This does not look like an unfinished game problem. An Unfinished Game problem must change its possible outcomes as the game progresses like the coin-toss game.

To make it clearer, I have two small bags each with one ball in. Each bag contains either a white or a black ball (with 50% probability).

If I show you that the first bag contains a white ball, what's the probability that the other bag contains a black ball? Obviously it's 50%

This more abstract example shows that trying to link the two events is ridiculous and they have no bearing on each other

I think the problem (and a source of a lot of the confusion) is actually is in the vague and imprecise sematics of the original question. Depending on how you interpret the question there are actually three possible (and correct!) results. For example, here are the three possible versions of the same question edited with slightly more precise semantics:

A) Let's say, hypothetically speaking, you met someone who told you they had two children, and [only] one of them is a girl. What are the odds that person has a boy and a girl?

Answer: 100%

B) Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person's other child is a boy?

Answer: 50%

C) Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the [overall] odds that person has a boy and a girl?

Answer: 66.66%

As ever in software engineering (and problem solving in general) one of the most important things before trying to solve any problem is to nail down what exactly is the problem you are trying to solve by asking precise questions so that you have clear requirements.

If either player wins, why do you have to divide the pot?

---

Either way, the sex of a child at birth is *independent* of any other child at birth. The "extra" information you speak of is not relevant at all - the fact that a couple has already had a child of a particular sex before is completely inconsequential to the sex of another child.

So why would it matter which order they were born?

---

After some thought it occurred to me that you could be suggesting that the probability of a heads occurring increases after every successive tail. This at first seems somewhat intuitively, but counteracts everything I know about (in my admittedly limited education) probability - and we seem to be taught that that intuition is wrong.

Put it another way: Does the chance of "your" lottery numbers coming up increase every time they don't come up?

--

It is possible I've completely misunderstood entirely. Please, enlighten me! I'm intrigued.

Well, I too will have another try to explain why some get 50 % and some 2/3:
The difference is whether
"you know the gender of one child, and it is a girl"
or
"you know there exists one girl in the family"

I do not agree that "and one of them is a girl" gives enough information to differentiate between these, but this is a question of language, not mathematics (which is why you don't use oridnary English when you do serious mathematics).

Gareth, when you change the example, you have to take care not to change the situation.

Your balls in bags example is completely different to the original question.

Try putting two balls in the same bag with the same probability for black and white (that is, each ball is selected with the same 50% probability). Now throw away any bag that has two black balls. What's the probability that a given bag has both a black and a white ball?

Gareth, your example with balls is incomplete because you are not counting that in this ball game you cannot have 2 black balls, just like we were told a priori that there were not 2 boys.

So, if I know there can't be 2 black balls, and I take 1 black ball out, it is 100% certain that the other one is white.OTOH if I take 1 white ball out, the other may be white or black, sure, but the chance of having at the end of the game 1 white and 1 black is 2 out of 3, 2/3. Capice?

@Sigh:

"You met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

It's a simple step to get to:

"You met someone who told you they had two balls, and one of them is a white. What are the odds that person has a white and a black?"

And this is identical to my example:

"I have two small bags each with one ball in. Each bag contains either a white or a black ball (with 50% probability).

If I show you that the first bag contains a white ball, what's the probability that the other bag contains a black ball?"

@Lerxst: There are two completely independent bags, the fact that you have already seen a white ball, and that you can only open one more bad means that it is impossible to see 2 black balls.

in the game gb and bg equivalents would be different
but in 'this context' they are the same.
the thing you're ignoring is that it's all based on context.

@Lerxst: Yes, the case that there are two black balls is ruled out by me showing you the first white ball.

@Sigh: It's you who is changing the question and the situation by talking about throwing away balls. My example is semantically identical to the original question. The reasoning which transforms the original question into a 3-state question is flawed.

This is not the Monty Hall problem

Gareth - it's not the same at all. You are talking about the 'first bag', which is like saying the 'oldest child'. You still have to a) explicitly exclude the possibility of two black balls, and b) not tell me which bag has a white ball in it.

I have two bags, each of which hold a randomly selected ball. What possibilities exist?

I tell you one of the bags definitely has a white ball. NOW what possibilities exist? What is the probability GIVEN WHAT YOU NOW KNOW that the other bag has a black ball? It's not Monty Hall, but it is conditional probability.

For those still insisting it's 50%, the analogy Lerxst did with names (Joe, Jane and Mary) might clear it.

@sigh: "a) explicitly exclude the possibility of two black balls,"

You look in only 2 bags.
Each bag only has one ball.
You have already seen one, and it's not black.

Possible outcome: (oh no not again..)

W + W
W + B

"b) not tell me which bag has a white ball in it"

The question says the there is [at least] one girl (white ball). Why can't it be told?

I don't agree at all with the "66%" (it's actually 67%, but I'll let that slide).

If the person has two children, there are _three_ possible combinations of gender: BB BG GG. If we rule out BB that leaves us with two options -- BG and GG -- which results in 50% chance of the other kid being a boy.

Sorry Jeff, but you fail.

The comment Sigh did about 2 posts ago is also spot on.

Come on guys, stop nerding, this is a holiday!!! :) You should be out skiing or whatever...

@Gareth (and others)

Suppose there are 100 families, all just got their first child. 50 of them have got a boy, and 50 a girl.

Next year, all 100 families get a second child. Of the 50 that already have a boy, 25 will have another boy, and 25 will have a girl. And of the 50 that already have a girl, 25 will have a boy, and 25 will have another girl.

So now there are 25 families with two boys, 25 with two girls, and 50 with a boy and a girl (in any order).

Now you're meeting one couple of parents. You know that they have at least one girl, so they definitely don't have two boys. They either are one of the 50 that have a boy and a girl, or they are one of the 25 that have two girls.

So, the chance that their other child is a boy, is 2/3 (50/75).

Fowl:
You're right, those are the only two options, but they are not equivalent in likelihood of existing.
It's like this (crappy ascii art ahoy):
+----+----+
| GG | BB |
+----+----+
| G+B |
+---------+
This is "all probability", where the chances of 2 girls are equal with 2 boys, and the 1 girl/1 boy option is twice as likely as either of those. No-one disputes this, I don't think.

When we eliminate BB as a possibility, "all probability given what we know" looks like this:
+----+
| GG |
+----+----+
| G+B |
+---------+

b) Why can't it be told?
Because we weren't told which child was the girl. Likewise, we can't be told where the white ball is. It's not knowing that makes the probability different. If we knew "the bag on the right" had a white ball, the probability of a black ball in "the bag on the left" would be 50%.

I'm with Gareth. The way the original question is worded does not (in my reasoning) indicate that order is important and so to me GB and BG are the same thus 50%.

Yvon: Thank you. Finally an argument that can't be argued against. Let's stop this now.

@ChrisP: You are right! I was trying to solve the wrong problem :)

results.Add("H", 0);
results.Add("TH", 0);
results.Add("TT", 0);

for (int i = 0; i < 10000; i++)
{
string result;

if (rand.Next(2) == 0)
result = "H";
else
result = (rand.Next(2) == 0)? "TH" : "TT";

results[result]++;
}

Jeff, ever heard about permutations? You're talking about permutations not combinations ;)

Can we agree that

"You met someone who told you they had two balls, and one of them is a white. What are the odds that person has a white and a black?"

is identical to:

"You met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

I'm only using this abstract "ball" form to get away from any preconceptions about people.

I tell you that each bag has a 50% chance of having a black or white ball. That's the same as the "any child has a 50% chance of being a boy or girl" assumption.

Does anyone really think that if I take a white ball out of one bag and give you the other, that there's *isn't* a 50% chance of the ball in your bag being black? That 50% probability was the very assumption we started the question with!

The context in which the "person I just met" told me they had a girl of course matters.

"Hi, I have 2 kids, the girl really loves his Barbies."

In this case, if the person did in fact have 2 girls, he/she would've never referred to either one by gender, but rather by name.
So the other one "logically" has to be a boy - 100%.
Statistically, it's 50/50.

On the flip side, Gareth tried to use the rock analogy.
I'll give it a try with sodas.

A person has 2 cans of soda in their fridge, they can be either a Coke, or a Pepsi. We know that one is a Coke. What are the odd the 2nd is a Pepsi?

This renders the whole "who's older than who" problem irrelevant.

possible combination of SEXES of two children is: BB, BG/GB, GG. order has absolutely no bearing and neither does 'unfinished' game

Like using bits you get 2,4,8,16 etc possibilites each time you add a digit, if you knew you value of one of these bits it would half the number of possibilities. You can't have 3 potential outcomes.

The unfinished game example is correct (since having one eliminates another possibility) but you still can't apply the same reasoning to children. Having one child does not eliminate the need to consider a second child for victory in the same way.

@Yvon. That is a different situation than the one presented.

Once again, when order does not matter, BG = GB and there are only three possibilities. BG (which equals GB), GG, and BB. With BB eliminated there is only GB and GG. 50/50. Since order does not matter, we already have G happening first, with no weight. We have the first (established) child, and we are not sure about the second.

This is basic *ordered set* theory versus *unordered set* theory. The fact that some teachers (or at least students) confuse the two is sad.

Gareth - your two questions are initially equivalent.

Repeat your example with 1000 people and tell me how many people got a black ball and how many got a white one. Remember, you must hand them a white ball in every case, and that white ball must have been paired randomly with another bag. That is, the outcome is not the same if you just hand every person the same white ball and pretend it was paired with the one in the bag.

Practicality: I have a three sided dice that says "GG", "BG" and "BG" on the faces. My result set is {"GG", "BG"}, but that doesn't mean the chances of each are 50/50.

Of course it is 50% in the example given.

BG and GB, in this case, ARE the same. The question asked is what are the chances of the other one being a boy when one of them is a girl. The fact one of them is already a girl does not affect the chances of the other one being either gender.

So why are BG and GB different.

2/3 is definitely not 66%, 66.66% nor 67%. Why do you stick to inaccurate decimal representation when you can use PRECISE 2/3? I don't get it.

This reminds me of the Monty Hall problem in that until I read that ONE explanation that clicks with me, I sway back and forth (thanks Yvon!).

Jeff, thanks as always, and keep up the thick skin.

@Gareth

You seem to be missing the crucial point that in your balls example you have explicitly opened one bag. The "first" bag as you call it.

This is different from the statement

You open "a" bag.

Consider the possibilities when you label the bags in this case left and right

Left Right
W W
W B
B W
B B

Clearly if as in your statement you tell me what is in the left bag then indeed the chance of getting a white ball in the right bag is 50%

However in the case where I tell you that in "a" bag (without specifying which bag) there is a white ball then you can see that there are three cases where that holds true.

Of those cases two have the mixed case and 1 is the identical.

As PHP code (1 is a girl, 2 is a boy):

<pre>

$iterations = 100000;
$has_two_girls = 0;
$has_a_boy_too = 0;

for ($x=0; $x < $iterations; $x++) {
$children = array();

$children[] = 1; // first child is born, a girl
$children[] = rand(1,2); // second child is born

if (in_array(2, $children)) {
$has_a_boy_too++;
}
else {
$has_two_girls++;
}
}

print "Has two girls: ".round(($has_two_girls / $iterations)*100)."%<br>\n";
print "Has a boy too: ".round(($has_a_boy_too / $iterations)*100)."%<br>\n";

</pre>

Please tell me why does this code print out two "50%" every time and regardless of whether the girl was born first or not.

Someone flips 2 coins that you can't see the outcome of. Possible outcomes: HH HT TH TT.

If he tells you the first coin was tails then the possiblilities are: TH TT (50-50)
If he tells you the last coin was tails then the possiblilities are: HT TT (50-50)
If he tells you a coin was tails then the possiblilities are: TH HT TT (67-33)

If you're not told /which one/ is a girl the the chances are: 67% boy, 33% girl.

A child has a 50% probability to be a boy (and more precisely around 51.2%).

But this number is calculated with MILLIONS of birth. In a GLOBAL human population there is about as many girls as boys, but in a particular family with only TWO children, your probabilistic problem is total NONSENSE.

I know families with five boys and no girl, three girls and no boy...

So the only good answer is : one child may be a boy or a girl, INDEPENDANTLY OF THE OTHER.

@sigh

You have a two sided "dice." Nothing about the question says that GB is more likely than GG.

Let's re-phrase this.

THE *FIRST CHILD* IS THE ONE ALREADY ESTABLISHED.

THIS IS WHAT WE ALREADY HAVE:

G?

(Question mark representing the unknown)

Again.
G?

What will the second one be?

We cannot have BG. It does not exist. We already established the first child. YOU CANNOT HAVE BG because the first child is G.

So the two ordered possibilities are:

GG
GB

@Lasse

That would be true if we are meant to guess the order of birth, or in your case the order of the coin toss results. Since we do not care about the order in this case.

So the difference between BG and GB is not one to be taken into account.

MichalT - You did not start with two 50% events, that's why. You have forced the oldest child to be a girl.

Practicality - if you want to ignore probabilities while calculating probability, I don't see how you can be helped. I am going to sleep.

So, the girl / boy question was just an unfortunate example for a probabability problem, ignoring monozygotic multiple births, relative gender birth rates and common spoken word conventions. Slightly disappointing.

@Lasse

The order is important BECAUSE it causes BG to be different from GB. They are therefore two separate scenarios each weighted equally.

If they were not then we would expect that 1/3 of couples have two boys, 1/3 of couples have two girls and 1/3 have mixed.

Clearly this is wrong...

@Sigh.

It does not matter if you force one of them to be the oldest, the age has no relation on the question. The question is, What are the chances that they have a Boy and a Girl. I don't think it matters in what order they are a boy and a girl so BG and GB are the same.

Oops not @Lasse but @Damien!

All you 50%ers baffle me. The means to get to the answer of the problem experimentally are all around you. Get a friend and a deck of cards. Deal him two cards from the deck. He looks at them and shows you a black one if he has one (if he has two red cards you restart). Then you make an even-money bet on the color of the remaining card. Whoever takes black in that game is going to hemorrhage money, because the probability of the second card also being black is 1/3, not 1/2.

Or do it with coins that are flipped where you can't see them. Etc etc.

Most people get this one wrong the first time they hear it, but to persist with the wrong answer when you can prove it to yourself in the physical world is baffling.

@Damien,

WIth your logic you make the chance of having a boy and a girl less than 50%!

@Alan

Yes, but we don't not care about the order, so BG and GB are not different.

In the coin toss example the order is important as it requires one of them to score successive times to win. So the order of the results are important.

@Sigh: like I said, it prints out two "50%" regardless of whether the girl was born first or not. If you swap:

$children[] = 1; // first child is born, a girl
$children[] = rand(1,2); // second child is born

for:

$children[] = rand(1,2); // first child is born
$children[] = 1; // second child is born, a girl

...the result is exactly the same (which is rather obvious). Try it.

The mistakes people are making when saying it's 50% is that they're answering the wrong question, and connecting the fact that gender's random in a faulty way to the information that one is a girl.

Let's go to the "balls" territory:

2 bags, 2 random balls (black or white). Now I look in both bags, and tell you "i've got at least one white one". That's not the same as opening one bag and looking at it (which is clearly 50%). By giving you information about the entire system (2 balls - 1 is white), I've given you information about the random probability of the other ball, since it's a part of the random system.

From the beginning, there's 4 combinations of equal probability - BW, BB, WW, WB. I open both bags, and tell you it's one of WB, WW, or BW, so the colour of the second ball in this system is now dependant on this piece of information of the entire system. Since the possibilities were already established to be equal, they're still equal and the chance is 1/3 the other one is also white. This is not the same as "what's the colour of the ball" when I first looked in the bag.

In the same way, the question "what's the chance these parents have a girl and a boy?" is not the same question as "what's the chance their third child will be a boy?"

When I modify your simulation for the B/G problem ($result="G", tossses=1,...) , the results are:
[Perl]# perl BoyGirl.pl
GB 5021
GG 4979
-----
Your first example enumerates ALL possibilities, those done and those not. The H/T problem (and Monti Carlo solution) only looked at the completion moves. Doing the same simulation to the B/G problem looks close enough to call 50%.

at least this discussion demonstrates why software is in the sorry shape it is in.

@Damien: The order is just there to show that the 2 coin tosses are different. Having a boy then a girl is different from having a girl and then a boy. The "coin toss" in this case is the conceptions and each conceptions is unique. Otherwise you are assuming that all boys and all girls are exactly the same.

BG mean the boy was born first and the gilr was born second.

GB means the girl was born first and the boy second.

@Damien

You DO care about the order as there are two events happening. Therefore 4 outcomes can happen. When assessing the chance you need to consider all paths through.

This is the crucial piece you are missing.

If we were awarding 5 points for a win and zero for a loss and had a large number of people competing over two rounds we would expect half the competitors to have 5 points. For the sake of argument say 100 people.

So take that group out of the main set. In the main set give the 25 people with 10 points a Blue t-shirt and the 25 with 0 points a Red shirt.

Now go to the group with 5 points and give them a yellow shirt. That's 50 yellow shirts.

Now we'll send the people with no points home.

So then we'll put you the other side of a curtain and send through the remaining people one by one. Before they come through you have to guess whether they have a yellow shirt on or not. Every time you guess right you get a prize. Everytime you get it wrong you lose.

You'll end up a lot better off if you guess yellow everytime as there are 50 yellows.

What's the definition of a yellow? Having got one right. If the cases were no different then you would not have so many yellow shirts.

It is this difference that causes the 2/3's when considering the question asked

@MichalT: What you're saying now though is that there was no randomness involved -- that the couple were destined by some fate to have a girl, making it impossible for them to have a boy.

Whether that's possible is a different discussion altogether... but in doing that, you've destroyed the random system we're trying to discuss.

What we have is essentially people who have tossed two coins, then told us one of the results (but not which one). That's not the same as only tossing one coin.

@ MichalT
Now you force particualar gender to the second child. Both childer should have the same probability to be boy or girl.

Updated code, maybe this will make it more obvious (though if you still stick to the "66%", I doubt it will).

<?
define('GIRL', 1);
define('BOY', 2);

$iterations = 100000;
$has_no_girls = 0;
$has_two_girls = 0;
$has_a_boy_too = 0;

for ($x=0; $x < $iterations; $x++) {
$children = array();

$children[] = rand(GIRL,BOY); // first child is born
$children[] = rand(GIRL,BOY); // second child is born

if (!in_array(GIRL, $children)) {
$has_no_girls++;
}
elseif (in_array(BOY, $children)) {
$has_a_boy_too++;
}
else {
$has_two_girls++;
}
}

print "Has no girls: ".round(($has_no_girls / $iterations)*100)."%<br/>\n";
print "Has two girls: ".round(($has_two_girls / $iterations)*100)."%<br/>\n";
print "Has a boy too: ".round(($has_a_boy_too / $iterations)*100)."%<br/>\n";

?>

The BG / GB difference still has me confused.
The only difference would be in order (boy older than girl, or girl older than boy).
You know one is a girl...but she must be either older (GB), or younger (BG) so at that point you're then get back to 50/50 as you eliminate 2 of the choices - BB BG GB GG

Assertion 1: Assuming fair coins, when flipping two coins, the outcome of one coin flip does not affect the outcome of the other flip.
Assertion 2: Assuming fair coins, when flipping two coins the chances of getting one heads and one tails is double that of getting both tails.
Assertion 3: Assuming a 50% chance of having a boy or girl at birth, the probability of having a boy and a girl is double that of having two girls.

The point here is that having two possibilities does not mean that they are each equally likely. Assertion 2 can be easily demonstrated, and follows from assertion 1; assertion 3 is the same as assertion 2, only changing to b/g logic instead of h/t.

Statement 1: Assuming a couple with no children, the probability that their first child will be female is 50%.

Statement 2: Assuming a pregnant couple with a single female child; the probability that their next child will be male is 50%.

Statement 3: Assuming a couple with two children, at least one of which is female. (3a) There are two possibilities: Both are female, or one is female and one is male. (3b) Per assertion 3, the latter is twice as likely as the former. (3c) Ergo, two times out of three the other child will be male.

@Alan

Well ok, If we do care about the order then I accept it's 2/3. I just don't get what we care about the order.

You guys who argue the 50/50. In what way is Yvons example incorrect?

Ah dam it, I just got it.

@MichalT:

Your code outputs:

Has no girls: 25%
Has two girls: 25%
Has a boy too: 50%

Now I tell you it's not option 1. Adjust the percentages of the remaining possibilities?

I asked yesterday if people who provided wrong answers would admit that they had been wrong.

Obviously it is not the case here :-D

Actually, let's suppose that the person talking is not a statistician, but an avarage English speaking person. Then oviously the probability of the second being a boy is 100%. Common sense - who but a statistician, a programmer, or somebody wanting to trick someone would expect such an answer.

@Mikael Hedberg: Please tell me how to change the code so that it arrives at a different percentage. I don't see what you mean.

I'd say it's 50%.

If you have two children, you have BB, BG, GB, or GG. One of them is a girl then it's not BB.

When you mention that one of the children is a girl it's 25% chance you are talking about a girl in BG, 25% a girl in GB and 50% (25%+25%) a girl in GG. (Four girls and you're talking about one of them with equal probability.)

Then chance to have a girl and a boy is 50% because it's either BG or GB, not GG.

I get it now. It's in the wording as usual.

If the question was:

1. Given a family with two children, at least one of which is a boy, what are the chances of the children being a boy and a girl?

Answer: 2/3

2. Given a family with one boy, if they have another child, what are the chances of it being a girl?

Answer: 1/2

Paul

http://en.wikipedia.org/wiki/Boy_or_Girl

RT -- the trouble is that you don't know which of the GB or BG choices to eliminate. Certainly if you knew which it was, it'd be 50% -- but how can you eliminate one of them without knowing which one to eliminate?

@Damien and other 50% people:

What Alan Wray said - this is about the number of ways to get the same outcome. There are FOUR possible ways to have two children (let's pretend twins do not happen for simplicity).
Three of them involve having a girl. And there comes the unfinished game part: once your first child is a girl, you have eliminated the possibility of having two boys.

If you talk to a person who has two kids there is a 75% probability that one of them is either
boy or girl. Once you KNOW the gender of the one child, you have eleminated the case of having two kids of the sex opposite to the named – hence you reduced number of outcomes from 4 to three, and two of them involve the kid of the opposite sex as named one.

The question should have said "at least one of them is a girl" to avoid the confusion.

MichaelT: Your last program produces correct results -- it shows that there is a 50% chance of it being one boy one girl, and a 25% chance of it being two girls. (Demonstrating my assertion 3).

However, unless you're arguing that there's a 25% chance of a couple with at least one girl having no girls, you're looking at a set that includes more than we need.

Try:

print "Has two girls: ".round(($has_two_girls / ($iterations - $has_no_girls))*100)."%<br/>\n";
print "Has a boy too: ".round(($has_a_boy_too / ($iterations - $has_no_girls))*100)."%<br/>\n";

To finish it up.

Given the sentence structure of the original question, I couldn't look past if I was being asked to decide the probability of whether I was talking to a child-bearing hermaphrodite.

That didn't seem terribly likely, but I was unable to put a number on it.

The key part of the question is "ONE of them is a girl". It doesn't say which one is first girl or boy. I think that is what is tripping people up because they are making the assumption that the first born is a girl, which is wrong. Based on that we get the following possibilities GB, BG, GG which 2 are correct ie. 2/3.

Nice question Jeff. So you are a contestant in a game show, and the game show host tells you there is a prize behind one of the three doors...... ;)

MichalT - Your program is wrong because it always assumes that the first born is a girl ($children[] = 1;), which isn't what the question was asking.

Gareth, your analogy is wrong because you are giving two pieces of information: that one of the balls is white and also you are telling which bag contains the white ball.

That is why the probability is different. In the original problem, the man just says that one of his children is a girl but does not tell which one.

@MichalT:

Ok, well in telling you "at least one is a girl", you need to remove the results that was "no girls". So you get this:

add after the loop:
$iterations -= $has_no_girls;

... and then remove the first print line.

I had a long post written explaining stuff, but I've deleted it. The 50%ers will not be moved.

I abandon this discussion with the following request: if you believe the odds are 50%, let's find a time to get together and gamble for lots of money.

And a quick question for Jeff: had this been posted on StackOverflow, would the right answer have been voted to the top? If not, what does that say about the SO approach?

Solution is flat out wrong, it assumes a dependency that does not exist. If I flip a coin, its a 50% chance of being heads, and a 50% chance of being tails. This is independent of, and completely regardless of the outcome of a previous coin toss.

This is exactly the same. Having one child that is a girl has absolutely no bearing on the sex of the other child, its still the same completely random 50% chance for each sex.

Since there is no dependency, this question really boils down to: What are the odds of a child being a boy? Just because we were given additional information does NOT make it relevant.

I don't think the palm of my hand has spent more time on my forehead for quite a while than it has when reading over the debates and responses to a very simple logic problem. At least now I know why that legacy codebase I always seem to inherit always seems to use 10x as much code as it needs to.

MarkG, you are wrong in your understanding of he question.
(This repsonse was mainly written for MichalT)

This is because you have defined the *first* child to be a girl. The question does not state this, it says *one of the children* is a girl.

Think about it with numbers, 0 is a boy, 1 is a girl:
a = randint(0,1)
b = randint(0,1)

Saying the *first* child is like saying the following:
a == 1
what are the chances of b being 0?
1 / 2

Saying *one child* is a girl is like saying:
a + b > 0
(a or b)
Either a or b. There are three possible situations where this is true:
a == 1 && b == 0
b == 1 && a == 0
a == 1 && b == 1
(I hope this is obvious to programmers, it's just the OR truth table)
What are the chances of the right hand column being 0?
Essentially:
(a or b) and (a nand b)

Python code:
from random import randint

girl=0
boy=0
for i in range(10000):
a=randint(0,1)
b=randint(0,1)
if (a or b):
if not(a and b):
boy+=1
else:
girl+=1
print boy
print girl

I think I understand what you guys mean. The difference is in how you understand the question, and the answer is either 50% (the probability of the other child being a boy) or 67% (the probability of an average person having a boy amongst two children, one of which is said to be a girl.) I'm not sure if one is more correct than the other, but now I see how one could argue that 67% is a correct answer.

As a way of saying "sorry", here's a related quote from "Waking Life" (one of my all-time favourite movies):

Creation comes out of imperfection. It seems to come out of a striving and a frustration. This is where, I think, language came from. I mean, it come from our desire to transcend our isolation and have some connection with one another. It had to be easy when it was just simple survival. “Water.” We came up with a sound for that. “Tiger behind you!” We made a sound for that. But when it gets really interesting, I think is when we use that same system of symbols to communicate all the abstract and intangible things we’re experiencing. What is “frustration”? Or what is “anger” or “love”? When I say “love” the sound comes out of my mouth and hits the other person’s ear travels through the byzantine conduit in their brain through their memories of love or lack of love. They say they understand, but how do I know? Because words are inert. They’re just symbols. They’re dead. You know? And so much of our experience is intangible. So much of what we perceive cannot be expressed. It’s unspeakable. And yet, you know, when we communicate with one another and we feel we have connected and think we’re understood I think we have a feeling of almost spiritual communion. That may be transient, but it’s what we live for.

I've just written some code to test this out, and in the procces (and result) I've come to the 2/3 chance (25% no girls, 50% boy+girl, 25% boy boy) conclusion.

code is here: http://pastebin.com/f70b3e253

>=1Girl | 2Girl | 1 boy
750556 | 250417 | 500139
749833 | 249842 | 499991
...

I'm still quite confused by the implications. It *feels* like I'm being told that past events influence future events.

The distinction between these two will now give me nightmares ;)

"1. Given a family with two children, at least one of which is a boy, what are the chances of the children being a boy and a girl?

Answer: 2/3

2. Given a family with one boy, if they have another child, what are the chances of it being a girl?

Answer: 1/2"

If only we had some sort of formal way of expressing mathamatical constructs!

Wait a minute... :P

@Lerxst Thanks for giving them names, it was the key to getting me to understand why BG and GB are different.

@MichalT:

Well, even if you make that difference, the question is still the same. "Is the other child a boy" is a weird but valid question, and the reason you get put off from the right track is the fact that you dont know which one the first child is, so it makes little sense to ask which one the "other" is.

In essence, the questions are the same, and the answer is still 2/3. The difference happens if they tell you "our oldest child is a girl" -- then the dependance on the overall system disappears and the answer comes out as 1/2.

I won't get into the argument of whether or not this is the Monty Hall problem. Each of you can decide for yourself.

But I do encourage everyone who found Jeff's blog entry interesting to read the entertaining and explaining "Monty Hall, Monty Fall, Monty Crawl" paper by Statistics professor Jeffrey Rosenthal which you can find at: http://probability.ca/jeff/writing/montyfall.pdf.

And you can play it online here: http://www.nytimes.com/2008/04/08/science/08monty.html?_r=1#

Why not just rephrase the question

"Let's say, hypothetically speaking, you met someone who told you they had two children, and AT LEAST one of them is a girl. What are the odds that person has a boy and a girl?"

Then there can be no argument about the semantics of the sentence.

Do people not read through all of the comments, or is there really a difficult mental block to overcome when talking about this? Someone way up at the top explained why this works the way that it does.

To all the 50%ers....
BG and GB are only the same if you don't mind attributing a different probability to (BG/GB). In other words, you can combine BG and GB if you agree that the combined item has a 50% chance of occurring. Why is this? It's explained way at the top.

The probability that a given child will be a boy is 50% (for the sake of argument.) So when the couple has their first child, the chances that they have a boy is 50%. When the couple has their second child, their chances of birthing a boy is still 50%. So the chances of a couple having two boys is 25% (0.5 * 0.5). The chances of a couple having two girls is also 25% (0.5 * 0.5). The probability that the couple have a boy and a girl is equal to:
1 - (chances of two boys + chances of two girls) or 0.5 (50% chance.)

However all of the 50%ers are considering the probabilities of two boys, two girls, and one of each to be equal. By the above, they clearly are not equal. That's the error that almost every single one of you is making.

At issue is the fact that we're not just asking a straight probability question--we're asking a question and giving you more information so that you can come to a more accurate answer. If I flipped a coin and told you that it did not come up heads side up, and asked what the probability was that this coin was tails-side up, would you say, "Well, the odds of a coin coming up heads is 50%, so the odds of it being tails in this situation is 50%!" Of course you wouldn't. You have extra information which allows you to eliminate some of the possibilities. It's the exact same thing here--extra information lets you eliminate possibilities, yielding an answer which is counterintuitive.

""1. Given a family with two children, at least one of which is a boy, what are the chances of the children being a boy and a girl?

Answer: 2/3"

This is only correct if you were asked the question without being told that there is already a boy.

To simplify things, think of a coin toss. If you asked the question what are the chances that I will flip a head 2 times in a row, then the chances are 1/2 x 1/2 => 1/4. But if you flipped the coin and got a head, then asked the question to what are the odds that you will get another head, it is now a 1/2 chance. The fact that you already flipped a head once doesn't change the probability on the second flip.

So if you are told that one of the kids is a girl, there is a 50% chance the the OTHER child is a girl as well, and a 100% chance that the kid you were told was a girl is that.

In the end there is a 75% chance that they are both girls and a 25% chance that there a boy and a girl.

If that still doesn't make sense think of it like this. If I win the lottery once are my chances even worse to win it again?

The thing that made it confusing for me is that there is a very big difference between knowing that "one is a girl" and that "the oldest is a girl". If you know that the oldest (or youngest) is a girl, then the probability of a boy and girl goes back to %50, which you would expect. Having that extra information really matters.

Each successive coin toss has 50:50 odds of coming up heads or tails (presuming that the coin is weighted evenly), irrespective of earlier tosses.

As for the boy/girl scenario: the mathematics must be based on biological premises.

Just my two cents :-)

BG can't occur because you have already established the first one is G.

So GB is not twice as likely.

If you toss a coin then its 50 - 50 heads or tails, if I throw the coin and get heads then there is still a 50 - 50 chance I will get heads or tails on my second throw, the first throw is irrelevant how can it at all effect the result of the second coin toss ? Its total BS that I am 66% likely to throw a tails on my second throw and unless someone can come up with a rational explanation of how its physically possible that the act of tossing the coin the first time can affect the second coin toss then i will continue to believe its total bullshit.

Practicality:
No! You have not established that the FIRST one is G, you have established that ONE of them is G. The question states that ONE was a girl, not that the FIRST or LAST was a girl. As such, you do not know whether or not we are talking about BG or GB, and both must be present in the final tally to determine the true probability.

Thanks ChrisP, you made it clear for me.

Matthew:
The coin has already been flipped both times, and the results recorded. I tell you that ONE of the flips was heads (but not which one.) That means that out of these possibilities:
HH HT TH TT
we can eliminate all of the outcomes in which none of the flips were heads. That leaves us with the following possibilities:
HH HT TH
So the odds of one of the flips (again, not a specific flip) being tails is 66%.

It's not one of the independent events affecting the other event, it's knowledge of one of the flips which lets us eliminate an outcome.

Try it yourself. Flip two coins. If there's at least one HEAD, record the flip. After 30, count how many times there was a TAIL. It should be about 20 times (according to statistics.)

Does anyone else wish that these comments had Stack Overflow style voting up/down?

@rick: "but it's a topic covered in high school Mathematics courses"

Not in my school. Atleast, not in the 80's.

And to those who said "will he/they admit when they are wrong?"

Hell yes. Happy to now the answer is out in the open.

I was wrong. Totally.

But as Jeff pointed out, I'm in good company. :)

I was initially convinced the answer should be 50% even in the face of Jeff's explanation. However I think what's missing from the discussion is the fact that in a large population of 2-children families there is a 50% probability of having mixed-sex children and a 50% probability of having both children of the same sex.

I think it is helpful to imagine meeting unknown parents out walking with one of their children, say a girl. That means they are not one of the 25% of families with 2 boys. They may either be one of the 25% of families with 2 girls, or one of the 50% of families with a girl and a boy. So the chances of them being a G/B family are 50/75.

Some of the contributors suggested that the birth order should be ignored. But it's not the order that is relevant, rather its the fact that in the overall population B/G and G/B each appear with the same frequency as GG or BB.

50% is the right answer.

The possible combinations of children is right: BB, GB, BG, GG

But when you know at least one is a girl you can't just take the BB out of the equation. Because if you do so you are taking the order in consideration in the GB BG, but not in the GG. So, when you know at least one is a girl, instead of "GB, BG, GG" (as stated in the post) we stay with:

"GG, BG" (if the order doesn't matter)

or

"Gg gG BG GB" (with order into account, being capital G the girl we know exists)

both give 50% chances of the other being a boy.

Thank the Lord we do not program in English.

@Yvon: Your restatement of the problem shows that if time is taken, English can get fairly exact.

@Gareth: Your 100% correct when looking at the problem the way you are.

@Jeff: If you were a project analyst at my company and proposed this vague of a problem, you would have probably been fired.

Everyone calm down on the holiday! With the very poorly written problem, it is the just like the picture that if looked at one way, its a pretty girl in a flowing dress. Looked at another way, its an old crone with a big warty nose.

@chrisP: You summed it up when you rewrote the problem to show all the ways it can be interpreted.

Lee

http://books.google.com/books?id=yqUmpHyBU0cC&pg=PA200&lpg=PA200&dq=%22permuations+on+unordered+sets%22&source=web&ots=7a3GcS2yu6&sig=JiSarrZEbWdB3v_9DLcMYfxcvV0#PPA200,M1

Read "unordered sets without replacement"

Note "the order in which the cards are dealt makes no difference in the hand"

Scroll up and read "unordered pairs without replacement"

Note "Thus, there is no basis for distinguishing between (A, B) and (B, A)"

So, once again order is irrelevant, and unweighted. If you want to include the order, then G has already happened.

Continue reading and notice that you have to divide the number of permutations by 2 to get the number of *combinations*

Originally there were *4 permutations*, but only *3 combinations*. We eliminated a permutation or a combination.

So the likelihood of getting 1 of the correct permutations (implying that order matters) is 1 in 3. The likelihood that you get the correct combination is 1 in 2.

This exact problem is why half the class failed discrete mathematics. It's only a permutation if you care about order. Nothing about the original information implied that these were ordered sets.

If you still think that because order is arbitrary that it doesn't matter, I would ask you to think this over for a few minutes.

Saying that GB is no different from BG is the same as saying that any of the sums resulting from the roll of a pair of dice between 2 and 12 are equally likely.

"Read older entries »" down, at the bottom of your blog, is working NOT as expected!

You're not serious, are you? You're using a 0.37% difference to say that they're not equally probable?
Do an infinite number of coin flips, and the difference will go to 0.

So the thing that makes the difference, if you care about order, then you have to say that there are two solutions with girls. One where the first girl is the one we already know about, and one where the second girl is the one we do not know about.

Ordered solution:
G1G2
G2G1
GB
BG

Or
Girl-mentioned. Girl-Unmentioned
Girl-unmentioned. Girl-mentioned.
Girl-mentioned. Boy-Unmentioned
Boy-unmentioned. Girl-Mentioned.

Unordered solution:
GG
BG

joao,

we don't know which of the girls is G, so you're double counting the GG case.

Put another way, we can take into account that the girl the parent is talking about is another random event, so consider the following 3 cases, each of which has a ~33% chance of occuring:

GB: 100% likely that the parent is talking about G
BG: 100% likely that the parent is talking about G
Gg: 50% likely that the parent is talking about G, 50% likely that the parent is talking about g.

Thus we have P("a boy and a girl") = 33%*100% + 33%*100% = ~ 66%
And, P("two girls") = 33%*50% + 33%*50% = ~33%.

It all works out, even if you think there are two GG cases, you just have to assign the probabilities correctly.

@mike.

It is precisely because you don't know which one of the girls is the one we mentioned that you DO have to double count GG.

If order matters there are two! solutions with girls. One where the girl we know about is born first and one where the girl we know about is born second.

Practicality:
If order doesn't matter, the odds of GB are still 50%--twice as high as either BB or GG.
If order does matter, then BG and GB ARE UNIQUE, each with an equal share of the probability, and you can't eliminate either one until you know for sure which child you were initially told about is the girl.

So GB (unordered) or GB+BG (ordered) is twice as likely as GG. How do you explain your answer (0.5 probability) if one of the outcomes is twice as likely as the other?

Remember, when you eliminate BB as an option, you don't get to arbitrarily assign the remaining probability to the GG outcome.

Practicality,

You're still not making a leap of understanding here.

What you're not seeing is that the information we're being given is incomplete information about the sum of the result of *two* random events.

Indeed, we're talking about the outcome of twice sampling into an ordered set with replacement, thus (B,B), (B,G), (G,B), and (G,G) are all different outcomes. (B,B) is the only case that is ruled out given the information in the problem.

Practicality:
We don't double-count anyone because we don't know which one of the children (first or second) was the girl. Look at the possible outcomes:
GG GB BG BB
We aren't double-counting BG and GB. We're separating out those possibilities as orders to account for the fact that having a boy and a girl is twice as likely as having either two girls or two boys.

In a population of 1000 couples with two children:
50% will have a boy and a girl
25% will have two boys
25% will have two girls.

You don't double the girls. You double the (unordered) BG because it's twice as likely to occur as either BB or GG.

In that 1000 couple population, if we eliminate all "two boys" outcomes (because we know that there's at least one girl), then you have:
500 couples with a boy and a girl
250 couples with two girls.

Thus, when Jeff posed the question, the probability that there's a boy is 500/750, or 2/3.

I groked that order matters thinking about the words "possible outcomes".

G 1st B 2nd and B 1st G 2nd were all possible outcomes of the scenario. You couldn't just combine them, one had to occur and not the other. The parent either had Boy or Girl first, thus, both situations had to be considered. Another way, without the 'one of the child is a girl' yet:

GG - 25%
BG - 25%
GB - 25%
BB - 25%

Since BB is impossible:

GG - 33%
BG - 33%
GB - 33%

And since the question was asked:

BG - 33%
GB - 33%

Ok, there are still lots of confusion on this, so I think I'll finally throw in some explanation. Yes these are disjoint sets, which make intuition say 50%, this is indeed wrong based on one thing; you don't know whether you were told about the first or the second child.

So as seen many time's already, we have the truth table as follows:
BB
BG
GB
GG

Now many of you say, well if we know the outcome of one birth, has no bearing on the other, which is absolutely true. However, we don't know which birth we are being given, the first or the second. This means we can only remove one combination, which is two boys.

BB - Not possible since first or second birth is a girl
BG - Possible, since we know one child is a girl
GB - Possible, since we know one child is a girl
GG - Possible, since we know one child is a girl

Now we take the resulting set, and 2 out of the 3 combinations have one child as a boy, and the other as a girl. This is why they need to be combined together.

If I told you that I have two children, and my first was a girl, we would get the following set:
BB - Not possible since first is a girl
BG - Not possible since first is a girl
GB - Possible, since we know first child is a girl
GG - Possible, since we know first child is a girl

If we take the now available set, the second option is either a boy or a girl, which is 50%, and what most people expect. This also works in reverse if we're given the second child.

The key piece of information which most people appear to have missed is that we simply don't know what we're given, the result of the first birth, or the second, or subsequently, the first coin toss, or the second. It&#8217;s a very important piece of information which cannot be assumed.

Practicality,

You may double count the gG case if you wish, but you must then assign a probability that the parent is talking about g or G, which is 50%. This cuts the resulting chance of gG in half, leaving you back at ~33%

mike, my point is, if order is taken into consideration (as it was in the post) then we have:

GB: 100% likely that the parent is talking about G
BG: 100% likely that the parent is talking about G
Gg: 100% likely that the parent is talking about G
gG: 100% likely that the parent is talking about G

If order is not taken into consideration, we have:

GB: 100% likely that the parent is talking about G
Gg: 100% likely that the parent is talking about G

In your example you considered order in the boy-girl case, but not in the girl-girl case, and that's wrong.

joao:
Good show. If you decide to care about order at all, it must be done uniformly for the math to be correct. Practicality should be eating his condescending words about his discrete class.

joao,

But we don't know that the parent is talking about G. We only know that the parent is talking about one of G or g. If we were introduced to G and were asked, what is the sex of her sibling, you would be absolutely correct.

Unfortunately, that's not the problem being stated here.

@Matthew, @Chris:

You are both making an invalid comparison. You have to throw two coins in any case. You can't assume that the *first* throw is heads - that is like assuming that the *oldest* child is a girl.

One of the consistent things that the 50%ers seem to be missing is that you have the assumption that both events have already happened independently, and each of those events has 50% chance of boy or girl.

What the 50% people are missing:

We know that *one* of the kids is a girl. We *don't* know which one. If we knew which one, the answer would be 50%.

You can't just use the set {2 girls, 2 boys, 1 of each}, or a subset of that, without accounting for what Jeff bolded near the end here: NOT ALL THE OPTIONS HAVE EQUAL PROBABILITY. *That* is why BG and GB are considered different; they each have the same probability as each other and as BB and GG.

----

Each birth has a 50% chance of being one or the other. If you know that the first birth is a girl, that doesn't change the chances of a girl in the second one. And vice-versa. BUT WE DON'T KNOW WHICH ONE IS A GIRL. That makes all the difference.

Sorry guys, I often disagree with Jeff, but this time he's right on. If you still don't get it, read the book he's talking about.

If sequence doesn't matter then: (BB BG GG).
If sequence does matter then: (BB BB BG GB GG GG)

The problem definition doesn't require sequence. This reminds me of sorting algorithms.

Hey Jeff, thanks for posting more again. A while back your blog got thin from all the work you were doing on the other site. You are a good writer.

BG and GB are only two cases if you add the cases G1G2 and G2G1, because if you care about order, then you must also care if the girl you know about came first.

So if you care about order, the original possibilities are:
B1B2
B2B1
G1G2
G2G1
B1G1
G1B1

(Note that you can't have a second boy without having a first boy, or a second girl, without having a first girl)

You have eliminated
B1B2
B2B1

The remaining possibilities:

G1G2
G2G1
B1G1
G1B1

Have of them are mixed.

*half (not have)

The big difference between whether the answer is 50% or 33% is if you know which of the two children is a girl.

Without the knowledge of which one is a girl, the answer must be derived by looking at all of the possiblities. However, if you know which one is the girl then you can remove her from the equation and have a 50/50 chance that the other is a girl.

The coin toss is entirely different because you don't know the order of the last two coin tosses and the outcome relies on the order in which the coins come to rest.

If you can remove the outcome of one, the probability changes.

How about this problem:

Step #1) A couple has two children, and they do not have two boys.
What is the probability that they have one boy and one girl?

Step #2) How is that different from the problem originally posed?

You should read The Drunkards Walk, it illustrates this problem and similar statistical problems quite well :)

Sorry, Jeff, but you're wrong about the boy/girl thing. B/G equals G/B because the order does matter.

As you said, there are four basic combinations: bb bg gb gg.

"Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?". The first one I know about is a girl, so the available possibilities are: gb and gg. bg is not available because the first one I know about is a girl, and the first one in that combination is a boy.

Also, you could put it this way:

"Let's say, hypothetically speaking, you met someone who told you they had two children, and I know the gender of one of them. What are the odds that person has a boy and a girl?".

If the first one you know about is a boy, the combinations are bb and bg, which makes it 50/50. If the first one you know about is a girl, the combinations are gb and gg, which makes it 50/50.

Also, as you know the gender of one of the children, the problem can be simplified to:

"One person tells me he has a child whose gender is unknown to me. What are the chances it is a boy?". That's 50/50.

You can also increase the problem.

"Let's say, hypothetically speaking, you met someone who told you they had three children, and two of them are girls. What are the odds that person has a boy and two girl?".

This problem is equivalent to yours, but with your way of solving it, the chances are:

bbb no
bbg no
bgb no
bgg yes correct
gbb no
gbg yes correct
ggb yes correct
ggg yes wrong

3/4. That's not correct.

The thing is that you're attaching the possibilities of having a boy to whatever he had before, and that's not true.

Thanks.

Joe:
These aren't sets with equal probability. In your example, couples have a 1/3 chance of having two boys, 1/3 chance of having two girls, and 1/3 chance of having one of each. That's simply not the case. Because each child is an independent event, you have a 50% chance each time of having a boy, and a 50% chance each time of having a girl. Thus, the probability of having two boys is (0.50 * 0.50) or 0.25. The same for girls. So with a 50% chance of having both children of the same sex, you must have a 50% chance of having one of each sex. If not, please tell me what the other possibilities are, and what their chance is.

The reason that we spell out the outcomes as GG GB BG BB is because then, each outcome has an equal probability; in other words, it's easier to glance at four equally probable outcomes and note the probabilities than, say, "(0.25)GG + (0.50)BG + (0.25)BB". Technically order doesn't matter, but we pretend that it does for the BG case because it means that we can deal with outcomes with equal probabilities.

In truth, order doesn't matter, and we're not dealing with equal weights, which is the real reason that the answer is 2/3.

Practicality:
"So if you care about order, the original possibilities are:
B1B2
B2B1
G1G2
G2G1
B1G1
G1B1"

You are suggesting equal probability of having 2 girls, 2 boys, and 1 of each. Do you know why that is incorrect?

The 2/3 crowd is getting confused by the element of time.

Before we knew that one was a girl, there were 4 possibilities. Once we know that one is a girl, there are 2 possibilities (one is mixed).

We didn't just eliminate one possibility, we are now presented with an entirely different situation. Before we heard about a girl, we had a situation where half of the outcomes were mixed.

After we heard about the girl we had a situation where the only unknown was the gender of the remaining child. A 50/50 scenario.

Once again "my statistics book" told me so, or "my statistics teacher" told me is a logical fallacy. Appeal to authority does not make your argument any more valid.

Everybody who thinks the probability is 50% needs to explain why they think the probability of a couple with two kids having a boy or girl in any order is 1/3. That is what they are saying when they say order doesn't matter and treating the three cases as having the same probability. Instead, the probabilities are really 25% for two boys, 50% for boy and girl, and 25% for two girls.

You can work the problem without ordering but you have to keep track of the different probabilities for each case. Knowing that one of the children is a girl means that the remaining cases are the boy and girl (50%) and two girls (25%). The probability of a boy is (2/4)/(3/4) = 2/3.

Doesn't this assume that the mother knows the sex of the second one? That's what I struggled with. WHAT IF SHE ONLY KNOWS THE SEX OF THE FIRST? Lets say the second is unborn, and she hasn't found out the sex yet. Then, the odds are 50%. The information that changes this is the fact that we ASSUME she knows the sex of both, which is more information that WE have.

It looks like we will be unable to convince you. That makes me very sad.

I urge you to try a real life experiment with coins, or at least to look back up at the scenario I gave with 1000 people. You should be able to find it easily with your browser's "find" feature. It clearly shows the probabilities.

I'm perfectly willing to entertain different explanations as to WHY the probability comes out that way, but there's no argument on why, and thus no logical fallacy to consider--it's simply fact.

@sadnessbowl

That is correct. Those are the chances. Do you somehow think that it is more common to have mixed children?

It looks like we will be unable to convince you. That makes me very sad.

I urge you to try a real life experiment with coins, or at least to look back up at the scenario I gave with 1000 people. You should be able to find it easily with your browser's "find" feature. It clearly shows the probabilities.

I'm perfectly willing to entertain different explanations as to WHY the probability comes out that way, but there's no argument on why, and thus no logical fallacy to consider--it's simply fact.

Even if someone wanted to view the birth order as important, you'd still have {B1B2, B2B1, BG, GB, G1G2, G2G1} as the possible permutations. If one is guaranteed to be a girl then you'd be have the subset as {BG, GB, G1G2, G2G1} or 50% of having a boy and a girl. If the order is not significant (which the problem NEVER states that it is), it would still be a base set of {BB, BG/GB, GG}, one girl guaranteed and you have {BG/GB, GG} 50% again. I can't figure out how they go from saying BG and GB are distinct, yet ignore the GG has having 2 permutations.

I wonder if everyone realizes that given a couple with two children, there's a 75% chance of having at least one boy, and a 75% chance of having at least one girl.

Oh my God.
Say a couple has 20 kids, boys and girls. What are the odds the 21st will be a boy? 50/50.
Regardless of how many kids there are, the next one will always be either a boy or a girl. 50/50.

John:
You are correct. But that's not the question that was asked.

If the person had said "the oldest is a girl" or "the youngest is a girl" then the odds go back to 50-50 on the other one.

This is easy

Odds of Having an All Same-Gender Family
If there are roughly even odds of having a boy or a girl with each baby, given the laws of chance we should still expect to see some all same-gender families, even in large families. Here is the number of all same-gender families we would expect to see, purely by chance:

Family Size Same-Gender Mixed-Gender
2 Children 50% 50%

http://www.in-gender.com/xyu/Odds/Gender_Odds.aspx

Practicality,

We've explained this ten ways to Sunday, provided several analogical problems, even provided code which you can run to empirically see that the only correct answer here is 2/3.

Now, what you keep missing, perhaps because there are so many explanations being made, is that the information we are given (the family has a girl) is covering two separate random events, the birth of child X and the birth of child Y. There are two children, two "flips of the coin," two random selections from a set {B, G} with replacement. What follows directly from that situation is that there are four possibilities with equal likelihood:

X=B, Y=B
X=B, Y=G
X=G, Y=B
X=G, Y=G

Once we are told that the family has no boys, we eliminate one and end up with:

X=B, Y=G
X=G, Y=B
X=G, Y=G

The ods of the family having one boy and one girl is 2/3. The odds of the family having to girls is 1/3.

I'm not sure how much clearer it can get than that.

If you're not seeing how independent events can have interdependent outcomes, consider rolling a pair of dice. The chances of getting a 7 are much higher than getting a 2 or a 12. For the same reason, the chances of a family having a boy and a girl is higher than having two boys or two girls.

Are you getting it now?

blah,

Good point!

@Practicality:
Yes, it's twice as likely to have mixed kids as to have two girls, and twice as likely to have mixed kids as to have two boys.

Again you're assuming we know *which* child is a girl. We don't. If we knew that then the answer would be 50%.

#!/usr/bin/python

import random

possibilities = (
("Boy", "Boy"),
("Boy", "Girl"),
("Girl", "Boy"),
("Girl", "Girl")
);
counts = {};

for key in possibilities:
counts[key]=0

for i in range(0,1000000):
counts[random.choice(possibilities)] += 1;

for key in possibilities:
print key, ": ", counts[key]

Oh, I get what you are saying. With the odds of the three categories. That confuses the issue though.

Once again, we are not in a situation where we have eliminated 1 of 4 possibilities, we are in a situation where we have 2 possibilities (the remaining child is a boy, or the remaining child is a girl).

It's a new situation, not dependent upon what the possibilities could have been had we not known about the girl.

There are not two permutations of GG if we agree that the first letter is the first one born. (It has to mean something and maintain that same meaning throughout the entire analysis).

@Practicality:
You're almost there. 50% same gender, 50% mixed gender.

Now, 50% of the same-gender families are all-boys, and 50% of them are girls. That means your gender percentages become:
50% mixed gender
25% two boys
25% two girls

In the question, we know that there's a girl, so that eliminates the two-boys option. *Of the remainder*, 2/3 are mixed gender and 1/3 are all-girls. QED.

Practicality,

"Once again, we are not in a situation where we have eliminated 1 of 4 possibilities, we are in a situation where we have 2 possibilities (the remaining child is a boy, or the remaining child is a girl)."

What piece of information presented allows us to remove more than one of the four possibilities? All we know is that the family has at least one girls (which is logically equivalent in the frame of this question to saying that the family does NOT have two boys).

@mike

"the chances of a family having a boy and a girl is higher than having two boys or two girls."

That is false from a chance standpoint.

Although the data says it is slightly more likely (52% vs. 48%).

Practicality:
When you were in school, did you throw out information you learned in previous classes when taking a new class? That's what you're doing here.

We know that there's a 50% chance of having a boy and a girl. We know that there's a 25% chance of having two boys, and a 25% chance of having two girls.

When you say that at least one of the children is a girl, we eliminate the 25% chance that there are two boys. That 25% chance must go somewhere.

For those that still persist in believing that 50% is correct, that order may or may not matter, or that it's a trick of wording, sit down and flip some coins. Don't think about it - do it! This was suggested by someone else yesterday, but it seems to have been ignored. Do the following:

1. Flip two coins.
2. If both come up TAILS, go to step 1.
3. Remove one HEADS.
4. Write down the side of the remaining coin.
5. Go to step 1.

If I've described this correctly, this should be the same as the GB problem, and you should have written down roughly twice as many TAILS as HEADS.

How does this map to the GB problem?

1. Family has two kids.
2-3. At least one is a GIRL.
4. What gender is the other kid?

Practicality,

Sorry, my phrasing wasn't quite clear, and you're right. What I meant to say was:

"the chances of a family having a boy and a girl is higher than having two girls."

Boy Girl and Girl boy are different because you're doing on the odds of that outcome. You're basically adding up the odds of them having a boy and then having a girl to the odds of them having a girl and then having a boy.

Boy Girl, Girl Boy is the same... but it's twice as likely to happen than BB or GG. That's why it's included. There's 2 ways they can have boy girl, whereas girl girl or boy boy there's only one.

English language is not math.

It's easy to hide behind tricks like this. The problem is not clearly specified. It's presented as a real world issue, but in the real world a couple which has a younger boy and an older girl and another one which has a younger girl and an older boy, both are said to have "one boy and one girl". In normal speaking English, nobody assumes the born order counts.

So it's not the same problem as the one with the coins.

Jeff is quite a pragmatic person, I'm kind of dissapointed on him insisting that BG and GB are different.

@Practicality:
Yes, there are two possibilities, but those two possibilities have different probabilities.

It actually is important to look at the entire universe of possibilities first, in order to determine the probability of each one. After that we use the additional information we're given to eliminate some of those possibilities -- but we still need to take into account the different probabilities in the initial universe.

You have "protected" sex with someone. Afterward, you test negative for chlamydia. Then you have protected sex (using a second piece of protection) with a second partner. Examining your two pieces of protection, you find that one was perfect, but the other was faulty. Then you find out that one of your two partners had chlamydia when you did the nasty, but you don't know which one it is, or which protection you used with which partner.

Assuming a 100% infection rate when the protection is faulty (and 0% when it's not), what are the odds that you have contracted chlamydia?

I can't believe you guys are still arguing this. Accept that you were wrong, look at the 500 different explanations of why this is so, and think about it. Or just try the damn thing yourself. It's a simple experiment.

Guys, could you please help me. I have 2 boys (seriously, Ryan & Alex). My wife wants to have another baby, but only if there's a good chance it will be a girl.

What are the odds that if we have another child, it will be a girl?

:-)

@Derek

Exactly! You have twice as many heads because ONE OF THEM IS THE GIRL WE ALREADY COUNTED!

You are counting the total number of girls and the total number of boys. Half of the girls you would count in your coin flipping scenario were actually in a mixed family.

BG and GB may be semantically the same, but lumping them together INCREASES the probability of their occurrence. This is the original problem:

1. "Standard" child distribution of 100 two-child families:
- 25 have GG
- 50 have GB/BG
- 25 have BB

2. Remove the families that don't have at least one girl:
- 25 have GG
- 50 have GB/BG

3. What is the probability within set (2) that the other is a boy?

The comments to this are making me crack up laughing: so many people debating about the correct statistical stuff, percentages, algorithm...

I agree with both ChrisP and David Sheeks in the above comments:

ChrisP:
[...] I think the problem (and a source of a lot of the confusion) is actually is in the vague and imprecise semantics of the original question. [...] As ever in software engineering (and problem solving in general) one of the most important things before trying to solve any problem is to nail down what exactly is the problem you are trying to solve by asking precise questions so that you have clear requirements. [...]

David Sheeks
"[...] So, the girl / boy question was just an unfortunate example for a probabability problem, ignoring monozygotic multiple births, relative gender birth rates and common spoken word conventions. Slightly disappointing. [...]"

ROTFL :-)

As for the programming examples, if you are counting the number of heads, you are in error. You have to count the number of times the tosses are mixed versus the number of times they are the same. You will find they are 50/50.

@Gigi:
BG and GB are different solely because there are two ways of having both a boy and a girl, so that option is twice as likely as the other two. While there are legitimate quibbles about the vagaries of English in the problem as posed, that's not one of them.

"@Derek

Exactly! You have twice as many heads because ONE OF THEM IS THE GIRL WE ALREADY COUNTED!

You are counting the total number of girls and the total number of boys. Half of the girls you would count in your coin flipping scenario were actually in a mixed family.
Practicality on January 1, 2009 10:44 AM "

He's removing one HEADS. That takes care of the HEADS in mixed groups. It takes care of the issue of having "at least one HEADS" as in the problem statement. If you perform this experiment enough times, the HEADS:TAILS ratio will give you an idea as to the correct probability.

@Practicality

Please re-read the 5 steps, and actually try it if you don't agree with the proposed results. Don't count the H removed in step 3, since that's one of the given conditions of the original problem. The problem boils down to: if you flip two coins and at least ONE of them comes up H, what are the odds that the other comes up T? Nothing about FIRST or SECOND, or LEFT or RIGHT.

I repeat, if you're in the 50% camp, actually try it.

@Practicality: (on counting heads)
"You have to count the number of times the tosses are mixed versus the number of times they are the same."

Nope. The question is "given at least one A, what's the probability of the other one being B." The question is not "what's the probability of one A and one B in all sets of two."

Practicality,

Exactly! I think you're getting it now!

Now, when you remove the TT combination as a possibility, you'll find that there about 2 of 3 tosses will have one T and one H.

Thus, given two coin flips, and the knowledge that at least one of them came out T, then there's a 2/3 chance the flip lead to exactly one T and exactly one H (in any order).

Everybody please do Derek's problem of flipping the coins and see that you have twice as many tails as heads. Then understand that those represent individual coins and not outcomes of both coins.

Then understanding that the fact the the first coin is already a tails as a given, means that half of those tails are the tail we already know about.

So half of your tails would be a tail matched with another tail (included in the same half). And half of your tails would be a tail matched with a head.

Practicality,

You keep packing more information into the statement of the problem then was actually there. We are not given _which_ one of the two children the parent was referring to. We only know that they are not both boys (i.e., one is a girl).

This is why the question maps directly onto (and into) the coin toss problem.

You're given the information that one of them is a girl - then the chance you would bet on that person having a girl would be 100% -

In other words, while the order here is arbitrary, *you* don't get the pick the order, the parent gets to pick. And that makes all the difference.

Once again, you are taking a ratio of heads:tails. But that is not what we are counting! We are counting mixed:same.

I'm getting the feeling that Practicality is trolling us.

Let's repeat the problem statement:
"Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

This is the same as saying, "If a parent tells you that he has two children, and that one of them is a girl, what is the probability that the other is a boy." It's the same because it's a given that one of them is a girl.

So it's perfectly acceptable to count the ratio of boys to girls. This is because if the other child is a boy, then the family is mixed. If the other child is a girl, then the family is not mixed.

There is a 2/3 chance that the family will be mixed, because there is a 2/3 chance that the other child will be a boy.

sadnessbowl,

I think you may be right WRT the trollness of Practicality. :-/

Is there anyone else here that still thinks that the answer to the question as posed is 50%?

Let's assume there are 4 couples with the following children assuming birth order: BB, BG, GB, GG

Show me what is wrong if I assume their children are BB, BG, BG, GG (no order preservation).

We still have 4 boys, for girls, each couple has the same childrens, even if each child is distinct (we actually have B1+B2 B3+G1 G2+B4 G3+G4)

What I'm saying is that B3+G1 == G2+B4 == BG, the same way that B1+B2=BB, even if B1 != B2.

Again, English language is not math. Mathematicalize everything and clearly state the problem (you have this set of things, what is the probability .....) if you want to reach an agreement, because everybody interprets the English statement their way. That's why math was created, so that the same truth is valid for everybody, because there's no way to misinterpret (if you know math and don't make a mistake)

@sadnessbowl

Caught you! You aren't counting the other child. You are counting the total children.

@mike

From my perspective you would be trolling me. Aside that such a statement is not relevant to the conversation.

Ummmmmmm

this is not correct. It is 50/50. There is no ordering. The correct analogy is, if they have decided that they have gambled on 3 coin tosses and are tied. there is one more coin toss (is the other kid a boy or a girl) and its chances are 50/50.

This is the most horrifying example of people not having any common sense when it comes to probability and never knowing when to apply which set of probability rules.

Gigi,

I'm not sure I see the point you're trying to reach. This has been explained to death, both in how it maps to a probability question as well as how that probability question is correctly answered.

Surely you see from your example that if we eliminate the couple that had no girls (i.e., the couple revealed to us that they have a girl), that 2/3 of the remaining families have one boy and one girl.

Practicality, I'm astonished at your continued failure to understand this. It's been explained a thousand times. Why don't you simply run the numbers, as suggested? It's a trivial matter to write a simple program to solve this, similar to the one Jeff posted. Run it over and over and over again, and it will unambiguously show that the 50% solution is wrong, because you are not weighting the mixed-gender outcome correctly (despite earlier linking to - and quoting - a page which exposes your error). Just run the damn numbers instead of arguing based on a flawed premise, will you?

I agree with the poster that said "no wonder our industry is in such a mess".

John,

You have reframed the question by assuming you which child is being revealed. Once you do that, you are indeed looking at the result of just one randeom event. But you can't do that since you don't get to pick which child is being referred to.

@Gigi. You have a point.

Anyway a correct sequence would be this. (heads represents girl)

1. Put one heads down.
2. Flip and see how many are heads versus how many are tails (we will see 50/50.

That would be if the girl is born first.

Then.
3. Flip and see how many are heads versus how many are tails (we will see 50/50)
2. Put one heads down.

That would be if the girl is born second.

Flip to see if the girl we already know about is born first or second. (50%). And yes it would apply to all cases, since if it was a case where it was heads/heads, then the girl we already know about can still be born first or second.

(the second 2 should have been a 4 there)

Are we not programmers? Write a small program examining the situation. Here's a Perl script. You can download the official Perl interpreter for Windows for free from http://activestate.com. Don't worry. I kept the code readable.

#! /usr/bin/env perl

$NUM_OF_RUNS = 100000;

$boyGirl = 0;
$girlGirl = 0;
$boyBoy = 0; #Not even looking at!

for($count = 0; $count < $NUM_OF_RUNS; $count++) {
if (int((rand(1) + .5))) {
$firstChild = "boy";
} else {
$firstChild = "girl";
}
if (int((rand(1) + .5))) {
$secondChild = "boy";
} else {
$secondChild = "girl";
}

if (("$firstChild" eq "boy") and ("$secondChild" eq "girl")) {
$boyGirl++;
} elsif (("$firstChild" eq "girl") and ("$secondChild" eq "boy")) {
$boyGirl++;
} elsif (("$firstChild" eq "girl") and ("$secondChild" eq "girl")) {
$girlGirl++;
} else {
$boyBoy++; #Not looking at this
}
}

$total = $boyGirl + $girlGirl; #We aren't counting boy/boy pairs.
print "Number of Runs: $NUM_OF_RUNS\n";
print "Number of Times One Child is a Boy: $boyGirl\n";
print "Number of Times both Children are Girls: $girlGirl\n";
print "Total number of legitimate situations: $total\n\n";

printf ("Ratio of Boy/Girl in Situation: %4.2f\n", ($boyGirl / $total));
printf ("Ratio of Girl/Girl in Situation: %4.2f\n", ($girlGirl / $total));

===

Here's an example of the output:

Number of Runs: 100000
Number of Times One Child is a Boy: 49834
Number of Times both Children are Girls: 25100
Total number of legitimate situations: 74934

Ratio of Boy/Girl in Situation: 0.67
Ratio of Girl/Girl in Situation: 0.33

The main thing to remember is that we are not looking at each birth as a separate coin toss, but a sequence of two coin tosses. There is a first child and a second child. Notice we run this test 100,000 times, but we are only looking at 74,934 of those times.

The situation would be different if the question said "The oldest child is a girl, what is the possibility of one of the children being a boy?". In that case, we are only considering the sex of the second child. That would be 50%.

To the 50% people:
This is a solved mathematical problem. The answer is 2/3 as stated. This would not be Jeff Atwood being wrong, but a lot of mathematicians being wrong. Complains about the wording are fine, the question could have been better expressed, but instead of refuting the correct answer, do try to understand it.
please.

Practicality,

you have to flip both coins at the same time and look at the combined result afterward if you want to model this problem correctly.

Again, YOU don't get to pick which coin is heads, you only get to know that one of the two is heads.

Jeff,
The problem is that this is not statistical sampling. No matter how many children a person has, the next child is equally likely to be a boy or a girl (population or genetic statistics not withstanding).

I am sorry, but although the documentation may say 50% is wrong, it is absolutely and immutably correct.

Mike,

Let's assume there is a party and the four couples are invited there.

And the host says: "All the couples with a boy and a girl" please step forward".

What I'm saying is that the couples with B3+G1 and G2+B4 will both step forward.

Do you imply that only the B3+G1 couple will step forward?

If the problem was stated in terms of fruits, apples and oranges, there would be no "order" to talk about. I'm saying that people using English language treat B and G as apples and oranges, they don't assume order is important. If order is important, it must clearly be stated, otherwise you say something, but you imply another thing. This is deceiving.

I'm curions what would a court rule, if lets say somebody bet millions of dollars on the 50% probability answer, and then it was asked to pay.

You only have to count one child if you assume that we're only dealing with the set of parents who have at least one girl.

Draw a venn diagram of two circles with some overlap. Populate it with dots representing couples. 50% of the dots should go into the overlapping area. 25% of the dots should go into each circle, but not into the overlapping area.

Pick a random person from the entire diagram, and ask them if they have a mixed family. The odds will be 50/50.

NOW PICK A RANDOM PERSON FROM ONLY THE CIRCLE THAT INCLUDES FAMILIES WITH AT LEAST ONE GIRL. Note that this is subset that the problem describes. Ask the random person if they have a mixed family. Because we have eliminated every family composed of only boys, the odds will no longer be 50/50.

We have:

1st B / 2nd B - 1/4
1st B / 2nd G - 1/4
1st G / 2nd G - 1/4
1st G / 2nd B - 1/4

Then we take the BB out, it gets

1st B / 2nd G
1st G / 2nd G
1st G / 2nd B

Now comes the disagreement.. how do we distribute chances now?

Option 1

1st B / 2nd G - 1/3
1st G / 2nd G - 1/3
1st G / 2nd B - 1/3

Option 2

1st B / 2nd G - 1/4
1st G / 2nd G - 2/4
1st G / 2nd B - 1/4

I think option 2 is the right one. Why? because the girl can be both the first or the second.

In my above example, one circle should have been labelled "has at least one girl" and the other circle should have been labelled "has at least one boy."

Gigi,

What I'm saying is this:

At the party, someone says "all the couples with at least one girl step forward!" Three couples will step forward. Now, someone else says "all the couples with one boy and one girl step forward!", two of those three will step forward. This is yet another way of getting to the 66% answer.

The casinos must love you, Jeff, because you just fell for the Gambler's Fallacy: <a href="http://en.wikipedia.org/wiki/Gambler%27s_fallacy">http://en.wikipedia.org/wiki/Gambler%27s_fallacy</a>

Nobody's questioning the math or the code posted AFTER the set and it's possibilities have been established.
That's what the 2 sides can't agree on.

Is it: BB, BG, GB, GG, or,
Is it: BB, GG, BOTH

Noooo!!!!!!

Its only 2/3 if you know that all the mixed parents will tell you they've got a girl. In reality, half the mixed parents will tell you they've got a boy.

Its like if Monty Hall's producers stopped telling him where the car was. Do you really think he can help you any more?

What are the chances of a hermaphrodite?

joao,

Why do you think it's just as likely that a family will have two girls than as it is for the family to have one boy and one girl?

Scot McPherson, this problem is not concerned with the gender of the *next* child, which as you say is 50%. The problem is concerned with probabilities of events THAT HAVE ALREADY OCCURRED, once certain other probabilities have been removed.

If I toss a coin and tell you it's not heads, will you still argue that there's only a 50% chance that it's tails? After all, by your reasoning, a coin toss ALWAYS has a 50% chance of being tails, even after I've already told you it's not heads.

Powerlord,

You are misunderstanding the question. There are indeed two random events, but the information given covers the result of both, therefore we must examine both when trying to solve the problem.

My answer yesterday was wrong. It was not the answer you were looking for. I still disagree with *everyone* here :)

If this is absolutely a question on a test designed to test your understanding of logic puzzles or statistics, then all the arguments above have to be taken into consideration.

If this is a real world question... if I meet someone and they tell me they have two kids, and one is a girl... they MEAN the other is a boy. They aren't testing my ability to determine statistical chances of the other gender. A few people might say that but really have two girls, and just be goofs.

That's still my answer, even if it isn't right!

Russ,

I like that analogy! After over 1000 comments, that's the first time I've seen someone explain the situation that way.

Jason B.
I agree :) That's why it's better to phrase these types of puzzles very carefully.

Mike,

Yes, because I've started with 4 couples ASSUMING ORDER. If we don't assume order, we only have 3 couples, only one with both a boy and a girl.

I wanted to say with that example that both those couples will feel they qualify for the "couple with boy and girl" condition, even if the boy was born first in only one of them. Those two couples will not ask "When you stated the condition of a boy and a girl, which wan is supposed to have been born first?" They will not consider order.

I totally agree with the math probabilities in the coins, abstracts objects, whatever problem.

What I'm saying is that this is a deceiving stated problem. They assume something which is not clearly stated and which in fact is opposed to what "normal" (I'm not implying stupid) people assume.

This is like inssurance companies selling "fire insurance" but only if the fire .............. and ........... and ........ which leaves you bankrupt after a fire while assuming everything is peachy before.

"Why do you think it's just as likely that a family will have two girls than as it is for the family to have one boy and one girl?"

Because we know at least one of the children is a girl.

@Powerlord

From the first sentence of your linked article to the Gambler's Fallacy:

"...likely to be evened out by opposite deviations in the future"

This isn't about future events, it's about events that have already occurred. Gambler's Fallacy does not apply. If I toss a coin, look at the result, and tell you it's not heads, would you be guilty of the Gambler's Fallacy to believe that the probability of it being tails is greater than 50%? Sure, future coin tosses are 50%, but we're talking about a past event for which some possibilities have been ruled out.

The probability of the sex of the child is not tied in with the coin flipping because that is not stated in the requirements.

The answer to the coin flip is easy, since the conditions of the original bet were never met the monies should be returned to the original bettor.

Gigi:
With one child:
Odds of having a boy is 50%.
Odds of having a girl is 50%.

With two children:
Odds of having two boys is 25%.
Odds of having two girls is 25%.
Odds of having one girl and one boy is 50%.

Your three couple situation will not work. Do you not understand why?

sadnessbowl,

I agree that in a given population of 2 children families you'll have twice more families with B+G than only with B ang G.

What I say is that even if those families are twice as many, they will "consider" themselves as only one group, the "with boy and girl group".

They will not consider themselves as two groups, "with boy and girl group" and "with girl and boy group"

odds of 2 boys: 25%
odds of 2 boys and being told one is a girl: 0%
odds of 2 boys and being told one is a boy: 25%

odds of two girls: 25%
odds of 2 girls and being told one is a girl: 25%
odds of 2 girls and being told one is a boy: 0%

odds of one boy and one girl: 50%
odds of one boy and one girl and being told one is a girl: 25%
odds of one boy and one girl and being told one is a boy: 25%

Overall answer to question: 50%

I think it's most clear if you don't consider that there either is or is not an order of birth, or a difference in age.

It really doesn't matter, as the parent could have adopted both children and at the exact same time.

The important concept that I think one is trying to illustrate by considering order is actually the concept that each child is independent from the other. There really are two children.

If someone is a child, that someone is either a boy or a girl. If there is a child, there is a 50% chance the child is a boy or a girl.

In crib A is a child, so that child is either a boy or a girl.
In crib B is a child, so that child is either a boy or a girl.

The question says there is a girl. So there is a girl in crib A, or there is a girl in crib B. At least. For in this example "OR" includes the possibility of also meaning "AND"--that is, if there is a girl in both cribs A AND B, it is also true that there is a girl in crib A OR in crib B.

It is not possible that there is a boy in crib A at the same time there is a boy in crib B. This is because there is a girl in either crib A or in crib B.

Three possibilities that satisfy the parent's claim that one of the children is a girl:
1. There is a boy in crib A, and the girl must be in crib B.
2. There is a boy in crib B, and the girl must be in crib A.
3. There is a girl in both cribs A and B.

In two of the three possibilities there is a boy and a girl.

The probability of there existing both a boy and a girl is 2/3.

Gigi:
That's fine, but you can't use that as an example. If you want to use real-world examples to represent statistics, you must take into account the probabilities.

Look at it this way: there are two types of people in the world--those who have won the lottery, and those who have not. If I come across two people at a party, and ask them which one of them had won the lottery, they'd both look at me like I was crazy because you can't have a representative sample of people without considering the odds.

In order for the party example to work, you must have a representative sample. Assuming four couples, that means having one couple with two girls (25%), one couple with two boys (25%), and two couples with one of each (25% + 25%). Otherwise, you're talking about odds adding up to only 75%, and that simply does not make sense.

@xooorx

Even if we ignore the bizarre way you've worded it, you have still drawn the wrong conclusion from your own data. Since we know that 'two boys' ISN'T the right answer, that 25% is eliminated. Therefore, your 50% answer is actually from a problem space of only 75%, since 25% of the possibilities have been eliminated. 50/75 == 2/3, which is the correct answer.

Putting BG/GB in the same basket is NOT A MISTAKE. You insist on the fact that we should use the information provided in the question to answer. The question DOES NOT say anything that could lead the reader into rightfully assuming that the order in which the children were born should be taken into consideration. Therefore, the answer IS 50%, and you are MAKING UP the additional information that the order is important to justify that 66% is the right answer.

For 66% to have been the right answer, the question should have been "What are the odds that THE FIRST (OR SECOND) CHILD is a boy?", not "What are the odds that the couple has a boy and a girl.". Heck, my aunt has a boy and a girl. Can you tell who was born first?

When I wrote "There really are two children." I meant also that there is not merely a pair of children to consider. There are two things to consider, and they should not be considered as one thing, i.e. a pair.

xooorx:
The problem says nothing about being told about boys. You're adding in parameters to the problem. You can't do that without changing the problem fundamentally.

The only reason to state "I have at least one girl" is to eliminate the subset of [couples with two children] where both children are boys. That's all, nothing else. Given the knowledge that this person has at least one girl, and that mixed families are twice as likely as families where you have two girls, the correct probability is 2/3 that the family is mixed.

David != david

:)

@Russ

Look closer. There are 2 cases where you're told its a girl. Each equally probable at 25%.

To get 2/3 you have to assume that *all* mixed parents reveal that they've got a girl.

For me, this has moved beyond math and code fragments to one thing ... does order matter or not? If yes, then 66%. If no, then 50%.

sadnessbowl,

Ok, you convinced me. In the real world the probability is indeed 2/3.

I sense that the 2/3 advocates are feeling superior right now. At the risk of being stepped on, I'm changing my answer from 2/3 to 1/2.

It's true, and it's been proven many times now, that if you sample families from the entire population (where BB, BG, GB, and GG occur with equal frequency), the mixed gender families are twice as common among families with girls.

However, I want to approach this problem like a gambler, so I need to be realistic about the sampling. If you are talking to someone (The question says, "you met someone...", not "you sampled the population of families in a non-biased way", and this is important.) and it is revealed to you that this person has at least one girl, is it still true that BG, GB, and GG each occur with equal frequency? I don't buy the simplistic explanation that BG and GB are "the same," and therefore they occur with equal frequency to GG.

However, I think that in this particular sample (the existence of a girl being revealed), there is an overrepresentation of GG which is double that of either GB or BG. This is because the situation wherein you meet somebody with at least one girl is twice as likely if they have two girls. As some people have pointed out before, you are not encountering a randomly chosen family so much as you are encountering a randomly chosen girl. As others have pointed out (I think EVERYTHING that can be pointed out has been so far; I'm just trying to summarize.) the answer to the question is different when any detail or specificity at all is attached to the "at least one" girl.

This specificity, in my opinion, is a much more realistic way of thinking about the problem. Anyone who says that 2/3 is the correct answer will agree that if someone gives the birth order of the "at least one" girl, then the answer is 1/2. Birth order, however, is an arbitrary ordering trait. We can choose to order siblings in any systematic fashion: who is taller, who is better at probability, whose name sounds more like "Mary." So if we meet someone who says, "I have two children. Let me tell you about Mary..." the probability changes, now that we know that in the classification scheme of Mary-soundaliking, this child is First (Adventures of Pete & Pete situations excluded). The relevant question is now "What gender is Mary's sibling?" And it's a 50% probability either way.

This is somewhat paradoxical, but it's true: as soon as you learn a single differentiating detail about a child that acts as the "proof" of the at-least-one-girl statement, you are dealing with a sample in which GG is doubly represented. In actual experience, such proof is the only method by which you learn that someone has at least one girl.Therefore, as the question was stated, I take 1/2 as the correct answer. The survey study that found an answer of 2/3 is right in its own way, but it uses a naive sampling method to represent the actual realistic situation.

David (capital D):
Right. Order doesn't matter. But probability does, and having BG is twice as likely as having GG. Having BG is also twice as likely as having BB. Due to the way the problem is phrased, though, we don't care about BB--we know that the person doesn't have BB.

Since having BG is twice as likely as having GG, the odds are 2:1 that the family is mixed. 2:1 odds equates to a 2/3 probability, or a 67% chance.

@David

"Putting BG/GB in the same basket is NOT A MISTAKE."

However, it would be a mistake to say that the BG/GB basket is not BIGGER than the BB and GG baskets.

If you have 100 two-children couples, statistically 25 of them will have BB, 25 will have GG, and 50 will have mixed. Eliminate the BBs, and the mixed-gender pairs account for 2/3 of what's left (50/75). Therefore, the answer is 2/3.

Really, given how Jeff has praised the readers of his blog as the best-of-the-best in the past, it makes me weep to see how many so-called software-engineers can't grasp this simple math. This is basically a truth table, replacing the TRUE/FALSE states with BOY/GIRL. Aren't truth tables one of the FIRST things CS students are taught?

@youfailatwordproblems

If you're telling me you've got at least one girl to HELP ME work out whether there's a boy in the family then yes the answer is 66%. If ALL parents with a girl say they've got one then the answer is 66%.

But in the question, the parent has just happened to reveal they've got a girl. This is TWICE as likely to happen if both are girls than if just one is.

@JamesR

If you think order does/does not matter, go flip some coins. Follow the 5 steps I outlined earlier.

sadnessbody:
No, the probability of having a combination of a boy and a girl regardless of what order the kids are born (50%) is absolutely not the same as the probability of having a boy THEN a girl (66%).

Again, the question does not mention or otherwise contains anything that indicates that the order is important, so the answer is 50%.

@Damien
> The question is, What are the chances that they have a Boy and a Girl [given that you know they have a girl]

Now change the question to: What are the chances that their next child is a boy, given that you know they have a girl.

The latter is clearly 50%. The heart of the confusion is that the latter is not the same question as the former.

That is, knowing that one of the children is a girl, of course.

You know, suddenly I can really see why puzzles like this are used during interviews. It's not so much the ability to think logically that's being tested, as much as the ability to read the original requirement. The amount of people trying to answer the wrong question is astounding.

xooorx:
That's why I said you fail at word problems. You're reading too much into it, assuming that there's a possibility of lies and such.

It's a word problem. It's there to convey concepts using English instead of math. Our job, then, is to translate the problem, and you failed at doing that.

@James R and 1/2ers

stop thinking about order.
stop thinking that a parent has two children.
stop thinking about this being a family.
forget the biological context.

there are two things that already exist.

describe each as either b or g.

bb bg gb gg are the four possible descriptions if you want to describe these two things at the same time.

until you cross off bb.

three descriptions.

two could describe this person's children.

one could not describe this person's children.

"No, the probability of having a combination of a boy and a girl regardless of what order the kids are born (50%) is absolutely not the same as the probability of having a boy THEN a girl (66%)."

I didn't say they were the same.

Out of 1000 couples, 500 will have a boy and a girl. 250 will have a boy. 250 will have a girl.

If I tell you that I have at least one girl, you can eliminate 250 couples right off the bat. Because I have at least one girl, I must either be in the set of 500 couples who have both a boy and a girl, or I must be in the set of 250 couples who have two girls.

Now look at the question. Understanding the above numbers, what is the probability that I have a girl and a boy. There are 750 total couples, and 500 of them have a girl and a boy. 500/750 = 2/3 = a 67% chance.

@xooorx

This a is math problem, not a psychology problem. I think we can all agree that Jeff didn't word the problem very well, exposing the specification to all sorts of weird distractions like "what percentage of couples with mixed-gender children would say 'I have a girl' as opposed to 'I have a boy'?"

It's irrelevant to the problem under discussion. The premise is that we know the couple has a girl and one other child. We assume that the source of that information is untainted by weird "what-ifs" and bizarre parental bias. Given this, the probability of that other child being a boy is 2/3.

Must ... resist being pulled in ... and adding no value ...

Yes, order matters in coins. But as I think about the question, I start to think in quantum physics or BTTF McFly fashion; the end result(s) fade(s) in and out. Of course order matters with children (although what if they were twins, born at the same time?). But I am thinking "beyond" the question, into the nature of how it was asked, hence I am swaying in and out in the question, not the reality of children. One last point on rereading ... I sound, uhhh, dumb.

Mother: XX
Father: XY

You are more likely to have a girl child period. Moral is, never ignore the real-world, even in a hypothetical question.

@Russ
Jeff is asking us to answer using the information provided in the question. There is not demographic data provided in the question. I do not know if there are more two children couples that have two girls, two boys or a combination of both our there. Therefore, I am making the only reasonable assumption which is there is a 50/50 percent chance that a child is a boy (or a girl).

And again, when you say that the BG/GB basket is bigger than the BB or GG basket, you are considering the order in which the children were born. Why don't you also consider the order in which two boys or two girls were born, that is, why not change BB into B1B2/B2B1?

Kevin:
What?

Each parent only donates a single chromosome.
Mother is guaranteed to donate an X.
Father has a 50% chance of donating an X, and a 50% chance of donating a Y.
Child has a 50% chance of being a boy (XY).

Not counting genetic abnormalities.

David:
He is not considering order.
You have one child. 50% chance it's a boy.
You have another child. 50% chance it's a boy.
25% chance of having two boys.

You have one child. 50% chance it's a girl.
You have another child. 50% chance it's a girl.
25% chance of having only girls.

What is the chance of having a boy or a girl? 100% - 25% - 25% = 50%.

That's where he's getting his information. You are twice as likely to have a boy and a girl in any order (50%) than to have EITHER all girls (25%-- 50/25 = 2) or all boys (25%-- 50/25 = 2).

I meant "What is the chance of having a boy and a girl? 100% - 25% - 25% = 50%."

Sorry, I was typing fast.

For those that want to throw order into the mix and double-count the BB and GG buckets, please tell me how you can give birth to your oldest child after your youngest!

@Russ

You have failed to understand what I was saying.

My point isn't about psychology its about the maths.

To arrive at 2/3 you need to specify that *all* parents of boy/girl will preferentially reveal they've got a girl. If you just randomly stumble upon the information that there's a girl its completely different, because that's twice as likely to happen in the girl/girl households.

Monty Hall is only 66% because he's preferentially opening a non-car door. If he doesn't know or care where the car is, he can't help you.

This is only 66% if the parent is preferentially revealing a girl. Which wasn't in the question.

@sadnessbowl:
Your question is "You have no child, you plan on having two. What are the odds that you will have two boys?", it is not "You already have two children, one of them is a boy. What are the odds of the other one being a girl/boy?" The two questions do not have the same answer.

I disagree greatly with this question, since it contains an unknown variable.

The order in which children are born is a relevant fact in normal life, but if you look at the original question its about making bets, and it was about trying to calculate the probability of 5 games if you only played 3 of them.

So, in this instance Jeff, a new variable not present in the original question is presented and thus caused a great deal of confusion in this newer version. Does the order matter? The question doesn't say. Hence any calculations are null since there is 2 parallel methods of calculation based on weather or not that fact SHOULD be taken into consideration.

If the order in which the children were born was to be taken into consideration, it should have been mentioned in the question, or a different example should have been given.

Either way, its amazing how heated people get over a simple question, look at all the comments on these two pages! Pretty cool.

@David

B1/B2 is utterly, utterly irrelevant, since it's still BB. Order is NOT relevant, what is relevant is that the BG bucket is bigger than the others.

Look, lets take this from first principles, OK? I'll number each step, and would be obliged if you would specify exactly which step you disagree with.

1. Assume the chances of having a boy or a girl is 50%. Therefore the chances of having two boys is 0.5*0.5 = 25%. Agreed?

2. The chances of having two girls is, likewise, 25%. This is a total of 50% (25% BB + 25% GG). Agreed?

3. That means that the probability of a mixed gender pairing is 50%, since there are no other possibilities. Agreed?

4. Before we go any further, can we at least agree this is incontestable? We're not counting hermaphrodites or anything else, so given the numbers above it is IRREFUTABLE that the (order-insensitive) BG bucket is 50%, agreed?

5. OK, now we've got 100 two-children couples. Statistically, 25 of those couples have two boys, according to the numbers above. Agreed?

6. We eliminate the BB couples, leaving 75 couples, of which 25 are GG and 50 are mixed-gender. Agreed?

7. We have now reached the problem space as defined by Jeff's original post. We know that all 75 couples have a girl child, and want to know the probability of the other being a boy. Agreed?

8. Well, of the 75 couples, 50 have a boy since 50 are mixed-gender pairs. This is DIRECTLY derived from the numbers above, which you must have agreed with to get this far. Therefore, 50 of the 75 couples have a boy as the other child, and 25 of the 75 have another girl as the other child. Agreed?

9. 50/75 == 2/3 == approx 67%.

> What is the chance of having a boy and a girl? 100% - 25% - 25% = 50%."

And now add: given that you know you have a girl.

That changes things, eliminating the BB case, and so you get 2/3.

@doublethink

Order doesn't matter in the BG problem. You're examining the state of the game after the random events have occurred. It doesn't matter whether they occurred simultaneously or in sequence.

Go back to flipping two coins. It doesn't matter if you flip one and then the other, or flip two at once. As long as you examine the game after both flips have been completed.

Two coins have been flipped. At least one is HEADS. What are the odds that one of them is TAILS?

@xoorx

You're still deliberately muddling the question with irrelevancies. For the sake of argument, cut out the parents completely.

You are told by an emotionless robot with no connection to anyone involved in the story that it knows of a couple with two children, one of which is a girl.

The probability of the other child being a boy is still 2/3.

xooorx:
Are you suggesting that parents neither know, nor care what gender their children are?

You have a fundamental misunderstanding of the problem. Don't think about Monty Hall.

Take the set of all people with two children. 50% of them have a boy and a girl. 25% have two girls. 25% have all boys. This is simply statistics, and requires no information from anyone to discern.

Now let's revisit the word problem.

"A person comes up to you and says that they have two children."
This puts them into the set that we care about--people with two children. Right now, if I were to ask you what the probability is that they have a boy and a girl, the answer is 50/50 because 50% of people have a boy and a girl, and 50% of people have two boys or two girls.

Now let's say that the person tells you that they have at least one girl. It doesn't matter if this experiment ever happens again--right here and right now, the person says that they have at least one girl.
This eliminates 25% of the entire set. Always. It doesn't matter if the next person you meet doesn't tell you this--what we care about is this particular instance. 25% of the entire set is eliminated. 50% of the original set has mixed children, and 25% of the original set has all girls. That's 2 to 1 that they have mixed children, or 67%.

Monty Hall has nothing to do with this problem. They're only similar in that they deal with probabilities, and that the answer is not what you'd first guess.

Although I didn't answer, I agree with a lot of the comments, Jeff. In the way you put the question, the answer was 50%.

@doublethink:
But the original question does not say that the order has to be taken into consideration! It's not about "order being a relevant fact in normal life", it's about answering the question given the data provided!

The question, for that matter, can be changed into something that has nothing to do whatsoever with boys/girs:
- You have a jar of 50 red tokens and 50 blue tokens mixed together,
- You have picked two tokens out of that jar,
- One of them is blue.

What are the odds that you have one blue and one red?
What are the odds that the first token you picked is red?

Note that I should really ass the following to the question:
- There is a 50% percent chance of picking either a blue or a red token out of that jar,
because otherwise you would have to make an assumption about that probability. In real life, it is common to accept the 50% chance of having a boy or a girl as reasonable.

Here's what all the programs and statisticians leave out: There are more people with one child and more than two.

What this means is that running a simulation using only the case where a woman decides to have two children is woefully incomplete in determining the probability that in the sample of those two-kid-havers that there is a perfect distribution of orderings.

So, you can't simulate this by generating pairs and determining the distributions. Unless you also take into account the probability of having one, two, three, etc., children as well.

It is easier instead to just generate a whole bunch of unique events, because that's essentially what they are and find the probability of girls and boys. That's your probability.

David:
No, you got the question wrong again. My question is the following:
Out of all of the couples in all of the world who have two children, what percentage of them have one boy and one girl? What percentage of them have two boys? What percentage of them have two girls?

The answers are 50%, 25%, and 25%.

But none of those is the answer to the question posted by Jeff, obviously.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

I think it IS worthwhile to think about Monty Hall; it demonstrates how people THINK about these problems. I, of course, fail. I am still stuck in the space-time where BG=GB, at 50%. I however WANT to think 66%, but my brain refuses. By gods I shall live with it, with 66% being my one true god. (BSG 01/16)

@David

"The two questions do not have the same answer."

EXACTLY RIGHT. The probabilities are altered by the fact that we have been given some information. This information allows us to eliminate the BB category, which is WHY the correct answer is 2/3.

How can you understand the problem so well without being able to see the answer?

@Russ

If the emotionless robot is programmed to always reveal a girl if there is one then its 2/3

If the emotionless robot is programmed to randomly reveal the sex of one child (and it said girl on this occasion) then its 1/2

@Russ
Sorry Russ, I can't pretend I know better than you, but I certainly believe that you are missing something. So let me return the question to you: how come you seem to understand the problem so well and still don't see that, given the way the question is formulated, the correct answer is 50%?

@David (again)

Why don't you actually TRY that experiment you've proposed, and tally the results? Note that in the original problem order IS NOT important, and the probability of the first token in any round being red is indeed 50%. However, if you ACTUALLY PERFORM your experiment you will find that, when you take your final results and cross out all the red/red combinations (equivalent to removing the boy/boy pairs), red/blue (order insensitive) combinations will account for 2/3 of the remaining results.

Jeff, you are wrong.

Remember how in the first post, you linked to Eliezer Yudkowsky's article on Bayesian Reasoning? Well he wrote a post on his blog where he dealt with this very problem. He says it's 50%.

http://www.overcomingbias.com/2008/10/my-bayesian-enl.html

@Russ:
I have tried it Russ, I have...
The question Jeff asked mentions a fact: there already are two children. With the jar analogy, there is also a fact: two tokens have already been picked. It's not about calculating the odds before the tokens are picked, it's about calculating the odds once they have been picked. In other words, it's Combination vs. Arrangement.

Brian G -- I disagree with your reading of the question (It specifies finding a random family, and not a random girl, IMHO), but I agree with the logic connected to it: If you meet a random girl with a single sibling, she has a 50% chance of having a brother.

If you meet a random family with at least one girl, it has a 67% chance of having one boy.

The question is whether your sample is "families with two children" or "girls with one sibling".

xooorx,

In the context of the question, any parent with at least one girl will report that they have a girl. No parent with a girl may say they do not have a girl. Further, no parent with two boys may report that they have a girl.

The key to this problem that you're missing is that we are meeting the parent, getting information about BOTH births at once. Thus parents with a boy and a girl are twice as likely to appear as a parent with two girls.

If on the other hand, you happen across a girl and she says she has a sibling, then yes, the odds are 50/50 that her sibling is a boy.

@Dave

"I certainly believe that you are missing something"

I numbered my steps for easy reference. Which step, specifically, is wrong?

@xooorx

"If the emotionless robot is programmed to always reveal a girl if there is one then its 2/3"

You're mis-stating it AGAIN. The FACTS you are given - the premise of the problem - is that you know the gender of one child and are being asked the probability for the gender of the other child. The WHOLE POINT of the problem is that when you are given information, the probabilities of the remaining outcomes changes.

If you're going to state that the answer is altered by the fact that you've been given a piece of particular information, when the entire point of the article is that given information alters the answer, then I really don't know what else I can say. You are in entire agreement with everyone that says the answer to THIS PROBLEM is 2/3, but seem to be arguing anyway.

I think you are missding the fact that the children already exist, just like the tokens have already been picked. As I said before: It's not about calculating the odds before the tokens are picked, it's about calculating the odds once they have been picked. In other words, it's Combination vs. Arrangement.

@Dave

"It's not about calculating the odds before the tokens are picked, it's about calculating the odds once they have been picked."

Yes, I know. And that's why the probability for a boy in this problem is 2/3, not 50% as it would be BEFORE conception. If you've really done your own experiment, why don't you post your summarised results here (red/red, blue/blue, and mixed)?

@Russ:
Sorry, it's the other way around.

@Russ

*I"M* mis-stating it? Its YOUR robot!!! How was it programmed? The answer depends on that.

"Its YOUR robot!!! How was it programmed? The answer depends on that."

No, it doesn't. The point of the robot was that it would be completely impartial. Your immediate response is to attempt to taint the robot with some sort of bias through its programming, which is dishonest misdirection. Forget the damn robot. We are dealing with a dramatisation of a mathematical problem, in which you are provided information and asked to draw a conclusion. Questioning the source of the information might be a valuable trait in general, but in this particular case it is simply ducking the issue.

Q: A couple has two children, one of which is a girl. This information came from a completely unbiased source and is not dependent on whether someone is 'more likely' to say they have a girl, or whether the boy misbehaved last night and is being punished, or the father died in a yachting accident before knowing that the mother was pregnant, or any other irrelevant BS. What is the probability that the other is a boy?
A: 2/3

Try the Monty Hall problem

http://en.wikipedia.org/wiki/Monty_Hall_problem

@David

As I said before, perform your experiment and post your results. Your own proposal will show that the answer is 2/3.

@David

As I said before, perform your experiment and post your results. Your own proposal will show that the answer is 2/3.

I think I'm using this next time I interview someone for a job.

xooorx and Jason are right. As the Overcoming Bias article points out, there's a difference between finding out that one of the children is a girl and being told that one of the children is a girl.

In the case where the parent offers up the information, a parent of a BG set is as likely to report that there's at least one girl as there's at least one boy. That cuts the BG bucket in half and brings the probability that the sibling is a boy down to 1/2.

To illustrate the difference between finding out clarifying information vs. revealing information, change the coin flip experiment to:

1. Flip two coins.
2. Randomly pick one of the two coins. If it's TAILS, go to step 1.
3. Remove the coin picked in step 2.
4. Write down the side of the remaining coin.
5. Go to step 1.

Here, you're still only considering sets containing at least one HEADS, but you're throwing away some HT sets because you revealed a TAILS in step 2. This is different than examining the set for one or more HEADS (asking the parent whether they have any girls).

@Russ

(lay off the cr@p about yachting accidents please, thats not my point and you KNOW it)

Of all couples who have a girl, 2/3 of them have a boy as well. Of course. Total agreement. If you randomly select a parent from the pool of all-parents-with-2-kids-and-ones-a-girl, i'll put my money on the other one being a boy, right next to your money.

But if you randomly select a parent from the pool-of-all-parents-with-2-kids, and THEN discover that one of those kids is a girl, it doesn't increase the likelihood that the other one is a boy.

It really really doesn't.

who has 2 kids, one of which is a girl, and one of them is a girl

Trevel,

That sounds like a reasonable disagreement to me. The most interesting thing about this problem is that it's only partially a probability problem. It's also a problem of human psychology and communication.

Some have accused Jeff of botching the question, but the vagueness of his delivery gave us three possible answers: 1, 1/2 and 2/3, none of which is exactly right. The resulting melee has been a lesson in how we as biological reasoning machines think about probability, and how we compare our thoughts to others. Of course, the sample is biased. Perhaps the general population would be less arrogant about the whole thing.

[ignore last line of previous post its just droppings :)]

You 2/3 people are reading more into the question then exists, the possible top combinations are not BB,BG,GB,GG they are BB,BG,GG then based on information provided(a girl) we have possible combinations of BG,GG. So 50% chance of a boy and 50% of girl.

This question is NOT the boy or girl paradox it is a question if the people have a boy/girl or girl/girl.

The Overcoming Bias argument is an interesting way to view the problem. More completely stated, that question is framed:
"given that you meet a parent of two and that parent chooses to reveal the sex of one of their children, and that sex is female, then what is the probability that the parent has one boy and one girl?" Of course 50%.

It's an interesting way to interpret the (arguably poorly stated) question. Unfortunately, it doesn't seem to be the argument being made by the 50%ers. I don't _think_ it's the most reasonable way to read the problem, but I can understand how someone might logically get there from what's written.

There is a big difference with the Monty Hall problem: the participant does not know what's behind the door picked by the host. And you can clearly see in the demonstration provided by Wikipedia that this non-knowledge is very important.

To me, the way Parade exposed the problem contains something that is misleading (there is no paradox): it tells you (the guy who's going to calculate the odds) that the host has picked a door behind which is a goat, but it doesn't say that the participant knows about it. It is therefore easy to use a piece of information that you shouldn't use and get to the conclusion that the answer is 50%.

In Jeff's question however, the very fact that one of the children is a girl is known and can be used to calculate the odds. Add that to the fact that the question does not say the order in which the children were born should be taken into consideration, and the answer can only be 50%.

@xoorx

"But if you randomly select a parent from the pool-of-all-parents-with-2-kids, and THEN discover that one of those kids is a girl, it doesn't increase the likelihood that the other one is a boy.

It really really doesn't."

Sigh. If I toss a coin, it's 50/50 heads/tails, agreed? If you THEN discover (because I tell you) that it's not heads, does that not increase the likelihood that the coin toss was tails?

This is just an extension of the same principle. With no other information available, the BB/GG/BG split is 25/25/50. When you THEN discover that one of the kids is a girl, 25% of the problem space is eliminated, thus increasing the probabilities for the remaining possibilities. The BG outcome (which is what we're looking for) now accounts for 2/3 of the remaining possible outcomes.

This is utterly, utterly standard statistics in use all over the world. See Rule 4 removals in UK horseracing, for example - when a horse is removed from a race, the win probability for all remaining horses goes up (and thus the odds shorten).

Or, if you want to do it the "Unfinished Game" way

Setup for the "game".
Parents, two kids. They tell you all about Cindy's ballet recital last night (they have at least one girl, no mention of age). They call for the kids. They come in and are introduced as the older and younger (honestly, it will work out with first/last, tall/short, weight, any ordering).

There are two possibilities for the older.

B - 50%
G - 50%

If it is B, we can end the game because the second throw has to be G (the known girl, Cindy).

If it is G, we need to make another toss:

GB - 50%
GG - 50%

So B has 50%, and G has 50% split evenly between GB and GG, making the individuals 25%. So:

BG - 50%
GB - 25%
GG - 25%

So it's not 50% like I originally thought or even 66.6...%
It will wind up being 75% boy and 25% girl. (If you work out the problem using the Unfinished Game as your model.

However, this is when the 66%'ers come out saying "What about BB?", it literally never happens when a couple with two children have at least one girl. Then they will trot out their programs and analysis for a whole bunch of two children couples. But in my last post, I show those analysis to be virtually meaningless, because they are forcing a certain distribution.

The program should really stop when either it generates a boy, or on the second iteration.

So, yes I've changed my mind, from 50% to 75%.

xooorx, you sure that all couples who have a girl, 2/3 of them have a boy as well?

If you run a simulation, just to tabulate of 2-children parents, which combinations they get (AFTER the fact that they had them, therefore not caring about order), isn't it as follows?:

Boy/Boy - 33%
Girl/Girl - 33%
Girl/Boy - 33%

If you rule out boy/boy, you only have a 50% chance of each girl sibling having a brother.

Just wondering.

David, you're not making the right 50% argument. If you want to figure in the probability that the parent _might_ have said "boy" instead of "girl," then you can reasonable get to 50%.

However, given that you are learning that the outcome of one (you don't get to pick which) of the two births resulted in a girl, then the chances that the other is a boy is twice that of a girl.

Again, consider the roll of a pair of dice. Order doesn't matter, but 7 is the still most likely outcome.

Here, one boy and one girl is the most likely outcome, and once the chance that two boys has been eliminated, you must see that 2/3 of the remaining families will have one boy and one girl.

Given that we happened across the parent arbitrarily, they will more likely be in the 2/3 of families with at least one girl that also have one boy.

@Dave

"the question does not say the order in which the children were born should be taken into consideration, and the answer can only be 50%."

The order in which the children were born is irrelevant, as I have previously shown. It doesn't really matter how often you repeat it, it simply isn't true. The relevant factor is that the boy/girl combination, in any order, accounts for 50% of the possible two-child pairings. It doesn't matter whether the boy or the girl was born first.

How's that experiment coming along?

One way to reframe the BG problem: You know you are going to have to children, at least one of which will be a girl. What is the probability that you will end up with a boy and a girl? Here is the possible set of birth sequences:

GB
BG
GG

Of these three birth sequences, two have boys, therefore the probability that you will end up with a boy and a girl is 2/3 (67%).

Hell, I hate this, brute force clearly shows odds being at 2/3, yet my brain refuse to understand :(

According to the original example, the answer is 1/2 not 2/3.

There are 3 possible SETS of children. These sets are "BB", "GG", and "BG"

Sets "BG" and "GB" are the same. The elements of the sets are same, thus the sets are identical. There is no distinction in the original problem to distinguish order in the sets and hence the order of the elements in a set are irrelevant. Treating "BG" and "GB" as different sets is an error. If we say a family has "a girl and a boy" or "a boy and a girl", we are not referring to a separate set of sibling combinations. We are simply describing the same set of siblings in a different sequence.

My friends has twins: a boy and a girl. Or I could say, my friend has twins: a girl and a boy. Both descriptions refer to the SAME SET of twins.

By enumerating the set "BG" as both "BG" and "GB" it makes it appear that the probability for the other child being a boy is 2/3. This is an error. The error is in using the same set twice.

Another way to look at this problem, the typically intuitive 50% way. is to see it in matrix:

the set of children looks like this:

G? or
?G

by treating both ?G and G? as valid descriptions we get:

GG
GB
BG

the 2/3 error comes by missing that a correct description of the problem is:

G? OR ?G

Thus it should be:

GG
GB or BG

not:

GG
GB and
BG

Whether the sibling set is described as "G?" or "?G" is not relevant in the example. Describing it as BOTH "G?" AND "?G" introduces an error of duplication.

The chance of ? being a B or G is 50%.

....

Now as a practical matter, sibling gender is independent of the gender of other siblings. A second sibling's gender is not conditionally related to the described sibling. Therefore a statistical description of the second sibling's gender, based on an initial sibling's gender is questionable.

An upshot of the 2/3 argument is the statement: All sibling pairs, where one sibling is a girl, are ~66% more likely to have the other child be a boy.

Does that actually comport with reality?

-calvin

@Jason

No, BB and GG are 25% each, and BG (the order of birth is irrelevant) is 50%.

A problem with these code simulations (and Jeff's solution) is that they are simulating a chance of getting a Boy/Boy, then throwing out that possibility. That is not what the original question is asking for.

We know that either the first or the second child is a girl -- and that is how I coded my simulation. It returns 50%:

static void Main(string[] args)
{
const int NUMBER_OF_TOSSES = 10000;

int boyGirlCount = 0;
int girlGirlCount = 0;

for (int i = 0; i < NUMBER_OF_TOSSES; i++)
{
// We know at least one child is a girl; It
// is either the first or the second.

Coin coin = new Coin();
if (coin.flip() == Coin.Sides.Heads)
{
// First child is a girl; Get second child.

Child secondChild = new Child();
if (secondChild.IsBoy())
boyGirlCount++;
else
girlGirlCount++;
}
else
{
// Second child is a girl; Get first child.

Child firstChild = new Child();
if (firstChild.IsBoy())
boyGirlCount++;
else
girlGirlCount++;
}
}

System.Console.WriteLine("Boy and Girl: " + boyGirlCount.ToString());
System.Console.WriteLine("Girl and Girl: " + girlGirlCount.ToString());
}

The real gem to come out of this and the preious thread, is how difficult people find it to clearly explain a concept or reasoning that may seem simple to them to those that are not thinking the same way. It took until Yvons comment in this thread for there to be an answer that that a good job of convincing a lot of people thinking it was 50%. A simple example rather than theory always helps.

@BA:
In your demonstration, you are playing the game, are are detailing every step of it. Jeff's question is about a game that has already been played.

It's like a hand of scrabble:
- You pick one letter. What are the odds it's an A?
- You pick a second letter. What are the odds it's an A?

The second answer is totally influenced by the first one.

It's not the same as saying:
- You have two letters,
- One is an A,
- What are the odds that the other one also is an A?

There is only one possible answer to this question, unlike the second question above.

Jason,

That's not the case, though. One Boy and One Girl shows up twice as often as two girls or two boys.

Boy/Boy - 25%
Girl/Girl - 25%
Girl/Boy - 50%

Think of it this way; you flip a coin and get a result. If you flip a second coin, it has a 50% chance of matching the prior result, and a 50% chance of differing; BB and GG are the "matching" results here -- totaling a 50% chance for both (25% each). GB is the non-matching result, and carries a total 50% chance.

Correction to "The second answer is totally influenced by the first one."

The answer to the second question depends on whether or not you picked an A as the first letter.

calvin:
Because we consider probability in the problem setup, we assume that standard probabilities apply to the likelihoods of having boys or girls. If you don't want to do that, and you want to suggest that the probability of having BG, BB, and GG are all equally likely, then we really ought to consider some other possibilities.

First of all, we should consider that some people have pets that they consider children. We must also realize that not all pets necessarily have a sex. We should consider, then, the rather likely possibility that the parent is considering an asexual plant to be one of his "children." There are a whole lot of asexual plants out there, you know.

As such, the probability turns out to be way less than 50% that the parent has both a boy and a girl.

Hmmm ... my theory now is that there is no amount of information that can be injected into the original question that would have decreased this thread length. I am serious ... I mean no devalue to the comments! This is a fun study of human nature, and I am seeing value in the Bayesian-brained and the "others". My theory is that the first is deep knowledge, but narrow, and the others are broad but shallow. Question to myself; which way of thinking provides more total-life-long value?

@calvin

It's not a case of duplicate counting BG and GB. Look at it this way. If you assume a 50% chance of a boy and 50% of a girl for any given pregnancy, then the chances of both children being male P(BB) is 0.5 * 0.5 (25%). The same applies for girls P(GG). Therefore BB + GG = 50%. Therefore, order-of-birth-independent BG pairings MUST be 50%, since 100% - P(BB) - P(GG) = 50%. This is indeed backed up by real world statistics.

In the previous post, most of the comments were in agreement with this fact: 50%, not 33%, of two-child families are mixed gender. That was not the problem. The problem was one of selection.

Did we select at random a two-child family with at least one girl? If so, we arrive at the 67% answer.

Did we select at random a two-child family, and then select at random one of the two children? If so, the correct answer is 50%.

@Russ

"This is just an extension of the same principle. With no other information available, the BB/GG/BG split is 25/25/50. When you THEN discover that one of the kids is a girl, 25% of the problem space is eliminated, thus increasing the probabilities for the remaining possibilities. The BG outcome (which is what we're looking for) now accounts for 2/3 of the remaining possible outcomes."

It depends how you found out that one was a girl.

If you said: "Do you have a girl? Either of them? At least one girl?" and they said "yes" then the answer is 2/3

But if you said "Toss a coin and if its heads, tell me the sex of your eldest, otherwise tell me the sex of your youngest" and they tossed and said "girl" then the answer is still 1/2.

In the first case you're getting extra information from a preferential girl revealer. In the second, you're not.

@Michael Perry
There is no random selection of the family. The family is given to us in the question.

@David

If the family was given, then the truth is already known. Maybe not to us, but certainly to the mother. There is on probability.

We can only discuss probability in light of random selection.

xooorx,

Are you saying the (arguably poorly phrased) question is closer to your the first way or the second way?

@Michael Perry
And the random selection we are talking about in this problem is that of the second child (second not in order, second as in "the other child"), and there is a 50% chance that it is a boy.

I think I have figured out why I am stuck at 50%; I have divided Yvon's 100 families into 3 buckets; b1 with 2 boys, b2 with mixed, and b3 with 2 girls. Eliminate b1, and you are left with b2 and b3, so 50% chose b2. I think if I severed that link between left and right brain, I shall argue this with myself until sweet death.

Unfortunately, the clever math is all wrong and pointless.
The question is
"Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

So we are TOLD that there is one girl. Not "one or two of them".
Not "at least one of them". But ONE.
Therefore starting with the four cases BB GB BG GG, we rule out
BB and GG and get GB BG.

Therefore the answer is 100%.
The problem is interesting just because the problem statement is awkwardly interpreted. This is a common problem in clever math word puzzles.

It may be clearer if the statements were explicit.
"How many children do you have?" "TWO"
"Is one or both of them a girl?" "YES"

@xooorx

"But if you said "Toss a coin and if its heads, tell me the sex of your eldest, otherwise tell me the sex of your youngest" and they tossed and said "girl" then the answer is still 1/2."

That's a totally different question, since it allows you to know the sex AND the order of birth, which slices the problem space in a different way. The question as posed is NOT dependent on the order of birth.

In the question as posed, you can only eliminate BB from the problem domain given the information 'one child is a girl'. If you know that the OLDEST child is a girl, you can remove both BB and GB.

So, yes, the answer to your problem is 50%. But that isn't the question Jeff posted. It has nothing to do with 'preferential girl revealers', it has to do with answering the wrong question. The answer to Jeff's question, if NO BIAS is assumed, is 2/3.

@mike:

I'm saying there's nothing in the question to suggest the parent is a "preferential girl revealer" so its closer to the second way.

@JamesR
You should not assume that you are wrong thinking that 50% is the wrong answer. You should not try to convince yourself that 2/3 is the right answer. As proved many times in these comments, the answer depends on how you read/interpret the question.

I am quickly jumping into Steve's 100% boat. Nicely done! We are too quick to start up the math&code when we really needed to understand how humans talk, and exactness is impossible.

@Russ

The answer to Jeff's question is 1/2

You need to ADD the stipulation that the parent is a "preferential girl revealer" to get 2/3 but, like you say, the question didn't mention anything about that.

@JamesR, @David

Well, if you want to stick with 50%, do so for the right reasons. The three buckets you have: BB, BG, and GG are not all of equal size. The BG bucket (b2) is twice as big as the other two.

@David

But those are all the possible outcomes for a game where the result has one girl. (If you want order to matter)

In your Scrabble bag case (also with the stipulation that there are only A's and B's in the bag and an infinite number of each (like with the sex of a child)), the answers don't change. It's 50% for the first pull and 50% for the second.

And 25% that they are both A.

@David

If the problem were to select any two children at random and reveal one to be a girl, I would completely agree with you. However, as stated, it is a selection of families, not of children.

Put another way that does not rely on probabilities and random selection, what is the percentage of two-child families with at least one girl that are mixed gender? Now we aren't choosing random families, we are examining all of them.

Put this way, do you agree that 2/3 of families just described are of mixed gender, with the other 1/3 having all girls? If so, we are on the same page.

That is 25% they are both A if one of them already is an A.

@xooorx,

Gotcha, you're taking the approach of interpreting the question in such a way that the parent may have decided to say "boy" rather than "girl."

I don't see that as an obvious option in the realm of the problem as stated, but I can see the logic of packing that possibility in and the correctness of your argument if you do.

As mentioned on the OB post linked above, it would have been better phrased "you meet a parent on the street that tells you she has 2 children, you ask if she has at least one daughter, and she says yes." I do think that this is the more obvious reading of the problem, but I can see your point that the other reading is possible.

@Michael Perry
No, it is not a random selection of family. The question says:"Let's say, hypothetically speaking, ---> you met someone who told you they had two children, and one of them is a girl &lt;---. What are the odds that person has a boy and a girl? "

You have met someone, the selection is already done. The only remaining unknown is what sex the other child is.

I upvote xooorx as having the most concise and convincing explanation of the problem.

Monty Hall only comes out at 2/3 because we know Monty is a preferential non-car revealer. If Monty sometimes opened the door on the car ("you lose!") then the answer becomes 1/2 it doesn't matter if you swap or not.

The fact that Monty is a preferential non-car revealer is part of the Monty Hall problem though. The fact that the parent is a preferential girl revealer is NOT part of the given parent problem.

@xooorx

No preference is required. You are GIVEN the information that one of the children is a girl. The first of your examples is EXACTLY the question Jeff asked, and you evidently agree the answer is 2/3. So why exactly are you polluting this thread with your different answer to a different question?

@xooorx,

of course, it is also not given that the parent might have said they had the option of saying "boy." All we really know is that there are two children, and at least one is a girl.

Assume some unkwong guy visits yuo, says he has 2 children, Chances thast his children are boy and girl order irrelvant is 50%;
everybody agrres with this.
Now he says one of them is Girl, that is additional information, because chances of BB now becomes zero. Chance of BG (order irrelevant) increase and becomes 2/3

EDIT:

of course, it is also not given that the parent had the option of saying "boy." All we really know is that there are two children, and at least one is a girl.

@Derek:
I want to stick to 50% because the sex of the "other child" is the only thing we don't know (that and the order in which the children were born, which is irrelevant since the question doesn't say it's relevant). All the rest is a fact: we have one family, they have two children, one of the two is a girl.

I have a question for you: I happen to have two children, one of them is a boy. What are the odds that my second child is a girl?

@David

Right, so I've removed random selection from the equation.

In a population of 1000 two-child families. 250 have two girls, 250 have two boys, and 500 are of mixed gender.

One of the mothers has walked up and said they have at least one girl. They are therefore a member of one of the 750 all-girl or mixed-gender families. The selection has already been made, but to which type of family do they belong?

You have the final word.

calvin

To answer your question about real world, based on USA numbers.
2 child family 25.8$ of all boys, 52.2 having mixed and 22% of all girls.

Now it gets weird for 3 children, as you have more boys the chance of having a girl decrease, and as you have more girls you increase the chance of a boy however at 2 girls you have a better chance of having another girl.

But overall the 50% rule says the same for all children with slight variance.

@Michael
I don't have an interest in having the "final word" Michael, only to share points of view (although admittedly I am trying to convince people I am right :))

In your last comment, you are again making a random selection of the family, since you;re asking: "The selection has already been made, but to which type of family do they belong?"

You do not have to ask yourself this question to answer Jeff's. You are talking to a member of a family that has two children, including at least one girl. That's all...

If you ASK if one is a girl... if you seek out the girl... if all parents with any girl at all blurt it out at the first opportunity... then OF COURSE the answer is 2/3. I've been saying this since, oooh, the year 2008.

But thats not in the question. It needs to be ADDED to the question to get 2/3.

Its not there.

The answer is 1/2

Posts like these last two made by Jeff are why I love Coding Horror.

Just sayin'.

xooorx:
The Monty Hall question asks very specifically about whether or not you get a prize. This is the same as asking whether or not the other child is a boy, regardless of which gender the parent volunteers.

Jeff's question, which I now realize was aptly worded, is whether or not there's a MIXED FAMILY--that is, whether or not there's a boy /and/ a girl. This means that it's irrelevant which gender the parent reveals. If the parent reveals that he has a boy, we eliminate the GG outcome and the probability is 67%. If the parent reveals that he has a girl, we eliminate the BB outcome and also get 67%.

It's a very different problem from the Monty Hall problem.

Preferential <gender> revealer doesn't matter if what we're looking for is one child of each gender. It matters (as you point out) if you're looking for a specific gender, and you can't guarantee that the parent always reveals the other one.

@xooorx
There is no need to ASK if one is a girl in this case, since, as the question clearly indicates, you are being TOLD it's a girl.

Being Pascal an authority in Probability, what are the savants backing up that BG and GB are different (when no specifications about B and G are given in the problem, thus making the distintion of their succession irrelevant to the problem)?

We already know the family has one girl and know nothing about the other child. Is it a B or a G? In a universe of 2 possibilities what's probability of each one of them?

It must be 66%, right?...

@xooorx,

Again, I see your point. If you wish to pack the option that the parent *might have said* "boy," then yes, the result will work out to 50%.

It's not clear in the question that this is an option that should be considered as biasing the information. What is clear is that there's a family with at least one girl.

I agree that both interpretations of the question require some additional assumptions (I assume the information is being given in an unbiased manner plainly as written, you assume that the information is being given with some bias). Either way, we must both make some assumptions about the question and its frame in order to solve it.

@xooorx

"But thats not in the question. It needs to be ADDED to the question to get 2/3."

Rubbish. The question clearly says "one of the children is a girl". You don't need any further information than that (therefore nothing ADDED) to calculate the result, which can be trivially supported with very simple tests even a child could perform.

Perform the experiment with coin tosses. Repeat the experiment with red and blue tokens drawn from a hat. Repeat the experiment with a simple computer program. It doesn't matter whether or not it's children, it doesn't matter whether or not the parent likes to say "I have a girl" and ignore the boy.

Perform the experiment, tally the results, pick an outcome. Remove the impossible outcomes, and calculate the probability of a mixed outcome from the remaining outcomes. It will be 2/3. THAT is the question Jeff asked. Bias doesn't come into it. When you have a pre-recorded probability distribution and add information that eliminates certain outcomes, you see the effect that is in play here. It has NOTHING to do with biased parents, order of birth, or any of the other misdirections you've added.

Throwing my explanation into the fray:

I think the hard part here is realizing that you're being given information about the outcome of the trial in question, and that knowledge affects the probability (I was having this difficulty at first).

Think of it like this: If you flip a coin, the probability of getting heads is 1/2. However, if you flip a coin and you SEE that it was heads, the probability of that trial being heads is 1. Information about the outcome changes the probability of that outcome.

That example is painfully obvious, but the problem at hand is very similar.

You're told that someone has two kids, and one of them is a girl. That is information about the outcome, hence the probability of the outcome will not be the same as if you had no information (which would be 1/2).

@Russ
No, that's not the question that is being asked. We are not being asked to simulate the birth of two children one after the other and measure/calculate the odds of obtaining a boy and a girl among all possible combinations. We are being asked to calculate the odds that the other child is a boy given that THERE ALREADY ARE two children AND WE ALREADY KNOW that one is a girl.

The only unknown is what sex the other child is, and the odds that it is a boy are 50%.

@hovenlad,

Once you've picked which child is the girl, then the other has a 50/50 chance of being a boy. HOWEVER, if you only know that one of two children is a girl, then you must take into consideration that *either* of the two children is a girl in question. The myriad explanations for why this is so are detailed to excruciating precision above.

Like Russ said, I flip a coin, look at the result and tell you it is Tails, will you still wager 1:1 that it's Heads?

Expand on that example: If I flip two coins, look at both results and tell you that I have at least one Tails, will you bet 1:1 that I also have a Heads down there, too?

@Russ

I have not once mentioned order of birth not ever.

I have only mentioned biased parents ("preferential girl revealers") in that you NEED biased parents to get 2/3.

@David,

Right, we know that one is a girl, but you are ASSUMING that you know which one is the girl! You can't do that because you aren't given that information. You are given that two children exist, at least one of them is a girl. Map this to the coin flip problem and try it out!

@Russ -- I posted my simulation code above, and it returns 50%. I think your coin toss simulation is incorrect by even giving it a chance of return results that have to be thrown out. Can you tell me what I have misinterpreted?

@mike
No, I don't! Which one is a girl is totally irrelevant. There is another child, and the odds that it's aboy are 50%, nothing more.

[ahem i did mention birth order just once]

@David

I clearly said tally the results BEFORE running the probability calcs. This is EXACTLY the same as the problem in hand - the children have already been born. THEN you run the calcs after adding the information that one child is a girl/one coin is heads/one token is red.

Hell, the red/blue token example is YOUR proposed experiment! Am I to assume you are now seeking to distance yourself from it, since I have repeatedly challenged you to actually run your own experiment and post the results, knowing they will prove you wrong?

@Trevel
"If you meet a random girl with a single sibling, she has a 50% chance of having a brother."

Are you sure about that?

I agree, there are only two mustually exclusive outcomes: she has a brother xor she has a sister.

But if you look at the population statistics among two-sibling relationships, it suggests another outcome.

The overall population is 25% 2 brothers, 25% 2 sisters, and 50% a brother and a sister. Since you're meeting a random girl with a single sibling, then you've eliminated the possibility that one sibling is a brother, since a girl can't be a brother. That leaves only the sub-populations, where the probability of sister-brother is twice that of sister-sister.

So if you meet a random girl with a single sibling, she has a 2/3 chance of having a brother.

It's counterintuitive because the act of meeting a random GIRL is itself an act of selection. That is, you don't perform the second test (has a brother?) until you have selected a girl. The second test (has a brother?) is conditional on getting a girl. Getting a girl immediately narrows the possible outcomes by eliminating all brother-brother pairs, 1.e. 25% of the overall population. If you'd said "If you meet a random PERSON with a single sibling", then yes, that person of either sex with a single sibling has a 50% chance of having a brother.

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@xooorx,

Right, the 66% crowd works on the assumption that the parent is only allowed to speak about girls. I personally believe this is the clearest way to read the problem.

You work from the assumption that the parent made a choice about what information to reveal. I understand and accept why you make this assumption, but I personally don't believe it is the best way to read the problem. I just wish the rest of the 50% crowd were as enlightened as you are in this matter.

@mike
And the coin toss example is IMO a very bad one. Each coin toss is necessarily independant of the previous one. There is always a 50% chance that you get either side of the coin, regardless of the number of times it was tossed or the outcome of each and every previous toss.

Jeff's question adapted to coin toss would read: "A coin HAS BEEN TOSSED twice, and the outcome was heads one of those two times. What are the odds that the other toss was tails?"

@Tony

It is not correct to eliminate BBs whilst the initial problem space is being generated. This problem is related to how the addition of constraints modifies the problem domain, but you're generating a problem domain with foreknowledge of the constraint. This is akin to the national census ignoring all BB pairings when counting childbirths.

To accurately model this problem, you must first generate an unconstrained domain. This will, in accordance with strict probability calculations, be approximately 25% BB, 25% GG, and 50% mixed. When you THEN add the information that BB must be eliminated, you find that the 50% mixed expands to become 2/3 of the available possibilities.

The key is that the information is being ADDED to the domain, not baked in from the start.

@Russ
As I said multiple times before, there is no game/experiment to run here. The babies are born, the token have been picked from the jar, the coins have been tossed... It's not about perfoming a calculation based on all possible outcomes, it's about calculating the odds based on one precise outcome, namely "A couple has two children, one of ahich is a girl"

@Dave

"Jeff's question adapted to coin toss would read: "A coin HAS BEEN TOSSED twice, and the outcome was heads one of those two times. What are the odds that the other toss was tails?""

Try this: A coin HAS BEEN tossed. It's not heads. What are the odds that it is tails?

@Tony Alderman,

Your code is broken because you only make one child. You need to make two children, because there are two children discussed in the problem. You fall into the trap of double counting the case where there are two girls: these are actually the same case because it doesn't matter which girl the parent was talking about.

Rather, if you want to count the instances correctly, you should instantiate two children for each family, and discard those that contain two boys.

@David,

Are you saying that the choice of sex of independently conceived children are more dependent than a two coin flips?

And yeah, you have perfectly translated Jeff's problem directly into the coin flip problem. Is there something you think is incorrect in your translation?

"it's about calculating the odds based on one precise outcome, namely "A couple has two children, one of which is a girl""

And the answer is that there is a 2/3 probability that the sibling is a boy. This has been shown, mathematically, over and over again. I presented a step-by-step breakdown of how to derive this result, and asked you do point out the specific step that you disagreed with. You haven't done so. YOU proposed an experiment involving red and blue tokens (which you now dissociate yourself from), and I asked you to run it, since it was an accurate copy of the problem at hand and the results would be equivalent. You refuse to do so. At this point I have no alternative to conclude you are trolling, since you have been on the cusp of the answer on numerous occasions and have wilfully avoided it.

@weevil

"So if you meet a random girl with a single sibling, she has a 2/3 chance of having a brother."

Not quite. Take 400 two-child families with an even distribution. 100 BB, 100 BG, 100 GB, and 100 GG.

That's 200 boys in BB families, 200 boys in BG/GB families, 200 girls in BG/GB families, and 200 girls in GG families. If you pick a girl at random from this pool of 400, there's a 50:50 chance she has a brother.

Note that this is different from picking a girl-containing *family* at random.

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@Russ
Russ, this is not the same problem at all! You're basically asking "A child was born, it's not a boy. What are the odds it's a girl?"

The question is TWO coins were tossed, and ONE OF THE OUTCOME was heads. What about the OTHER outcome?

This is a true story.

In an econometrics course in college the teacher's assistant was talking about conditional probability. The question that he posed to the class was similar to what has been covered here; you randomly pick from a series of bags, which have different quantities of balls in them.
Then you get to flip a coin once for each ball.

The T.A. was not a native English speaker so his phrasing was a bit unfortunate. "Given three balls, what is the probability of getting head?".

Someone in the back of the room piped up loudly "Not very good!".

(it's the funniest advanced math that I know and it really did happen)

i still find a problem with order, in your grouping you assume order (gb and bg or more clearly g1st b2nd and b1st g2nd) then you would also have to include b1st and b2nd, b1st and b2nd, g1st and g2nd, g1st and g2nd giving 6 groups in all bb bb gg gg gb bg. taking away the 2 bb combos would leave 4 groups 100% divided by 4 equals 25% giving the possibility hitting the bg combos 50%.

if you don't impose order to it then you would have bb bg gg. after taking away the bb you would have 2 groups. 100% divided by 2 equals 50%

i could be way off, and if so then this is all well beyond me.

p.s. love goo world! thanks!

The original problem wasn't phrased as an unfinished game, though. Clear requirements are important. You've just stumbled into a statistical trap.

One of the problems in statistics is that people often let information they have ("I have a daughter") influence the answer they are looking for ("Is the other child a boy or a girl"). If that's the question, then the answer is 50/50. The fact that the first child mentioned is a daughter is not relevant information.

Here's another way to phrase the same problem: There are two children on the other side of a door. They walk through the door, one after the other. The first child through is a girl. What's the odds that the next one is also a girl?

The combinations that these could walk through are:
boy/boy (25%)
boy/girl (25%)
girl/boy (25%)
girl/girl (25%)

The first two are already eliminated, leaving the last two combinations. To map back to the original problem, replace "going through the door first" with "being mentioned by the parent first".

The important part here is that you eliminate BOTH of the combinations starting with "boy", and not just the "boy/boy" one.

Unfinished game questions are subtle, Jeff. This looks like one, it smells like one - but it's not one. That said, a different phrasing _would_ make it an unfinished game. For example, given the "walk through a door" - if I predict that there are two girls, AND a girl walks through first, then my prediction is 66% likely to be correct. But if I don't make the prediction until AFTER the girl has walked through, my prediction is only 50% likely to be correct.

In the example given, Jeff, the prediction is being made afterwards. Sorry, but you lose this one.

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@David

It's entirely the same problem, but with fewer elements in the sequence. In both problems, you have a sequence of outcomes with a known probability (in the coin toss example the sequence length is 1), and then after the event some additional information is added which alters the probabilities by eliminating some of the possibilities.

Aside from the sequence length, it's EXACTLY the same problem.

@David,

Russ was using the example to show how knowledge about the result of a random event effect the odds of the result.

So, indeed, the question boils down to, I flip two coins and tell you that one of the two came up heads. What are the odds that one of the two came up tails?

That is precisely what Jeff's problem boils down to. It's not: "I flip a dime and a penny and tell you the dime came up heads."

@mike
I meant that the coin toss was a bad example to illustrate the 2/3 answer is the right one, since it actually perfectly shows, IMO, that the correct answer is 50%.

@mike
"What are the odds that one of the two came up tails" is not the same question as "Waht are the odds that THE OTHER TOSS came up tails" and therefore doesn't have the same answer.

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@fireears

"If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB."

The GB/BG thing is simply a convention when enumerating sequences. It is irrelevant to this problem. What DOES matter is that BB and GG each have probabilities of 25%, meaning mixed-gender pairs occur 50% of the time. It doesn't matter which gender is born first.

@David,

Dude... try the coin toss. Flip a pair of coins 20 times each. Record the results. Count up the number of results with one T. Now count up the number of results with one T and one H. The ration of the second number to the first number will be around 2/3.

@Russ
As long as you are going to believe that the sequence of events is of importance (which is just fine if that's the way you interpret the question), we won't be able to agree (which is fine too!)

To me, there is no sequence to consider, just the fact that there already are two children, one of which is a girl. It doesn't matter in was sequence/order it happended, because the question doesn't say it does.

> "What are the odds that one of the two came up tails" is not the same question as "Waht are the odds that THE OTHER TOSS came up tails" and therefore doesn't have the same answer.

Absolutely, and what we're saying is that you're not reading the question right. You keep trying to answer the second one. The problem as stated does NOT translate that way. There is a family with two children and all you know is that one of them is a girl, you don't know which one!

The coin games only come out 66% if you've got a preferential revealer.

If the rules say "You must tell me if either of them is tails" and they say "yes" then its 2/3 that the other one is heads.

But if the rules say "you must randomly tell me what one of the coins is" and they say "tails" then its 1/2 that the other one is heads.

@David,

Do you agree that it is more likely for a family to have one boy and one girl than to have two girls?

"Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

In real life? Roughly 100%.

Because in RL, the only reason to tell me that one of them is a girl, would be to indicate that the other isn't.

@mike
Dude, feel free to see this problem as something that has to be run, that is, flip a pair of coins 20 times and calculate odds based on the 20 outcomes.

The question doesn't say or metnion in any way that something has to be run. It is not about picking 20 couples, having them make two children, and measuring the probability that, among the ones that had at least one girl, the other child was a boy.

It is about considering that ONE couple, which had two children, one of them being a girl, and figuring out the odds that the other child is a boy.

you should not be using probability to determine whether the next child is a boy or a girl; it's an amazingly dumb method and not accurate. boy/girl is predictable based on the makeup of the mother and father; it's not a truly "random" outcome. analysing it is misleading, and assuming you know the 'true' answer while other responses are 'naive' is exceptionally arrogant.

@xooorx,

You are absolutely right, but of course we're picking "girl" or "heads" before the question is framed, so there's no choice.

You fairly call that an assumption of bias, whereas we call that an inherent feature of the problem statement.

"since it actually perfectly shows, IMO, that the correct answer is 50%."

Then SHOW YOUR WORKING! You have been repeatedly shown why the answer is 2/3 and how to derive it for yourself, step by step. Either identify the flaw in the 9 steps I posted earlier, or post your own alternative. It's getting pretty dreary explaining the problem to you over and over again in as many different ways as we can think of, only for you to go "no, it's 50%" without a shred of reason or calculation to back you up. You won't even post the results of your own equivalent experiment, which you yourself proposed and even claimed to have already run. Where's the results?

@mike
"Do you agree that it is more likely for a family to have one boy and one girl than to have two girls?"

In real life, maybe, although I have no idea and don't know the relevant demographic data (not in the US nor in other countries).

Related to this problem, it is totally irrelevant anyway! I am not making any assumption, I am JUST reading the question and using the data it provides: two children, one of them is a girl.

I can resist no longer. My summary of the answers:

If we take the question literally, the answer is 100%. The original statement was “…they had two children, and one of them is a girl”. If they were both girls, the original statement would have been “they had two girls”, and there would not even be a question.

(Of course, if we wish to be *really* strict about it, we should recognize that it is possible for a human to be born with neither sex, or with both of them. Then the answer is not quite 100%, but it is close.)

If we assume that the questioner meant that “at least one of them is a girl”, then we have a new question to decide: Does the order matter?

The question says nothing about order. It asks whether the person has “a boy and a girl”. We know at least one of them is a girl; there is a 50% chance that the other child is a boy. If order doesn't matter, then the answer is 50%.

If order does matter, as Jeff implies by the answer he endorses, then the question changes: “What are the odds that the person had a boy, then a girl?” (Or the other way around.)

The answer for this question is 1/3:

Boy, then boy: 0
Girl, then girl: 1/3 (again, assuming “at least one girl”)
Boy, then girl: 1/3
Girl, then boy: 1/3

Another, similar question is “What are the odds that the person had a boy and a girl, in either order?” *Then* the answer is 2/3, as you can see from the table above, as we collapse “Girl, then boy” and “Boy, then girl” answers into one and add their probabilities together. The answers are then:

Boy, then boy: 0
Girl, then girl: 1/3 (again, assuming “at least one girl”)
Boy, then girl/Girl, then boy: 2/3

@Russ

Exactly. It is irrelevant to this problem.

That specific path of argumentation that I chose to employ was based on an ancient technique of demonstrating the error of the counter-argument by deepening it (and ultimately giving it a route of logic) to arrive at the conclusion that no matter what reasoning is used, if it is used with logic, it will give one and only solution. In this case the "50% chances" one.

Personally, I came to the "50% chances" conclusion simply as a consequence of knowing there is a girl and another child. Is it boy or is it girl? 50% chances to each.

@David

OK, s/sequence/permutation/g

@David,

Well, this is about probability, so in order to empirically see what the math inexorably leads to, then you need to repeat the test which covers the two pieces of information presented:

1) create family with two children
2) discard families with no boys

What percentage of families have both a boy and a girl?

@mike:

There's nothing in the problem says the parent is a preferential girl revealer. Nothing. If that was "an inherent feature of the problem statement" then the answer would be 2/3, but its not so it isn't.

We NEED to know that Monty Hall never opens the door on the car before deciding that "swap" is the right answer, and we're told that, so "swap" is the right answer. We're given no such information in this problem. Its 1/2.

@Russ
I have been showing my working countless times already! Maybe I'm just not expressing myself correctly, or maybe you just don't understand. Let me try again:

- I give someone a coin,
- I ask that person to flip that coin twice, behind my back, without me seeing anything
- The person comes back to me and tells me one of the two outcomes (either one, it doesn't matter)
- I can then tell the person that the chances the other toss was the opposite is 50%, because one coin toss has no impact whatsoever on another coin toss.

This is exactly the same a Jeff's question:
- A guy I don't know tells me he has two children,
- He tells me that one of them is a girl,
- And then he says "Here is a riddle: what are the odds that my other child is a boy?"
- And I answer "Easy: 50%. Because you had a girl once (whatever the order) does not have any impact on you having a boy or a girl the next time (or the previous time) you make (made) a child."

This is it...

Everyone who's not xooorx and is still claiming 50% needs to go flip some coins.

@fireears,

But you don't know which one is the girl! If you fix one child, find out that it's a girl, then the other has a 50/50 chance of being a boy. If you don't know which is being revealed (as in the case of this problem), then you must account for this fact.

If you believe as xooorx does that the parent made a choice to reveal "girl" or "boy", then you have a reasonable argument for the 50% result, but it is not as simple as your logic would indicate.

@mike
"Well, this is about probability, so in order to empirically see what the math inexorably leads to, then you need to repeat the test which covers the two pieces of information presented:"

No, that's what YOU want to do. That's NOT what the question asks.

@fireears

"Personally, I came to the "50% chances" conclusion simply as a consequence of knowing there is a girl and another child. Is it boy or is it girl? 50% chances to each."

Not so. The probability distribution of births is 25% BB, 25% GG, and 50% BG (order insensitive). Once I have eliminated BB, that means mixed-gender pairs now take up 2/3 of the available possibilities (50 out of 75).

The 50% chance no longer holds when additional information has been provided that constrains the problem domain. 25% of the candidates have been removed, so the other probabilities expand to fill the gap. In the same way as if I toss a coin and then eliminate 50% of the possibilities (by telling you it landed heads) then the other possibility expands to 100% (it must be tails). If I tell you 3 of the lottery numbers, it's no longer a 14m/1 shot that you can guess the result.

@xooorx,

There was nothing in the problem statement to indicate that the parent had the *choice* to reveal that it was a "boy." You make that assumption, and further assume that the choice was made randomly. I do not feel these assumptions are the common sense ones to make.

@xooorx,

There was nothing in the problem statement to indicate that the parent had the *choice* to reveal that it was a "boy." You make that assumption, and further assume that the choice was made randomly. I do not feel these assumptions are the common sense ones to make.

There's another piece of information being ignored here. The person told you he has *only* two children. Odds are that this person stopped having children after the second child *because* he had just completed the perfect 1 boy / 1 girl family!

This happens a lot in reality! 1 boy / 1 girl families are more common than families with 2 boys or 2 girls.

http://www.bioone.org/perlserv/?request=get-document&doi=10.3378%2F1534-6617(2007)79[255%3AHSRASD]2.0.CO%3B2&ct=1

"We previously analyzed data from the U.S. National Health Interview Survey (NHIS, 1998 to 2002) on families with two biological children (10 years of age and younger) and found that the distribution of families with two boys, two girls, and one boy + one girl did not statistically conform to a binomial distribution regardless of the boy/girl sex ratio used. Using the best estimate of the sex ratio from the data, we found that there were significantly more families with opposite-sex siblings than families with same-sex siblings. No biological mechanism could explain these results at the time. In the present study we conducted an analysis of the first two children in sibships of size 3 from the same data source and found that there are significantly more same-sex sibships than unlike-sex sibships. Combining the two sets of data for the first two children produced observed numbers in close agreement with the expected numbers. A hypothesis of parental choice (family planning) appears to be strongly supported as an explanation for the discrepancies in the two sets of data individually. For example, parents who have a boy and a girl (either order) as their first two children are more likely to stop having children (“stopping rule”) than are parents whose first two children are of the same sex."

@David,

When you don't believe the concise mathematical arguments, we must fall back to showing you with empirically evidence. Personally, I understand the math that is being discussed here.

Can you provide a step-by-step mathematical proof of why it must be 50%? I'd be happy to read it and show you where you're wrong.

The original problem wasn't phrased as an unfinished game, though. Clear requirements are important. You've just stumbled into a statistical trap.

One of the problems in statistics is that people often let information they have ("I have a daughter") influence the answer they are looking for ("Is the other child a boy or a girl"). If that's the question, then the answer is 50/50. The fact that the first child mentioned is a daughter is not relevant information.

Here's another way to phrase the same problem: There are two children on the other side of a door. They walk through the door, one after the other. The first child through is a girl. What's the odds that the next one is also a girl?

The combinations that these could walk through are:
boy/boy (25%)
boy/girl (25%)
girl/boy (25%)
girl/girl (25%)

The first two are already eliminated, leaving the last two combinations. To map back to the original problem, replace "going through the door first" with "being mentioned by the parent first".

The important part here is that you eliminate BOTH of the combinations starting with "boy", and not just the "boy/boy" one.

Unfinished game questions are subtle, Jeff. This looks like one, it smells like one - but it's not one. That said, a different phrasing _would_ make it an unfinished game. For example, given the "walk through a door" - if I predict that there are two girls, AND a girl walks through first, then my prediction is 66% likely to be correct. But if I don't make the prediction until AFTER the girl has walked through, my prediction is only 50% likely to be correct. The reason that the first prediction has the higher chance of being right is because it has already missed one of the points where it could have been proven wrong.

In the example given, Jeff, the prediction is being made afterwards. Sorry, but you lose this one.

To compare to the coin tossing, take the position where Ted has two coin flips to win the game, and needs one head - he has a 75% chance of winning. But add the information that one of the toss is definitely tails, and his odds drop down to 50/50 - because the other coin toss is not related.

In the same way, when you tell us that one of the children is a girl as part of setting up the problem, then odds that the other is a girl is only 50/50.

Some of the respondents arguing for the 2/3rds are assuming you've chosen this woman at random from a statistically even group. You may have meant that, but you didn't say it. For all we know, you have a magic filter that detects women with two children where one is a girl - for example, you are in a clothing store, and you see a woman with a five-year-old child looking at unisex baby clothing. All we can go buy is the information in the problem, which gives a 50/50 split.

@David

"- I give someone a coin,
- I ask that person to flip that coin twice, behind my back, without me seeing anything
- The person comes back to me and tells me one of the two outcomes (either one, it doesn't matter)
- I can then tell the person that the chances the other toss was the opposite is 50%, because one coin toss has no impact whatsoever on another coin toss."

That is not, by any stretch of the imagination, 'showing your working'. Your last bullet is just an assertion that you've jumped to without filling in the gaps. For pity's sake, go and TRY this experiment! Do it 100 times and tell us how often the 'other' coin is heads when the person tells you a coin is tails. I promise you that when you tally up the results, you'll find that when you are told a coin has come up tails, 2/3 of the time the other coin will be heads. GO AND TRY IT. You have now proposed two experiments in support of your position, and BOTH experiments will prove you wrong.

Let us put 1million guys who have 2 kids in a (large) room. And ask them the gender of their first kid.
We expect half of them (.5 million) to say boys, and another .5 million to say girls.
According to Jeff that would mean .5million *. 66 * 2 would have a boy and a girl (and therefore would be eligible to say that they have two children and one of them is a girl.
That means that only 33% of the kids are boy/boy and girl/girl. I don't think so.

The Monthy hall question is about dependent events. What is behind a door has a bearing on what is not behind the other door.

Whereas gender is random - not really, but in the context of this question we assumed it is- and independent. One of your kids gender has no bearing on the other one.

Since Jeff has only one child yet, he does not know this yet. So I forgive him.

Math is hard. Let's go shopping!

Math is hard. Let's go shopping!

Not double-posting is hard, too. Let's shop more!

@mike

I'm reposting my repeatedly posted original post so that you can read what a compelte reasoning to the solution of this problem may be. And I think that it's the only language all of us with different positions can work out from - to try to argument with the argument basis of the counter argument, and not allienate it with new arguments. In a less meaner tone that I seek, I'd take out the word "illusion" and replace it with "conclusion". Here it is, The Post:

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@Russ:

*YOU* go try it. If you're *always* told when one of the coins was tails then you'll get your 2/3.

But if you're always (randomly) told what one of the coins was, then being told "tails" doesn't make the other one 2/3 heads.

@mike

I'm reposting my repeatedly posted original post so that you can read what a compelte reasoning to the solution of this problem may be. And I think that it's the only language all of us with different positions can work out from - to try to argument with the argument basis of the counter argument, and not allienate it with new arguments. In a less meaner tone that I seek, I'd take out the word "illusion" and replace it with "conclusion". Here it is, The Post:

There's a family with two children, one of them is a girl.
If this 66% illusion is based on GB and BG being accounted as different because G and B may be first or last (second) in the combo, then there are two sorts of GG and BB. So,

GG GG BB BB GB BG. BB and BB go out since we know there must be at least one girl. Which leaves,

GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG).

@Russ
There is NO experiment to make, because there is ONLY ONE draw, for which we already know part of the outcome! For the Nth time, this IS NOT about calculating the odds against multiple outcomes, it's about ONE precise outcome. And I HAVE tried it: my wife tossed two coins ONCE, told me one of the outcomes, and I was right to assume there was a 50% chance that the other toss gave the opposite result.

And I too know my math. I am NOT discussing whether what you say is right or wrong given what YOU want to do with this problem, I am only saying that the question that was asked is about A SINGLE coin toss, not multiple of them.

@gavkar

"That means that only 33% of the kids are boy/boy and girl/girl. I don't think so."

Wrong. The boy/boy combination has been ELIMINATED from the possible outcomes, and girl/girl represents 33% of what remains.

@Robert

"In the same way, when you tell us that one of the children is a girl as part of setting up the problem, then odds that the other is a girl is only 50/50."

Wrong. The probability of a child being born a girl is indeed 50%. But we are dealing with an already existing probability distribution that has been distorted by additional information eliminating some of the possibilities. The question is simply asking us to consider the mixed-gender-pair population. Initially it's 50%. Then we eliminate 25% of the alternatives (the BB couples). Mixed-gender-pairs now account for 2/3 of the population, so there's a 2/3 chance that the couple we are talking to is part of that mixed-gender-pair population and has a boy.

The question is ill posed, because the anwser is highly dependent on how the couple told you that they have one girl.

If they tell you "we have two kids", you ask them if at least one of them is a girl and they answer "yes", then indeed the probability of the other kid beeing a boy is 2/3.

If they tell you "we have two kids, and one of them is a girl", then by all rules of rational discourse the probability for the other kid being a boy is 1.

Things start to get interesting if the talking member of the couple is a mathematician (so the rules of ordinary discourse do not apply) and he says "We have two kids, and at least one of them is a girl". Now you might say that this is equivalent two the first case. But you could also go bayesian (see the link given by jason above) and look at the possible conversations with couples containing at least one mathematician that have two kids. Say you have 100 couples. On average, 25 of these will have two boys and tell you "We have two kids, and at least one of them is a boy", another 25 will have two girls and tell you "We have two kids, and at least one of them is a girl". The remaining 50 couples will have one boy and one girl. Assuming no gender bias on behalf of the speaking mathematician (ha!), 25 of these will tell you "We have two kids, and at least one of them is a boy", while the remaining 25 will say "We have two kids, and at least one of them is a girl." So out of the 50 couples that tell you "We have two kids, and at least one of them is a girl," 25 will have two girls, and 25 will have a boy and a girl. So in this case, the correct answer is 1/2.

@David

"my wife tossed two coins ONCE, told me one of the outcomes, and I was right to assume there was a 50% chance that the other toss gave the opposite result."

HOW DO YOU KNOW? How do you know the opposite result wasn't 66%? How do you know the same result wasn't 33%? If you knew ANYTHING about statistics, you'd know that a single sample is meaningless. By that logic, if I do the lottery once, I can either win or lose, so it must be 50/50. See how nonsensical that is?

All of those have one thing in common: each has exactly *one* unknown outcome, (X).

If you're advocating that P(Boy+Girl) = 2/3, you're correctly answering the wrong question ("What is the probability of a currently-childless couple having one boy and one girl?").

The girl in the problem *as stated* is a red herring; the question

A couple has one child: one is a (X).
A couple has two children, one is a girl, the other is a (X).
A couple has three children, one is a girl, one is a boy and the third is a (X).
...
A couple has 436 children, 218 are girls, 217 are boys and one is a (X).
A couple has 437 children, 218 are girls, 218 are boys and one is a (X).

The single unknown flips one of two ways, so P(Boy) = 1/2.

problem is "what is the probability that the one unknown is a boy

P(X is a boy) = 0.5

"GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG)."

GG and GG collapse to the same thing, because we don't know which of the two girls the parent was referring to. In the GB and BG cases the parent must be referring to the unique G.

@xooorx

"*YOU* go try it. If you're *always* told when one of the coins was tails then you'll get your 2/3."

In the question Jeff asked, we ARE always told. That's the POINT, we have a problem domain constrained by additional info. You keep coming back to the source of the data. It doesn't matter. It's out of scope. We are given the premise and challenged to work from it. With the information we are given, the answer is 2/3.

If someone asks you to calculate from how high an egg can be dropped onto concrete without breaking, do you run off and question Einstein's credentials before accepting gravity as a premise in the calculation?

"GG GG GB BG, so the probability of the family having a boy and a girl is 50% (2 combos possible out of 4: GB or BG)."

GG and GG collapse to the same thing, because we don't know which of the two girls the parent was referring to. In the GB and BG cases the parent must be referring to the unique G.

(Gak. Sorry for the double post.)

All of those have one thing in common: each has exactly *one* unknown outcome, (X).

If you're advocating that P(Boy+Girl) = 2/3, you're correctly answering the wrong question ("What is the probability of a currently-childless couple having one boy and one girl?").

The girl in the problem *as stated* is a red herring; the question asks for the probability of a single unknown. It wouldn't matter how many other children the couple has:

A couple has one child: one is a (X).
A couple has two children, one is a girl, the other is a (X).
A couple has three children, one is a girl, one is a boy and the third is a (X).
...
A couple has 436 children, 218 are girls, 217 are boys and one is a (X).
A couple has 437 children, 218 are girls, 218 are boys and one is a (X).

The single unknown flips one of two ways and isn't influenced by prior outcomes, so P(Boy) = 1/2.

All of those have one thing in common: each has exactly *one* unknown outcome, (X).

If you're advocating that P(Boy+Girl) = 2/3, you're correctly answering the wrong question ("What is the probability of a currently-childless couple having one boy and one girl?").

The girl in the problem *as stated* is a red herring; the question asks for the probability of a single unknown. It wouldn't matter how many other children the couple has:

A couple has one child: one is a (X).
A couple has two children, one is a girl, the other is a (X).
A couple has three children, one is a girl, one is a boy and the third is a (X).
...
A couple has 436 children, 218 are girls, 217 are boys and one is a (X).
A couple has 437 children, 218 are girls, 218 are boys and one is a (X).

The single unknown flips one of two ways and isn't influenced by prior outcomes, so P(Boy) = 1/2.

@Russ

"If someone asks you to calculate from how high an egg can be dropped onto concrete without breaking, do you run off and question Einstein's credentials before accepting gravity as a premise in the calculation?"

No.

But I do ask what planet we're on.

I've read some (who has time to read them all) of the arguments above, and I still come to the 50/50 conclusion. I suppose I must be mentally defective in some way.

"Let's say, hypothetically speaking, you met someone who told you they had two children,"

OK, so we have child 1 and child 2, the possible sexes are:

c1 | c2
B | B
B | G
G | B
G | G

" and one of them is a girl."

OK, since we don't know which child is a girl (but DO know that one of them is a girl) lets look at the two options:

If child 1 is a girl, we are left with:

c1 | c2
G | B
G | G

If child 2 is a girl, we are left with:

c1 | c2
B | G
G | G

" What are the odds that person has a boy and a girl?"

In either case, even odds, 50%.

Big upvote to Russ for patience! :)

@mike

thank you. so instead of GG and GG we have one and only GG

same goes to BG and GB, they collapse in the same thing.

so, GB and GG.

50% probability of the family having a boy and a girl.

@Gak

"If you're advocating that P(Boy+Girl) = 2/3, you're correctly answering the wrong question ("What is the probability of a currently-childless couple having one boy and one girl?")."

Er, what? You have this backwards. The correct answer to the question "what is the probability of a childless couple having BG children" is NOT 2/3, it's 50%.

The key point in the question as stated is that since we already know they have a girl, that eliminates all the BB couples from the possible outcomes. Look at it this (macabre) way - if 50% of children are boys and 50% are girls, and all the boy-only families are killed, does that make it more or less likely that any given (already-born) child is a girl? It makes it MORE likely, because there are now more girls than boys, without in any way changing the 50% probability of either gender being born.

@Russ

No, we don't know that we are always told whether there's a girl in the mix. We know that in this ONE instance, the parent told us there's at least one girl. Whether the parent randomly picked a gender to reveal, or whether the parent always reveals only girls, is the entire difference between 1/2 and 2/3 probability here.

Your forgot to combine the odds.

Mark,

You should collapse your two GG cases, because in either case, the parent would have reported that they have a girl (it doesn't matter if it was c1 or c2). If you don't combine them, you'll end up double counting them.

@Russ

The "gravity" of the information that one child is a girl depends on whether its coming from a preferential-girl-revealer or a tell-you-the-sex-of-one-of-my-children revealer.

@mirror 3

You need to go flip some coins. BG and GB do collapse, but into a bucket that's twice as big as either BB or GG.

@mirror 3

Nope. BG and GB don't collapse because they are distinct. The parent _had_ to report the unique G in either case. In the Gg case, you double count it: once for when G is reported, once when g is reported.

But none of that is really important if you consider Russ's argument:

Any particular family of two has a 50% chance of having one boy and one girl. Any particular family of two has a 25% chance of having two girls. Likewise for two boys. Therefore, we get to a 50/(50+25)% chance of having one boy and one girl *given* that we know the family has at least one girl.

Another way to look at the boy/girl problem is to consider this instead:

Mr Smith played the lottery twice. Each time he had a chance of 1:1,000,000 to win. Thus, winning twice, the odds are 1:1,000,000,000,000. If I told you he won at least once, you'd say he was lucky. But if I told you he won twice, you'd say 'impossible', because while it is possible to find some people like Mr Smith that won once, you never heard of anybody winning twice.

The proof Jeff gives is really half-assed. He tells us that some famous mathemetician made the same mistake. So was it for exactly the same question? Maybe he made some mistake about something similar but not exactly the same. If it was the same question, could you not give us the name of the person that proved him wrong and his proof on how he proved Mr Pascal wrong.

Jeff's "proof" is based on this single sentence:
"We know that one of the children is a girl. This rules out one of those possible combinations of two children (BB), so we're left with: ...".

Really. Is that the proof? Let us say we forgot about this guy and we see him again. And this time he says he has a boy and he has two kids. I guess the chances that the other one is a girl is 66%. So this guy has a ~133% chance to have a boy and a girl. The randomness of the event and the randomness of our encounter with the observer of the event has not changed a bit. Why assume that we have to operate from a smaller set of possibilities.

If you are convinced that we need to operate from a smaller set of possibilities you would think that the answer is 2/3, but if you factor the randomness of meeting with this guy, that we had 75% chance of meeting with a guy who has two kids and one of them is a girl, then it is 1/2 chance that the other one is a boy.

So it is not about the math, but about about how poorly Jeff asked this question, and then how more poorly he answered the question.

And then Jeff goes on a tangent and speaks about half finished games.

But he has spoken so authoritatively about his answer I guess we have to take his word for it.

@ xooorx

You didn't include the odds of the parent being a preferential-girl-revealer with two boys.

You also didn't include the odds that the parents are aliens which a 3rd sex that occurs at 20% frequency.

Seriously, where do people come up with this stuff?

@xooorx

"But I do ask what planet we're on."

Well, that pretty much sums it up, doesn't it? "Xooorx, see how far you can drop this egg." "What planet are we on?"

Can you really defend this as a reasonable position? I think you're being deliberately obtuse and pedantic at this point. The original question was a (badly-phrased, admittedly) simple math analogy with a premise and an interesting conclusion. I fail to see any value in this relentless misdirected attack on the premise. It's a PREMISE, the point of it is you accept it for the sake of argument and work from there. The value is in identifying the statistical principles it reveals, not the motivations of the imaginary characters invented to add a bit of colour.

@ Russ,
if 66% has a boy and a girl, and an another 33% has girl/girl, that means there are no boy/boy in the population.
I think in haste you did not really read my post carefully.

Cheers.

The bug in the original spec invalidates the answer. You said (paraphrasing, because I really couldn't be arsed looking it up) that the parent in question has two kids and one of them is a girl. Leaving out aliens, hermaphrodites and other such coin-on-the-edge complications, that means the other one is a boy, because one is not equal to one-or-more in the English language.

Debug your spec, or don't complain about the implementation.

I had sex with 3 women last night and the babies were born this morning.

Damn if Jeff isn't correct!

@gavkar,

Exactly, the boy/boy combination has been excluded in the population because we know that there must be at least one girl.

Another variation of the problem, offered in the hopes that it will help break someone out of the log jam...

Anna brought two children to meet you today, one of them is her daughter Beth. Later, you meet a girl who tells you that Anna is her mother - what is the probability that this is Beth?

After you have figured out the answer to this, determine if this problem is analogous to the original.

@gavkar

"if 66% has a boy and a girl, and an another 33% has girl/girl, that means there are no boy/boy in the population."

YES! That's it! We know they have a girl, so the boy/boy combination is not present in the population of possible outcomes!

@charles
"But if I told you he won twice, you'd say 'impossible', because while it is possible to find some people like Mr Smith that won once, you never heard of anybody winning twice."

YOU might say "impossible", but I would only say "improbable".

Google is your friend. Keywords: two-time lottery winner

http://www.lotterypost.com/news/166011

@Russ
Are you actually asking my how I know that flipping a coin gives a 50% chance of heads vs. tails?

Let me try yet something else:
- I tell you I'm going to have one child. What are the odds it's going to be a boy?
- In two years, I tell you again that I am going to have a child. Put aside the fact that you know that I already have a boy. What are the odds that it's going to be a boy?
- Now, does knowing that I already had one boy before change anything? Absolutely not! The probability is still 50%.

@ Russ and @Mike
I guess you guys are pretty committed to your answer and refuse to bother reading posts in their entirety.
My post was not about the question Jeff presented. I was talking about a separate hypothetical experiment. Could you point to me how are the boy/boy combination is eliminated in my hypothetical case?
How do "We know they have a girl" or , " that there must be at least one girl" ? Half million people said they have a boy and did not say anything about the gender of the other kid. How do YOU know that the other one has to be a girl?

Stages of grief that you thought it was 50%:
Denial Anger Bargaining Depression Acceptance

@Russ
And by the way, if the lottery had only two numbers, each of them being of equal 50% probability of being selected, then yes, absolutely, you would have a 50/50 percent chance to win/lose each time you play.

@ steve
Oh, you got us. You must be so smart. I bet your other sibling is 66% chance stupid.

@steve
"Stages of grief that you thought it was 50%:
Denial Anger Bargaining Depression Acceptance"

Let me fix that for you.

Denial
Restating problem with different premises
Coding
Debugging
Questioning which planet the egg is on

@Russ

I'm being pedantic yes. I'm pedantically insisting that the question requires additional constraints added to it (that parents are biased in favour of revealing that they have a girl if they have one) before the answer comes out as 2/3. Thats the equivalent, in the egg problem, of being told that oh by the way we're on the moon, and it affects the answer.

Without this constraint the answer is 1/2

I've been thinking all day about this and I was wrong and Jeff is right. Sorry, Jeff. Too many testosterone, I guess.

@Russ
Putting aside your rudeness, sure, there are no statistics without multiple samples. But what do you see in the question that suggests there are multiple samples involved? If you answer is "nothing, but for the question to mean anything we need multiple samples", then you are not answering the question, no matter how right your math is. You are answering your own version of the question.

++++ to ChrisP for his comment regarding the imprecise wording of the original question.

I understand (and did umderstand) the probability concepts from both of Jeff's posts, but by the wording of the original question, my answer is still 100%.

"If you are convinced that we need to operate from a smaller set of possibilities you would think that the answer is 2/3, but if you factor the randomness of meeting with this guy, that we had 75% chance of meeting with a guy who has two kids and one of them is a girl, then it is 1/2 chance that the other one is a boy. "

Right, if you start throwing out information provided in the problem statement, then you end up with the wrong answer!

Of course, the problem states that the family has at least one girl, so we know that the family is not one of the 25% that has two boys. Therefore we know that there's a 33% chance that both of the family's children are girls. (and 66% chance that there's one of each)

Russ, no. The relevant sample set is not "all families", or "all families with two children", but "all families with two children, where one is a girl".

You can't bring in the probability of "families with two children where both are boys", because that assumes you are starting with the set "families with two children". You aren't.

Look at the progression. You start with a woman. She is in the set of all women. You find out that she has children. That narrows the set further - to women with children. Some women have 0 children; we can ignore them. Some women have 1 children, 2 children, or more.

We then narrow that set down further: the set of women with two children.

Finally, we narrow the set down to the information given: the set of women with two children, one of whom is a girl. This will be split 50-50, girl/boy for the second.

You can't go upwards to include probabilities from the earlier sets - you don't know the odds involved. Perhaps there was filtering involved (e.g. you walk up to a woman in a store, who has a female child at her side). Perhaps there wasn't. Without more information, you can not go upwards. If you do go upwards, why stop at the first level up (the set of women with two children)? Why not go up to the set of all people in your country, in which case you need to start with "well, what are the chances that this random person I bump into is a woman?" and "what are the chances she has children?" and "what are the chances she has exactly 2 children?"

I've seen people using assumptions from the general American population. Where in the problem is that stated as part of the sample? Maybe the woman is at a party for spouses of Israeli fighter pilots - who, due to the nature of their job, have about an 80% chance of having girl children. That would be pertinent information, but it's not stated as NOT being excluded. So you have to factor that chance in. It's not high if you happen to be in the US, but if you live on an Israeli Air Force base, it's pretty likely.

The answer is 50/50, _for the problem as stated_.

Look at the coin toss: Harry needs to toss the coins, and needs at least one head. What's the chance he wins? 75%. But that's not the question asked by Jeff with the girl/boy. In the girl/boy scenario, the relevant question would be: "Harry needs to toss the coins twice, and needs at least one head. He tells you later that one of the coins was a tail. What's the chance he won, assuming that the coins were definitely tossed twice?". It's 50/50.

You want combinatorial examples? Okay: two children, A and B. They can be ordered:

A/B, B/A

with a 50% split.

You are told that A is a girl. B has a fifty/fifty chance of being a girl or boy (when taken in isolation). This means we have these options:

Girl(A)/Boy(B) - 50% * 50% -> 25%
Girl(A)/Girl(B) - 50% * 50% -> 25%
Boy(B)/Girl(A) - 50% * 50% -> 25%
Girl(B)/Girl(A) - 50% * 50% -> 25%

They are all the valid combinations, and it results in an even 50/50 chance that child B is a girl or a boy.

Just another way of branching this problem.

Say the 2nd child is not born yet but inside mother's womb.
2 scenarios:

S1: The question is asked BEFORE the child's gender is known to the parent

S2: The question is asked AFTER the child's gender is known to the parent

I would say S1 gives 50%, S2 gives 2/3 (or 66.67%). Yea?

Actually you are wrong on all counts.

Each couple's probability of having a boy or a girl is different. You assume all couples are equally likely to have a boy or a girl. The couple's hormones at instance of conception decides whether they are predisposed to one gender or the other.

The right answer is not enough information:) If you want to have answer approaching any kind of accuracy. You need a little more than probability theory. Think of King Henry VIII. Who was cursed to have all female children.

As with all programming you should know your domain before writing any code. Maybe thats the real lesson here. The last 100 comments were theoretical fluff.

http://www.sexratio.com/facts.htm

@David

RE: no statistics without multiple samples

Likewise, you can't determine the probability of a single event without considering the pool from which it's been drawn.

@David

"Are you actually asking my how I know that flipping a coin gives a 50% chance of heads vs. tails?"

No, and you know damn well that that is NOT the problem area being discussed. This is a case of outcomes that have already occurred, not a question of a particular outcome occurring. You KNOW this because you've proposed two experiments that would confirm the 2/3 result, but you refuse to run any of them enough to produce statistically-valid results.

Maybe a different approach is in order, because I'm sick of explaining the math to you over and over and over again. Think of this as a categorisation problem. You have a problem domain divided into three unequal parts (I hope the ascii art comes out):

----------------------
| | |
| BB | |
| | |
| | |
|---------| BG |
| | |
| GG | |
| | |
| | |
----------------------

This represents the distribution of gender pairs in two-children families, in strict accordance with your beloved 50% boy/girl probability. Note that BG represents mixed-gender in which order is NOT important.

Now, if you meet a couple at random and they say they have two children, there is a 50% chance they are in the BG region, 25% chance they are in the BB region, and 25% chance they are in GG.

Now, if the tell us they have a girl, we can REMOVE the BB region:

------------
| |
| |
| |
| |
|---------| BG |
| | |
| GG | |
| | |
| | |
----------------------

Now, based on the information we have received, which constrains the problem domain, we know the couple are in either GG or BG. However, BG now represents 2/3 of the available solution space, so there is a 67% chance the couple is in that domain. In order to be in that domain they must have a boy in addition to the girl, so this is equivalent to saying there is a 67% chance the other sibling is a boy.

It has NOTHING TO DO with changing the likelihood of a particular gender being born. It's a categorisation problem.

Now seriously, if you don't follow that I give up, since even my cat is rolling its eyebrows at this now.

@Robert

""Harry needs to toss the coins twice, and needs at least one head. He tells you later that one of the coins was a tail. What's the chance he won, assuming that the coins were definitely tossed twice?". It's 50/50."

Not the same problem. If the first throw is heads, Harry doesn't need a second throw. He's already won.

Dammit, the whitespace got trimmed. Try here: http://pastebin.com/f44ef811c

Jeff, here's the heart of the problem. You stated:
"Our question contains additional information:

1. The person has two children.
2. One of those children is a girl.

We can use that information to come up with a better, more correct answer. We know this person has two children. What are all possible combinations of two children?"

You then proceed to use half of that additional information to set up your sample space: "the set of women with two children". You picked that half due to your own bias.

What if you had picked the other half? "the set of women with at least one child, who is a girl"? What sort of breakdown would you give?

You can't use only _part_ of the information. You need to use all of the information to set up the problem. This is the essence of the Monty Hall problem.

In the classic Monty Hall problem, you have three doors. You pick door X, and then the presenter picks a door (door Y) that does not have the prize. That is information, and it gets factored in to show that door Z has a 2/3 chance of having the prize. This is because the presenter KNOW that door Y does not have the prize.

In a modified Monty Hall problem, you have three doors, and you pick door X in your head (without revealing your answer). Another contestant - with no more information than you - picks door Y. It does not have the prize. What's the odds that the prize is behind door Z? 50/50. Because in this problem space, we need to include the chance that the other contestant picked the door with the prize - the very option we leave out in the classic version.

Jeff, here's the heart of the problem. You stated:
"Our question contains additional information:

1. The person has two children.
2. One of those children is a girl.

We can use that information to come up with a better, more correct answer. We know this person has two children. What are all possible combinations of two children?"

You then proceed to use half of that additional information to set up your sample space: "the set of women with two children". You picked that half due to your own bias.

What if you had picked the other half? "the set of women with at least one child, who is a girl"? What sort of breakdown would you give?

You can't use only _part_ of the information. You need to use all of the information to set up the problem. This is the essence of the Monty Hall problem.

In the classic Monty Hall problem, you have three doors. You pick door X, and then the presenter picks a door (door Y) that does not have the prize. That is information, and it gets factored in to show that door Z has a 2/3 chance of having the prize. This is because the presenter KNOW that door Y does not have the prize.

In a modified Monty Hall problem, you have three doors, and you pick door X in your head (without revealing your answer). Another contestant - with no more information than you - picks door Y. It does not have the prize. What's the odds that the prize is behind door Z? 50/50. Because in this problem space, we need to include the chance that the other contestant picked the door with the prize - the very option we leave out in the classic version.

@Derek
"Likewise, you can't determine the probability of a single event without considering the pool from which it's been drawn."

Absolutely. But then again, the only probability we are talking about here, given the way the question is formulated, is that of a newborn being a boy or a girl. And I'm saying it's 50% because that's the common accepted probability, not because I decided it was the right number.

Russ

It. Depends. How. You. Came. By. The. Information.

If you're reliably informed in 100% of cases when there's a girl in the family then yes.

If you happen to find out that one child is a girl then no. Because thats twice as likely to happen if both children are girls.

My cat has been jealously glaring at the laptop for hours.

ed, that's my point. Changing the problem changes the answer.

"I toss a coin twice. At least one of the coin tosses was Tails. What's the chance that the other is Heads" (50/50)

"I toss a coin twice. What is the chance that at least one of the coin tosses _will be_ Heads" (75%)

These are different questions, with different answers. The girl/boy question shown by Jeff is in the former category.

Robert,

You can't separate out the two births. Just because you know one of the woman's children is a girl doesn't mean you know *which* one.

Its only 2/3 if you KNOW that any parent with a girl will tell you they've got a girl.

Really really really really really.

Really.

Really.

If they just randomly revealed the sex of one of the children (<--- and I think thats what happened in the original question) then its 1/2

Monty Hall is only 2/3 because you KNOW he's not going to open the door on the car.

If he just randomly opens a door and it happens not to be the car, its 1/2

Mike,

You don't need to separate the birth order. The combinatorial example I gave showed _both_ possible birth orders.

In my first post, I didn't mention birth order - I said "order they were mentioned first". The first one mentioned was a girl, the second one - well, it's 50/50.

@Robert

"The relevant sample set is not "all families", or "all families with two children", but "all families with two children, where one is a girl"."

Not quite true, although this may be due to the way Jeff worded it. The fact that one of the children is a girl is presented as *additional* information, so the original set is in fact "all families with two children". Then, when we learn that one is a girl (which is explicitly described as additional info in today's post) we are asked to estimate the probability of a male sibling.

I understand your point about how many levels to go up - I happen to think that this is a bad way to present this problem since it is prone to all these details, none of which are relevant to the problem we are *supposed* to be focussing on. I much prefer the coin tossing alternative, as nobody wastes time arguing whether the coins came from a town known for circulating counterfeit coins or whether they've got mud on one side or whatever. Well, xooorx probably would, but he's special.

@ Mike
I did not make a prediction on if it is 50% or 66% on the original blog post, so I have no committed interest in defending my position. I simply did not have a position.

I guessed correctly, though I did not write about it, that Jeff would say 66%, his link to the Bayesian post was too much of a giveaway in that he though the set of possibilities were narrowed by the information.

I did not feel that was necessarily the correct answer since I did not feel the question was asked precise enough to come up with that conclusion.

I can see that you think that the question clearly presented: "Right, if you start throwing out information provided in the problem statement, then you end up with the wrong answer!" and would be justified under that assumption that the answer Jeff gave is correct.

I guess I expect a little more from the person who asks the question to present the context a little better to avoid the confusion. If Jeff's point was to remind us about his other posts and tell us a story about how math was evolved, that question though was the worst kind. We are all discussing the question, mostly because of its presentation, instead of talking about the points -at least what I assumed to be his points- that he was trying to make.

As far as I can tell everybody here has a disagreement on the semantics of the question and not the math. People are trying to re-word the question to better tell what they were thinking they were asked.

Congratulations on correctly understanding the question and then correctly answering it.

@Robert
"I toss a coin twice. At least one of the coin tosses was Tails. What's the chance that the other is Heads" (50/50)

Please show your work.

Here's mine:

HH at 25% frequency
HT at 25% frequency
TH at 25% frequency
TT at 25% frequency

Of the pairs where at least one toss was tails, what is the chance that the other is heads? Heads occurs in 2/3 of the remaining outcomes, so the conditional probability can't be 50/50.

So either show me your work, or show me where my work is wrong.

@xooorx

"Its only 2/3 if you KNOW that any parent with a girl will tell you they've got a girl."

Look, how about if you make the announced sibling non-gender-specific?

"You meet a couple, and they tell you they have a child of a particular gender. How likely is it that the sibling is the opposite gender?"

I know that's a bit clumsy, but it eliminates bias and STILL works out to 2/3 regardless of the gender of the first child and any biases on the part of the parent, as long as you accept that the parents aren't outright lying.

If you think this still comes out to 50%, I'd love to see your reasoning.

Robert,

No child was mentioned! You can't say that one was mentioned first, because the information that it is a girl covers the set of both children. You're being given partial information about the result of two random events. Thus, you don't get to order the children.

Russ, from the post yesterday:

"you met someone who told you they had two children, and one of them is a girl."

This is information presented at the same time - prior to drawing up the sample space and assigning probabilities.

The only sensible sample space to set up is "women with two children, one of whom is a girl". Otherwise, re-ordering the question to be:

"you met someone who told you they had a daughter, and that they have two children"

would change the answer without changing the information. That would be wrong.

I am a mathematician and you are wrong. I am a statistician working with frequentist probability and Bayesian problems and I have done genetics too. I can say categorically that you are wrong.

Without wading through all the comments, Gareth was right "This example is rubbish". Or in my words "woops buggered that one up.".

You are mixing up frequentist probability with Bayesian logic. Bayesian mathematics is used as information is progressively revealed, or where probability isn’t conceptually applicable – i.e. the Monty Hall problem (search Wikipedia for that). THAT is an example of what you are trying to explain. The division of the "pot" is also Bayesian logic.

In essence you picked the wrong example.

Your question simply asks whether the chance that the sex of the second child is Male or Female, as this is the only artefact that has a bearing on mixed gender siblings.

The answer is 50/50. Therefore the chance of BG is 50/50.

As many others have already stated, if you are going to say that the boy could be the 1st or the 2nd child, then why did you exclude a sister being the 1st or 2nd child and therefore represented twice????

You know what they say, to err is human but to really stuff up you need a computer (and apparently doing it publicly helps).

gavkar,

the intent of the question is quite clear, even if it was presented poorly. There is an interesting argument about whether the parent had the option to mention that they have a boy, but the usual interpretation, that there are two children, one of which is a girl is pretty clear. Most of the 50% arguments have presented incorrect math and information that contradicts what was presented in the question (for example, that we know which child was being spoken about).

@gavkar

"My post was not about the question Jeff presented."

Then why did you say "according to Jeff that would mean .5million *. 66 * 2 would have a boy and a girl"? If you're talking about a different problem, it's hardly fair to use it to critique Jeff's solution to the original problem is it?

For those who believe that this problem is not susceptible to the additional information, consider these problems, the only difference between them being the information given:

1) I have two kids. What are the chances that the youngest one is a boy?
A: 50%

2) I have two kids, the oldest of whom is a girl. What are the chances that the youngest one is a boy?
A: 50%

3) I have two kids, at least one of whom is a girl. What are the chances that the other one is a boy?
A: 66%

4) I have two kids, both of whom are girls. What are the chances that the youngest one is a boy?
A: 0%

5) I have two kids, both of whom are boys. What are the chances that the youngest one is a boy?
A: 100%

In each case after the first, the additional information changes the answer by eliminating possibilities. In the second case, the possibilities that are eliminated are irrelevant; in the other cases, they do have an impact.

I had a hard time getting it, but I think the comments involving the age are confusing if not irrelevant. I even tried adding the possibility of twins born at the same time to finally understand that focusing on the order is absolutely silly ; the order is only useful in representing the possibilities. The kids could as well be ordered by first name.

I think one of the most helpful comment is from Lerxst, because his example doesn't involve the age. But still, what's really important is already in Jeff's post : the game is already finished, and that's the only thing that matters to understand it.

I finally got it when I tried this simulation : just imagine someone tossing two coins, and then telling you one of them is heads. What are the odds that the other coin is tails ?

- generate 1000 random pairs of booleans
- only keep pairs where there is at least a 1. obviously you have about ~750 pairs. The odds of having a 0 are calculated from this subset.
- in this subset, count the pairs where there is a 0. And, hooray, you get about ~500 pairs !

@Russ

"You meet a couple, and they tell you they have a child of a particular gender. How likely is it that the sibling is the opposite gender?"

50%!!!!!!!

FFS!!!!

I pondered on the problem yesterday, and followed through to a wonderful page about Bayes theorem. And, as soon as I hit there, I understood why I was getting a different answer. It was simple: the description of the problem was "loaded".

The thing is, there is a very common problem used in teaching probability. It goes like this: A woman is pregnant with her second child. What is the probability of that child being a boy, given that her first child was a girl? The answer to that being 50%. This is a very different problem from the one you posted, but because it's a common problem, and because it was very similar to the one you proposed, it was the one that came to my mind.

KW: look, if you say a couple have a daughter and they only have one child, then that couple's first born is a girl. If you say a couple have a daughter and they have two children, then either the firstborn or the secondborn is a girl. Since the INFORMATION you have is different, the chances are different too.

Reminds me of a problem we did in discrete math this semester. Very famous problem that deals with odds that aren't so obvious:

Reminds me of a problem we did in discrete math this semester. Very famous problem that deals with odds that aren't so obvious:

Monty Hall Problem: http://en.wikipedia.org/wiki/Monty_Hall_problem

Credit to xoorx for persistently advocating the most sensible position (IMO). This follow up post of Jeff's is somewhat unbalanced, in pitting the valid 66% argument against an invalid version of the 50% argument. The deeper understanding comes with the realisation that both answers can be correct, and the question was incompletely posed (I think Rhys - 4.20am in the first post comments - was the first to clearly state this).

@Robert

I accept that the question was poorly worded. I still stand by my own interpretation, however, and that's been the basis of my comments. If you were right and there was only ONE set to consider - that of "families with two children, one of which is a girl" then what would be the additional information which forms the whole structure of the challenge? If you look at Jeff's solution, he clearly intended this to be an artefact of constraining the problem space from all possible child pairs to just those child pairs that include at least one girl, since that's our additional info. Of those remaining pairs, 2/3 also contain a boy, which leads to the result.

Ed, here's your mistake:

@Robert
"I toss a coin twice. At least one of the coin tosses was Tails. What's the chance that the other is Heads" (50/50)

Here's Ed's:

HH at 25% frequency
HT at 25% frequency
TH at 25% frequency
TT at 25% frequency

However, Ed, you assume that the HH is a possibility. It's not. The information _given at the start_ says that I have a tails.

This means my options are:
T/x (50% chance) and x/T (50% chance).

That covers all of my sample space. Because order is not relevant, I can collapse that to:

T/x

x has a 50/50 chance of being heads or tails. Therefore, by making the prediction after KNOWING that there is one tails, the chance of my prediction being correct is 50/50.

We have to remember that the question is about "what is the chance that my prediction (two tails, or two girls) being correct?" This depends on when you make the prediction.

If you make the prediction before you are told one of the children is a girl, you have a 2/3rds chance of being right. If you make it after, you have a 50/50.

If Monty Hall reveals the door without the prize before your guess, you have a 50/50 chance of being right. If he reveals it AFTER you make your guess, your guess has only a 1/3rd chance of being right. It's all about the timing.

Here, we are told that the woman has a girl BEFORE we make the prediction.

Just clarifying my earlier statement...

Assume we are talking about coin tosses:

This is frequentist probability:
HH at 25% frequency
HT at 25% frequency
TH at 25% frequency
TT at 25% frequency

So if you already have an H the chance of getting mixed is…
HH at 1/3 frequency
HT at 1/3 frequency
TH at 1/3 frequency

Therefore 2/3 chance of getting mixed.

But this is Bayesian:
Original frequencies are not longer relevant. We have 1 H, what are the options we have for another coin toss?
H – 50%
T – 50%

What if we don’t know whether this is the first or second throw? Then these are all the options:
We started with H:
HH
HT
We ended with H
HH
TH

So there is a 50% chance we will get HH and a 50% chance it is mixed.

This can be confusing, so please do a Wikipedia search on the Monty Hall problem.

@Robert
The only sensible sample space to set up is "women with two children, one of whom is a girl". Otherwise, re-ordering the question to be:

"you met someone who told you they had a daughter, and that they have two children"

would change the answer without changing the information. That would be wrong.

You say that it would change the answer, but does it? Show your work. Here's mine.

In the population of women with two children, one of whom is a girl, here are the birth-rate frequencies.

GB 33%
BG 33%
GG 33%

2/3 of the possible outcomes have one girl and one boy, so the probability is 2/3 for the other child being boy.

So either show your work, or show me where my work is wrong.

Ben, your example is wrong.

3) I have two kids, at least one of whom is a girl. What are the chances that the other one is a boy?

This assumes you start in the set of two children. You don't. You start in the set of two children, one of whom is a girl. It's 50/50.

You can stand by your interpretation, Russ, and I will agree with you that your interpretation is what Jeff means. But it's not what he said.

He even breaks it down clearly: He gives two pieces of information above, and then proceeds to set up his sample set using one of them. That's wrong.

"50%!!!!!!!

FFS!!!!"

Show. Your. Working.

Look, the initial set of permutations is BB, GG, Mixed, where P(BB) and P(GG) are both 25%, and P(Mixed) is 50%.

If the given gender is B, we eliminate GG leaving us with BB, Mixed. Since P(Mixed) == 2P(BB), it's twice as likely the couple are in P(Mixed).

If the given gender is G, we eliminate BB leaving us with GG, Mixed. Since P(Mixed) == 2P(GG), it's twice as likely the couple are in P(Mixed).

The math works REGARDLESS of which gender is chosen as a starting point. Therefore, if you meet a random couple and flip a coin to decide whether they tell you about a boy or a girl (thus eliminating any potential for them to be biased), you still get the same response.

Now, as I asked before, please show your work.

@philip

"This can be confusing, so please do a Wikipedia search on the Monty Hall problem."

Please understand that Monty Hall was a biased informant. He was well known never to open the door on the car. This was important information that contestants could use to increase their chances of winning the car.

@Robert

I'd be interested in your response to ed's post. Even if you eliminate BB from the starting set, B is still present in 2/3 of your sample set. You either start with a full set and add information to constrain it, or you start with the already-constrained set. The result is the same either way.

Ed, you keep making basic mistakes. The options are:

G/b - 25%
G/g - 25%
b/G - 25%
g/G - 25%

where the uppercase is used to distinguish the child designated as "girl" without specifying the birth order.

Making order irrelevant, this collapses down to:
G/b - 50% and G/g - 50%.

If you want to use your notation, it's
In the population of women with two children, one of whom is a girl, here are the birth-rate frequencies.

GB 25%
BG 25%
GG 50%

@Russ

boy/boy: (25%) <--- sibling is same
girl/girl: (25%) <--- sibling is same
boy/girl: (50%) <--- sibling is other

I'm making that same sibling 50% of the time. How are you getting on with it?

@xooorx
Yes - and as more information is revealed the probabilities change.

MORE INFORMATION was reviled - one child is a girl. Therefore the probabilities change. Everyone is using the "base" probabilities (see Jeff's original example where he just struck one out).

My point is that with the new information we can't just strike one out and say “that’s the answer”, we must reconstruct what we know the probabilities to be now. And we do that by looking at all of the current options. This is what the Monty Hall Wikipedia link does very well. This is what Jeff does not do.

Russ - I've now given the working twice, once with coins, once with genders. The answer is still the same.

By the way: you (and others) seem to be mixing up permutations and combinations. Children are not interchangable. You have to use combinations. The large set of combinations are:

B/b - 12.5%
B/g - 12.5%
b/B - 12.5%
g/B - 12.5%

G/b - 12.5%
G/g - 12.5%
b/G - 12.5%
g/G - 12.5%

Uppercase is used for the "child that will be identified". When you identify a child as a girl, you blow away the other half.

boy/boy: (25%) <--- sibling is same
girl/girl: (25%) <--- sibling is same
boy/girl: (50%) <--- sibling is other

Great. Now you flip a coin, it comes up heads, so you ask the couple if they have a boy. They say yes. Eliminate girl/girl since you now know that is an impossible classification. The couple must have either boy/boy, or boy/girl. It is twice as likely (by your numbers, above) that they have boy/girl. Hence, 2/3. Where's the bias? Where's the possibility for misinterpretation? It's clear as a bell.

Is anyone actually reading all these comments? :-)

I think the fundamental problem with people understanding these types of statistical questions is that intuition simply doesn't work.

Being able to solve problems like this requires formal training (e.g., a university-level statistics class), and further requires rigorously following then rules when setting up the problem. It's all in how you set up the problem.

Even then, it is enormously difficult to to get some of these things right. I got the Monty Hall Paradox wrong when I first read about it, but then when I read the explanation (slowly) I understood it perfectly. Then I send it out as a brain-teaser in an e-mail to some friends at work, and the explanation actually made some of the people angry.

Philip,

You count HH twice, but it will only be reported half of the time: i.e., P(reported first H) = 50%. Put another way, we don't know where the H occured: in the HT and TH cases, we know with certainty. In the HH case, we can only guess: 50/50, so you have to multiply the chance of HH appearing by the chance that the H that was reported was the first (or second) depending on your case.

I get the argument that says, given the presenter chose to randomly report either the first or the second coin's value that the other's value is independent. If you want to argue that, I'll concede that's a possibility. Is that what you're trying to get at? Because I don't buy what you posted as the "Bayesian" approach.

I preferred much more the OB discussion: http://www.overcomingbias.com/2008/10/my-bayesian-enl.html

@Robert
G/b - 25%
G/g - 25%
b/G - 25%
g/G - 25%

where the uppercase is used to distinguish the child designated as "girl" without specifying the birth order.

I didn't say anything specific about birth order. I'm just talking about the relative frequencies of 2 independent events.

There is no G/g vs. g/G distinction. They are one and the same for frequency of occurrence. That is, the frequency of occurrence is not dependent on "I'm mentioning my older daughter" vs. "I'm mentioning my younger daughter".

If you want to use your notation, it's
In the population of women with two children, one of whom is a girl, here are the birth-rate frequencies.

GB 25%
BG 25%
GG 50%

You're asserting that having a firstborn girl doubles the chances for another girl? Then the events aren't independent, and that's a different question.

I strongly urge you to run the experiment using a statistically significant sample. Seriously.

@Russ:

Yes! If I *ASK* the couple if they have a boy its 2/3! I'm turning them into preferential boy revealers. I know that they'll look at both of the kids and, if either is a boy, they'll say they they've got a boy.

But if they just happen to reveal that one of the kids is a boy its different. They're no longer preferential boy revealers, they're just people that told me what sex one of their kids is. The answer becomes 1/2.

Russ - we've said it before. It depends on when you make your prediction. Here, you are predicting that they are "boy/girl" before you ask the question, and therefore leaving out the chance that you are wrong.

Yes, if you predict "boy/girl", and THEN find out that one is a boy, you have a 2/3rds chance of being right. But that's because you had a 25% chance of being flat wrong before.

Here's another example: I have 5 boxes. One of them has a coin. Pick it and you win. 20% chance, right?
You pick, I get rid of 3 boxes, and one of the two remaining has the coin. You should swap - your original pick is still just out of 5 boxes, but the other box is one that survived 3 eliminations. That's Monty Hall.

BUT: if I put down 5 boxes, tell you one of them has a coin, and then take away 3 boxes BEFORE you pick, what's your chances? It's 50/50.

Here, the two pieces of information (woman has two children, one is a girl) are both present at the start. Jeff's working makes that clear. Using only half the information present is flat out wrong.

@Robert

It doesn't matter WHICH child is identified. b/G and G/b are not distinct. When you state you have a girl, you can ONLY eliminate BB (25%). You can't eliminate B/g (and it's 12.5% weight) just because the boy wasn't identified, since B/g is still a valid solution.

When you eliminate one of the same-gender sets, the ENTIRE mixed-gender set is still in consideration, and forms 2/3 of the remaining problem domain.

@ Russ
""My post was not about the question Jeff presented."

Then why did you say "according to Jeff that would mean .5million *. 66 * 2 would have a boy and a girl"? If you're talking about a different problem, it's hardly fair to use it to critique Jeff's solution to the original problem is it?"

Really? Do I need to explain this?

I will take the easy road and will ask you to explain how you eliminated the boy/boy combination from my hypothetical experiment.

Good luck.

Another question for the 2/3'rd crowd, and it has probably already been asked -- How is Jeff's question different than if I ask "Let's say I flipped a coin nine times, and got "heads" each time. What are the odds that the tenth flip will be tails?"

I would say the odds are 50%.

The domain of previous possibilities don't matter in this case because the outcome has already been given with the problem.

(At this point in this long discussion, I'm in the "it's a bad question" camp -- There's not enough info in Jeff's question.)

@xooorx

"But if they just happen to reveal that one of the kids is a boy its different."

Show. Your. Work. If they just 'happen' to reveal one child is a boy, it still eliminates the GG domain and leaves the BB domain and the BG domain. The BG domain is still 2/3 of the remaining solution space, so the answer is still 2/3.

HOW is it different for someone to 'just happen to reveal' something as opposed to being asked a question determined by an unbiased coin?

@Robert

"Here, you are predicting that they are "boy/girl" before you ask the question"

What? I did no such thing.

"Yes, if you predict "boy/girl", and THEN find out that one is a boy, you have a 2/3rds chance of being right. But that's because you had a 25% chance of being flat wrong before."

No, I would have had a 50% chance of being wrong if I predicted BG, since it could have been either BB or GG. You're not making any sense here at all. The 2/3 answer is dependent on the answer having been received, since the 2/3 answer can only be reached once either BB or GG have been eliminated. You can't reach that prediction beforehand.

@gavkar

"Really? Do I need to explain this?"

Yes, you do. I concede that I thought you were talking about the original problem when I mentioned eliminating BB. I accept that that doesn't apply in your example. However, your statement ".5million *. 66 * 2 would have a boy and a girl" makes NO sense if you HAVEN'T eliminated either BB or GG, since the .66 figure comes from that elimination. In other words, you are using numbers specific to Jeff's problem in your own different problem, and complaining that they don't work. That's a complete logical fallacy.

@Tony

"What are the odds that the tenth flip will be tails?"

That's a totally different question. No-one is arguing that the laws of probability suddenly change and that the probability of a future even has been affected. This problem is NOT discussing future events, it's discussing past events and using information to progressively improve a categorisation.

When we say the answer to this puzzle is 2/3, we are NOT saying that the chances of the couple having a boy have magically changed to 67%.

What I think is interesting, and what I would love to know more about, is why does my brain snap to %50 even though I know that isn't the right answer. I can feel my brain going to %50 and I have to nudge it away. Why is that? The brain does some amazing pattern matching all the time, so my guess is we quickly think of gb and bg as being the same because of some mental efficiency that gives us an evolutionary advantage.

That in turn probably helped long ago ancestors from getting killed, but reading articles on card shuffling algorithms and unfinished games isn't helping my procreation chances right now...

All this argument over the boy-girl problem just shows that people are better at math than they are at understanding questions properly.

The answer is either 50-50 or 2/3, depending upon your interpretation of a question that is ambiguous in a mathematical sense. The question does not explicity state if it means "one or more girls" or "exactly one girl", so it is therefore impossible for you to know if you are understanding the question in the same way as whoever asked it.

It is very simple :

If it is "one girl or more" then the possibilities are bg, gb, and gg, and hence there is a 2/3 chance the other is a boy.

If it is "exactly one girl" then the possibilities are bg and gb, and hence there is a 50-50 chance the other is a boy.

Since this information isn't given, the statement of the problem is incomplete, and whether your answer of 50-50 or 2/3 is correct or not is simply a matter of opinion.

This is why we have systems-analysts and programmers, and someone who is good at one job isn't necessarily good at the other.

@Tony Alderman
Another question for the 2/3'rd crowd, and it has probably already been asked -- How is Jeff's question different than if I ask "Let's say I flipped a coin nine times, and got "heads" each time. What are the odds that the tenth flip will be tails?"

The difference is that the person with one child is not asking what the odds are for an unconceived child. Or the person with two children is not asking what the odds are for a 3rd unconceived child. Those would be analogous to the "unflipped coin" question you're asking.

Jeff's question is about two independent events that have already occurred, and whose outcomes are known to the questioner but not to you, where each event has two possible outcomes of equal probability. The question is this, if the questioner tells you only one outcome of their choosing, what is the probability that the other outcome is the opposite outcome?

You can work it out with coins or children, it doesn't matter, as long as the two events are independent and of equal probability.

@Russ

The parent that 'happens' to reveal that one child is a boy also eliminates half of the BG domain - the half that would 'happen' to reveal that one child is a girl.

See my second coin flip example from 1:07 PM (PST) to see why.

@Russ:

The difference is this:

If a parent with 2 girls just happens to reveal the sex of one of the kids, they will say "girl" 100% of the time.

Whereas if a parent with one boy and one girl just happens to reveal the sex of one of the kids, they will only say "girl" 50% of the time.

This exactly cancels out the 2:1 ratio of parents with one of each.

You NEED to know that the parent is a preferential girl revealer to get 2/3. You NEED to know that Monty never reveals the car to get "swap".

@Tim

"If it is "exactly one girl" then the possibilities are bg and gb, and hence there is a 50-50 chance the other is a boy."

If they have exactly one girl and it's only 50/50 that the other is a boy, what other options are there?

If they have exactly one girl, it's 100% certain that the other is a boy (hermaphrodites etc notwithstanding)

*sigh* Russ, here's another go for you:

There are two children, A and B. Independently, they have a 50/50 chance.

Agirl | Bgirl - 25%
Aboy | Bgirl - 25%
Agirl | Bboy - 25%
Aboy | Bboy - 25%

I tell you that one of the children is a girl. Is it A or B? Dunno - 50 50 chance. This means by breakdown is:

(where A is the one identified)
A GIRL | B girl - 25% * 50%
A BOY | B girl - 25% * 0%
A GIRL | B boy - 25% * 50%
A BOY | B boy - 25% * 0%

A girl | B GIRL - 25% * 50%
A boy | B GIRL - 25% * 50%
A girl | B BOY - 25% * 0%
A boy | B BOY - 25% * 0%

Collapse the probabilities left, and reset to a total of 1, and you get:
A GIRL | B girl - 25%
A GIRL | B boy - 25%
A girl | B GIRL - 25%
A boy | B GIRL - 25%

Or a 50/50 chance that the second child is a boy or a girl. Right now.

This someone had children. This someone has no children. The odds that this someone has a boy and a girl is 0.

@Russ

I showed my working hours ago.

odds of 2 boys: 25%
odds of 2 boys and being told one is a girl: 0%
odds of 2 boys and being told one is a boy: 25%

odds of two girls: 25%
odds of 2 girls and being told one is a girl: 25%
odds of 2 girls and being told one is a boy: 0%

odds of one boy and one girl: 50%
odds of one boy and one girl and being told one is a girl: 25%
odds of one boy and one girl and being told one is a boy: 25%

This gives 50% as the overall answer.

You have to add the stipulation that the parent is a preferential girl revealer to get to 2/3

@xooorx

"Whereas if a parent with one boy and one girl just happens to reveal the sex of one of the kids, they will only say "girl" 50% of the time."

Fine. So, WHAT DIFFERENCE DOES IT MAKE? As long as they don't lie and say they have a girl when they actually have 2 boys, it makes no difference. If they say girl, there's a 67% chance the sibling is a boy. If they say boy, there's a 67% chance the sibling is a girl. It doesn't matter at all which one they mention.

If the parent with two girls says girl - which she always does - there's a 67% chance the sibling is a boy. OK, in this case it's the 33% chance that comes up, but it doesn't alter the odds even slightly.

Look, let's say you're right. A non-biased parent 'just happens' to reveal she has a girl, but 50% of the time she would have said she had a boy. Show how that affects the outcome of 67% chance the other sibling is opposite. I haven't seen anyone post anything that gets even close to demonstrating this without introducing irrelevancies about which child got identified or which child was born first (yes, I know you haven't based anything on age).

Show me. Cos I'm simply not seeing it.

@Russ

The chance of an "opposite" sibling is 50%.

I'm in the 50% camp. I wrote some code to try and verify my thoughts as to why this is so, and goddamnit, the results disagree with me! This is javascript code. Anyone with firebug installed can run this code by pasting the code into the firebug console, and hitting the "run" button. This code seems to demonstrate that ~66% is the correct answer.

var y=0; //number of y chromosomes found
var x=0; //number of x chromosomes found
var r=0; //control counter
var a, b;
var iterations = 100000, i=0;

for(i=0; i<iterations; i++) {
//c is a control, that i'll use to verify that my coin flip code
//has a roughly 50% probability.
c = Math.floor(Math.random()*2);
if(c===1){
r++;
}

//We know the guy has two children, a and b;
//0 represents girl, 1 represents boy.
a = Math.floor(Math.random()*2);
b = Math.floor(Math.random()*2);

//we know that one of the children is a girl
// we don't care which one, or in which order
if(a === 0 || b === 0) {

//at this point, if we've got two girls, adding a and b will give us 0
// but if we have a boy and a girl, adding them will give us 1.

if((a+b)===0){
x+=1;
}
if((a+b)===1){
y+=1;
}
if((a+b)>1){
throw("two boys? impossible!");
}

}
}

var twogirls = (x/(x+y));
var onegirloneboy = (y/(x+y));
var control = r/iterations;

console.log("probability of two girls:"+twogirls+" probability of 1 girl, 1 boy:"+onegirloneboy+" control:"+control)

@xooorx

I saw that working, and I don't see that it holds. The likelihood of a parent telling you about the boy or the girl does not matter, since I've shown elsewhere that the maths holds true REGARDLESS of whether it's a boy or a girl.

@Robert

You still haven't explained why it matters which child is chosen. As I said before, you can't eliminate the 12.5% for b/G if the parent identifies the boy. You can only eliminate the same-sex pairing - the ENTIRE mixed-gender domain is in consideration at the end.

@Russ
Just curious, do you have children? It doesn't really matter anyway, but let me ask you, how likely are you to have a boy as your first child? And if you had a second child, how likely would you be to have another buy?

@mike and all others.

Okay - lots of comments and lots of misconceptions from lots of people.

This is my last comment.

I have tried to put all of this as simply as I can, as have others, and now we have a Mexican standoff.

All I can suggest is that people stop listing frequencies and instead create all options (all combinations).

Jeff Original Options was about “combinations” but combinations are weighted. That is when flipping two coins there the break-up is:
BB – 25%
GB – 25%
BG – 25%
GG – 25%
Total = 100%

That is, the chance of having a child of any gender is ½. The chances multiply, so the chance of having two boys is ½x ½ = ¼. The chance of having a girl and then a boy is ½x ½ = ¼. The chance of having a boy and a girl is also 25%. Therefore we can look up the list and see the chance of mixed children is 50%.

BUT by the same logic, the chance of having another Girl AFTER having a Girl is 50% (see in the list above – half the entries that start with “Girl” end with “Girl”). The chance of having a Boy AFTER having a girl is 50%. The chance of having a Girl before a Girl is 50% (half the entries that end with G start with G). The chance of having a Boy before a Girl is 50%. So knowing that one child is a girl, the chance of mixed siblings is 50% and the chance of all girls is 50%

Knowing that one of the siblings is a girl means we don’t use the original percentages as they are, we look at them within the context of the information given.

You can’t fiddle around with the original frequencies – they have no bearing on the current population.

THIS is the break-up of all possibilities of births – each equally likely:
Younger Girl Older Boy – 25%
Younger Boy Older Girl – 25%
Younger Girl Older Girl – 25%
Younger Girl Older Girl – 25%
Total = 100%

So what’s the chance the other child is a boy? 50%. What’s the chance the other child is a girl? 50%.

I’ve spent enough time trying to explain this. If you don’t agree with me, that’s fine. If you do. That’s fine.

Just remember – half the people you know are below average.

whereisthespacebar?

@David

No, I don't have children. If I did, there's an approximately 50% chance I'd have a boy. If I had a second child, there'd be a 50% chance of it being a boy. The chances of me ending up with two boys is therefore 0.5 * 0.5 = 25%.

You're right though, none of this really matters since it's not what the question is about.

@Russ
So you'd have 50% chance to have a boy (or a girl for that matter) each time.

Now say you already have two children. Or let's say that I already have them since it's actually true. I was subject to the same rule, right? I'm telling you I have a boy. What are the odds I also have a girl?

Doesn't it depend on whether the game is unfinished or not?

Consider following examples.
(A) A couple has one child, it is a girl. What are the chances that they will have a boy and a girl after their second child is born?
(B) A couple has two children and one of them is a girl. What are the chances that they have both a boy and a girl.

In A, the game is half finished. The chance that the second child will be a boy, thus resulting in a boy and a girl, is 50%.

In B, you're projecting the outcome at the beginning of the game. You have the 3 possible combinations of GG, BG, and BB. The chance of having a boy and a girl is 2/3.

So, is the game unfinished, or not? The title of the article (The Problem of the Unfinished Game) suggests it isn't, but the example given (you meet a couple who tells you they have two children) is of a finished game.

The logical term for GB= BG is the Symmetric Postulate.

**That was in response to axequalsb

& now my link functions.

I'm having difficult posting. I'm a 50%. Here's some javascript code that you can run in firebug's console. This is similar to the algorithm I've been trying to post, which shows 66% probability, but I've got an idea from the talk of "preferential girl revealer", and as such this code now reveals a 50% probability.

var y=0; //number of y chromosomes found
var x=0; //number of x chromosomes found
var r=0; //control counter
var a, b;
var g;
var iterations = 100000, i=0;

for(i=0; i<iterations; i++) {
//c is a control, that i'll use to verify that my coin flip code
//has a roughly 50% probability.
c = Math.floor(Math.random()*2);
if(c===1){
r++;
}

//We know the guy has two children, a and b;
//0 represents girl, 1 represents boy.
a = Math.floor(Math.random()*2);
b = Math.floor(Math.random()*2);

//guy reveals the gender of one of the children, randomly
g = Math.floor(Math.random()*2) ? a: b;

//the guy has revealed that the gender of one of the children is a girl
if(g === 0) {

//at this point, if we've got two girls, adding a and b will give us 0
// but if we have a boy and a girl, adding them will give us 1.

if((a+b)===0){
x+=1;
}
if((a+b)===1){
y+=1;
}
if((a+b)>1){
throw("two boys? impossible!");
}

}
}

var twogirls = (x/(x+y));
var onegirloneboy = (y/(x+y));
var control = r/iterations;

console.log("probability of two girls:"+twogirls+" probability of 1 girl, 1 boy:"+onegirloneboy+" control:"+control)

@Russ

One of us is going to wake up tomorrow and, ten seconds later, think "oh... hang on..."

I give it 2/3 probability that its you not me :)

Too drunk to continue now. Thanks (<--- really).

G'night

@Robert
(where A is the one identified)
A GIRL | B girl - 25% * 50%
A BOY | B girl - 25% * 0%
A GIRL | B boy - 25% * 50%
A BOY | B boy - 25% * 0%

A girl | B GIRL - 25% * 50%
A boy | B GIRL - 25% * 50%
A girl | B BOY - 25% * 0%
A boy | B BOY - 25% * 0%

You got your percentages wrong. It should be this:

(where A is the one identified)
A GIRL | B girl - 25% * 50%
A BOY | B girl - 25% * 0%
A GIRL | B boy - 25% * 100% -- different
A BOY | B boy - 25% * 0%

A girl | B GIRL - 25% * 50%
A boy | B GIRL - 25% * 100% -- different
A girl | B BOY - 25% * 0%
A boy | B BOY - 25% * 0%

Remove the zero probabilities and you get the remaining probabilities:
A GIRL | B girl - 12.5%
A GIRL | B boy - 25%
A girl | B GIRL - 12.5%
A boy | B GIRL - 25%

Renormalize to 100 and the outcome is 2/3 boys.

Why are the boy/girl combinations given 100%? Because in those cases there is 0% chance that the parent can identify the boy as being a girl. In other words, the parent is FORCED to reveal the one girl they have, with 100% probability.

@bglick
In B, BB is not a possible outcome since you already know that there is a girl. Taking the BB combination into consideration is therefore a mistake that leads to the wrong result. The probability in both A and B is 50%.

The 2/3 answer is correct if nothing is known about the outcome. In this case, you do not know the gender of either of the children.

@bglick
In B, BB is not a possible outcome since you already know that there is a girl. Taking the BB combination into consideration is therefore a mistake that leads to the wrong result. The probability in both A and B is 50%.

The 2/3 answer is correct if nothing is known about the outcome. In this case, you do not know the gender of either of the children.

Philip, if you really are a mathematician you should take a refresher course. This is NOT about changing birth-rate probabilities. It's about categorisation.

There are 3 possible categories here - these parents have either two boys, two girls, or one of each. There must always be a 100% chance that the parents are in one of the available categories. As you point out, the chances of them having one of each is 50%, two boys is 25%, and two girls is 25%. 50+25+25=100.

When we find out that they have a girl, the boy/boy category is eliminated and therefore we have to increase the chances that they are in one of the other categories, since they MUST be in one of the remaining categories. In order to maintain the correct ratio, we end up with mixed=67%, girl/girl=33%. Since membership of the mixed category would require a boy (we already know they have a girl), the probability of the sibling being male is 67%.

This is nothing to do with the gender probability at birth. It's a case of narrowing down the possible permutations of events that have already occurred, given that some possibilities have been eliminated. It's simply a variation of "20 Questions".

I just realized that the someone you meet is crazy. "Someone" is a singular pronoun, whereas "they" is a plural pronoun.

@Russ:

Yes, 2/3 of all parents with one girl have a boy also, and 1/3 another girl. But only 1/2 of the ones with a boy/girl will decide to say "I have at least one girl", rather than "I have at least one boy". And of course those with two girls always say "girl", so you get 100% of that. So that adds up to 50%.

What else is there?

@Russ

No I don't need a refresher I obviously need a simpler way of explaining this so that others can get the correct answer too. I get what you are trying to explain but you are wrong. I don't get why you can't understand what I am trying to say.

Yes it is about categories - but the categories are weighted. That is GG is twice as likely as BG because it includes both the chance that the identified girl is younger and older.

@russ and @ed,

Yeah, you're right about my tenth coin flip question -- It's a different question. I was off on a bad tangent. Thanks

@russ and @ed,

Yeah, you're right about my tenth coin flip question -- It's a different question. I was off on a bad tangent. Thanks

the difference between my 50% code and 66% code is that for 66% you need to assume that there's a 100% chance that the guy would have revealed that he has a girl, if he has one.

In the 50% code, he reveals the gender of one of the children, randomly. I then count the probabilities based on the cases where he happened to reveal that one of his children is a girl. This results in a 50% probability the other child is a boy.

@David

"I was subject to the same rule, right? I'm telling you I have a boy. What are the odds I also have a girl?"

67%.

That doesn't mean that you MUST have a girl - there's a 33% chance you're in the BB category, and 33% chances come up one third of the time, which is pretty often.

It also doesn't mean that there was a 67% chance that your second child would be a girl once the first was revealed to be a boy. This isn't about gender probabilities at birth, it's about categorisation.

I tried to represent this visually earlier on. If you take 1000 two-child people, including you, then statistically 250 will have 2 girls, 250 will have 2 boys, and 500 will have one of each. The gender probability of each child is irrelevant - these approximate numbers can be produced over and over again with any decent random number generator.

You've told me you have a boy, so I can eliminate the 250 GG's since you can't be one of those. You must be one of the 250 BBs, or the 500 BGs.

Clearly, with no further information to hand, you're more likely to be one of the 500 than one of the 250, simply because it's a bigger proportion of the remaining people.

This doesn't affect the probabilities in play when you had your kids, and it doesn't affect the probability for the gender of any future child. It just means that, based on the information you've given me, there's more chance you're in the group of 500 than in the group of 250.

A probability of 2/3, in fact.

This is a probability of you being in a certain group after filtering has been applied, not a probability of bucking the trends of nature.

@David

"I was subject to the same rule, right? I'm telling you I have a boy. What are the odds I also have a girl?"

67%.

That doesn't mean that you MUST have a girl - there's a 33% chance you're in the BB category, and 33% chances come up one third of the time, which is pretty often.

It also doesn't mean that there was a 67% chance that your second child would be a girl once the first was revealed to be a boy. This isn't about gender probabilities at birth, it's about categorisation.

I tried to represent this visually earlier on. If you take 1000 two-child people, including you, then statistically 250 will have 2 girls, 250 will have 2 boys, and 500 will have one of each. The gender probability of each child is irrelevant - these approximate numbers can be produced over and over again with any decent random number generator.

You've told me you have a boy, so I can eliminate the 250 GG's since you can't be one of those. You must be one of the 250 BBs, or the 500 BGs.

Clearly, with no further information to hand, you're more likely to be one of the 500 than one of the 250, simply because it's a bigger proportion of the remaining people.

This doesn't affect the probabilities in play when you had your kids, and it doesn't affect the probability for the gender of any future child. It just means that, based on the information you've given me, there's more chance you're in the group of 500 than in the group of 250.

A probability of 2/3, in fact.

This is a probability of you being in a certain group after filtering has been applied, not a probability of bucking the trends of nature.

@Philip

Why does it matter if the identified girl is younger or older? The order in which the children were born is not a factor in this problem.

It's kind of ironic that Jeff's point about the unfinished games 'not being equally likely' can equally well be applied *in support* of the 50% argument.

When I'm told 'there's at least one girl', BB is excluded, just as in the 66% argument. The remaining options are BG, GB, and GG. However *if you will allow that I might just as likely have been told there's at least one boy*, these three options are no longer equally likely. BG and GB are then each half as probable as GG, which leads to the 50% result.

Maybe this will help to reconcile the two sides of the argument (he said optimistically).

This is an apples and oranges problem.

The boy/girl question is order-independent problem; there's nothing about birth order here, so it actually is 50% because each outcome is independent. Knowing something about one child doesn't tell you anything about the other. This is a combinations problem.

The coin flip question is different because order matters: The first person to guess right three times wins: It could take 3-5 tosses to decide a new gam. Each player has different conditions to win in the unfinished game above, and the game will be decided in 1-2 tosses. This is a permutations problem.

The difference between combinations (order independent) and permutations (ordered) is why the B/G question is actually 50% and the coin flip game is a much more complicated affair, as examined above.

so the question you've gotta ask, and whose answer isn't revealed, is for each of the combinations GG, BB, GB, BG... what is the probability that the was going to reveal that he has a girl.

if the answer is this:

GG: 100%
GB: 100%
BG: 100%
BB: 0%

then the correct answer to the original question is 66%. if however, the answer is

GG: 100%
GB: 50%
BG: 50%
BB: 0%

then the correct answer to the original question is 50%.

Q.E.D. None of the answers are incorrect when the original question is this stupidly ambiguous.

@Russ

Because BG indicates the boy before the girl. GB indicates the girl before the boy. If you want to take "younger/older" out of the picture, you ahve to do it with these categories too. In that case you are left with:

Category1: MixedSiblings
Category2: BothGirls

On the other hand, if you insist on having BG and GB, where the item to the left is younger and the one to the right is older, then you MUST have "Younger/Older" options for girl pairs too.

@Russ
Thanks for answering, and I must admit I was hoping you'd answer that.

I was asking you a question about me, not about other people. What makes you think that I was asking you what category I belong to? What makes you think I was asking you a trick question that involves comparing my family with others? What in my question as I have formulated it is unclear, especially with the extra steps I took before to make it even clearer where I was going?

I have asked you a very simple question about me (and my wife, oviously), and your answer essentially says that somehow my reproductive system is tied to other people's reproductive system. This is so wrong... You transformed something simple into something complicated.

The thing is, your calculations are right. You're just not answering my question. Which, incidentally, is exactly the same as the one Jeff asked.

Something we can all understand...

Program one, faulty:
Take two variables, put in the first one "girl", choose the next one randomly. You come out with 50/50. Why? Because you knew the older was a girl, that's not in the question.

Program two, Jeff is wrong:
Take two variables. At random choose one of the variables. Put a girl in it. Now, in the remaining variable, choose at random either a boy or a girl. Results I get:
GG total:
5027
GB total:
2466
BG total:
2507

"What are the odds that person has a boy and a girl?" GB and BG are both included, so it's still freakin' 50/50 (minor random variance in my results).

Program 3, faulty, but Jeff is right:
Take the first variable. Choose at random either boy or girl and put it in. Take the second variable, do the same. Now, if you happen to have a BB, which simply isn't possible, choose one of the variables at random and replace it with a girl. This gives us... 66%.

Why is Jeff Atwood wrong in the end? Because the program that gives his results does not reflect the story he told. Program two accurately reflects the story we were told.

Ladies and gentlemen, your intuition is correct, killing children is bad. You can go home now.

My question for you Jeff is... why the hell did you go off talking about poker and write a program for that? Why not write a program for the question at hand?

The problem I keep seeing here is people seem to be introducing additional components to the initial problem, to show it as being correct. This is showing up mainly with the combinations of GB and BG. In the initial problem, the order of the children is inconsequential, and by introducing it as a factor you are changing the original problem to meet the answer, rather than changing the answer to solve the problem.

If you're trying to abstract it into a more understandable concept, try numbers:

"You have two prime numbers, one of which is 3. The sum of these numbers is either 4, 5 or 6. What are the odds that the other number is 2?"

Well, the remaining outcomes are only 5 and 6, as 4 cannot be obtained without a second 2. The answer then is 50%, as there are only 2 possible choices to meet the criteria put forth by the problem. Would you try to say that the answer is really 2/3 (66.66...%)?

In this, a boy is equivalent to the number 2, and a girl would be 3. The outcome here is either 4, 5, or 6, hence the constraint in my version, as with gender that constraint is implicit.

The problem I keep seeing here is people seem to be introducing additional components to the initial problem, to show it as being correct. This is showing up mainly with the combinations of GB and BG. In the initial problem, the order of the children is inconsequential, and by introducing it as a factor you are changing the original problem to meet the answer, rather than changing the answer to solve the problem.

If you're trying to abstract it into a more understandable concept, try numbers:

"You have two prime numbers, one of which is 3. The sum of these numbers is either 4, 5 or 6. What are the odds that the other number is 2?"

Well, the remaining outcomes are only 5 and 6, as 4 cannot be obtained without a second 2. The answer then is 50%, as there are only 2 possible choices to meet the criteria put forth by the problem. Would you try to say that the answer is really 2/3 (66.66...%)?

In this, a boy is equivalent to the number 2, and a girl would be 3. The outcome here is either 4, 5, or 6, hence the constraint in my version, as with gender that constraint is implicit.

@Philip

I've been trying to avoid the whole BG/GB thing as it muddies the water. I agree with you that it's clearer to use 'mixed' as a category.

However, if you do that then all categories are not equal. BothBoys and BothGirls are 25% each; MixedSiblings is 50% of the population.

This is easily shown. Your chance of a boy is 50%, so your chances of BothBoys is 0.5*0.5 = 25%.

Same goes for BothGirls.

BothBoys + BothGirls = 50%, so MixedSiblings MUST be 50% otherwise we run out of categories before reaching 100%, which makes no sense.

So, once we eliminate BothBoys, we are left with MixedSiblings and BothGirls. However, MixedSiblings is twice as likely as BothGirls, since the original weights were 50% and 25% respectively.

Scaled up, the new proportions are roughly 67% and 33%. Therefore, 67% of the remaining population must have a boy, otherwise the wouldn't be in MixedSiblings.

Therefore, if I pluck a random person from this group there is a 67% chance they have a boy.

None of this depends on which child was born first, or which child was identified. It's simply that there are three initial groups, and when it's revealed that one of the children is a girl, we are left with two groups, one bigger than the other. The solution to the question is simply the probability of being in the group that represents MixedSiblings.

Jeff simplified the question so much that the answer changed. Ooops.

Jeff simplified the question so much that the answer changed. Oops!

The wonderful thing about math is that you can compute the same thing two different ways and two get different answers!

wait a minute...

The wonderful thing about math is that you can compute the same thing two different ways and two get different answers!

wait a minute...

The wonderful thing about math is that you can compute the same thing two different ways and two get different answers!

wait a minute...

The wonderful thing about math is that you can compute the same thing two different ways and two get different answers!

wait a minute...

The wonderful thing about math is that you can compute the same thing two different ways and two get different answers!

wait a minute...

@xooorx

>If the emotionless robot is programmed to randomly reveal the sex of one child (and it said girl on this occasion) then its 1/2.

***Thank you*** - I've finally understood where the 50%ers and the 2/3ers differ. We're counting different things:

2/3ers are counting those families with both sexes where the revealed child is always a girl. (2 in every 3 cases)

50%ers are counting those families with both sexes and the revealed child is either sex (2 in every 4 cases).

The problem doesn't state that this is to be repeated in large amounts, (even though that's how most probability 101 is taught), so I assumed it was always girls that would be revealed.

I can sleep now.

@David

"What makes you think I was asking you a trick question that involves comparing my family with others?"

Because this is a PROBABILITY question, and is therefore based on populations. I know nothing about you other than what you have told me, i.e. you have two children and one is a boy, and asked me what the probability is that your other child is a girl.

Statistically, you MUST fit into one of the three groups I identified, and the only way I can answer your question is by trying to work out which group you are in. It's nothing to do with your reproductive system being affected by others, it's a question of you being in a particular category and me trying to identify that category to answer your question.

With no information about the gender of your children, I'd conclude that there's a 75% chance that you have a girl, since statistically only 25% of people with two children have two boys. Again, NOTHING to do with your personal biological system, just a question of probability distribution over a statistically-significant population.

Once you TELL me you have a boy, I can now pin down your category more precisely, since now there are only two possibles instead of three. It's still nothing to do with mystical influences on your reproductive system.

"I was asking you a question about me, not about other people...You're just not answering my question. Which, incidentally, is exactly the same as the one Jeff asked."

You're contradicting yourself. Jeff's question was about a non-existent woman, a theoretical person plucked from an imaginary (but statistically-valid) population. Your question was the very antithesis of statistics since it focussed on the personal and not the general. Statistics, by definition, is about populations not individuals. 'You' are only meaningful in this problem as a single data point in a population, you are not statistically significant on your own.

Jeff, sort your comment system out, it's continually whinging about temp files.

@xooorx

>If the emotionless robot is programmed to randomly reveal the sex of one child (and it said girl on this occasion) then its 1/2.

***Thank you*** - I've finally understood where the 50%ers and the 2/3ers differ. We're counting different things:

2/3ers are counting those families with both sexes where the revealed child is always a girl. (2 in every 3 cases)

50%ers are counting those families with both sexes and the revealed child is either sex (2 in every 4 cases).

The problem doesn't state that this is to be repeated in large amounts, (even though that's how most probability 101 is taught), so I assumed it was always girls that would be revealed.

I can sleep now.

@xooorx

>If the emotionless robot is programmed to randomly reveal the sex of one child (and it said girl on this occasion) then its 1/2.

***Thank you*** - I've finally understood where the 50%ers and the 2/3ers differ. We're counting different things:

2/3ers are counting those families with both sexes where the revealed child is always a girl. (2 in every 3 cases)

50%ers are counting those families with both sexes and the revealed child is either sex (2 in every 4 cases).

The problem doesn't state that this is to be repeated in large amounts, (even though that's how most probability 101 is taught), so I assumed it was always girls that would be revealed.

I can sleep now.

Let me explain it since I had it wrong.

The odds of the first child is 50/50 and the same for the second one.

BB GB BG GG

Now, we must also select either the first or second child BEFORE we know what the sex is of the one selected. I will use capitals for the child that is selected. Since we can end up selecting either the oldest or youngest, we end up with 8 possibilties. (4*2).

Bb bB (Family #1)
Gb gB (Family #2)
Bg bG (Family #3)
Gg gG (Family #4)

Then we are told that the one selected is a girl. So we must remove all possibilities where the selected child is a boy. So no capital B's allowed. We are left with this:

Gb (Family #2)
bG (Family #3)
Gg gG (Family #4)

If the older sibling was chosen, there's a 50/50 chance that the other one is a boy. Same thing if the younger sibling was picked. Seemingly in 50/50 possibility.

But the problem is that Gg and gG is the SAME family. Picking a person does NOT create a new family. This is what's called the intersection when discussing probabibilities. So every time you get two girls, you can only pick either the youngest OR the oldest (both of which point to the SAME family). That means that Gg and gG are half as likely as the other families since the other families always must pick the ONE girl ALL of the time. The families with Gb and bG are obviously different since the oldest sibling has different sex.

IOW, what are the odds of picking a certain girl in each family?

Gb (Family #2) (1/1) 100%
bG (Family #3) (1/1) 100%
Gg gG (Family #4) (1/2) 50%

Family #2 and #3 are at 100% because there is only 1 girl to choose from. You'll get 100% chance of picking that one girl if you must pick a girl.

Family #4 can pick either the oldest or youngest. But it's still the same family.

The total is 100%+100%+50%+50%.
The odds of the other child being a boy are 100%+100%.

That makes 200%/300% or 2/3.

I certainly haven't read through all the comments in the two threads, but I could see from what I did read is the lack of understanding of mathematical statistics. Read an undergraduate text, Feller would be a good place to start. Bayesians assert that events are not independent, while real math stats do, when they are.

I read, years ago, that orgasmic women have a higher boy birthing rate. Other than that (if true), the sex of a child is not dependent on the sex of any others.

Bayesians are still not taken seriously. Graham's spamming filter is not really statistical, just as Bayes is not really a statistician.

Whoops, 8 versions of the same damn comment

What are the odds that a randomly chosen couple that have 2 children of which at least one of the children is a girl has a boy and a girl?

@Russ
No, it IS not a probability question. For some reason, YOU want it to be a probability question, probably because that's the question you want to answer.

My question was absolutely not a probability question beyond the simple fact that I asked you what was the odds that MY second child would be a boy. I have not mentioned anything but me and my children. The answer to my question is 50%, nothing else.

There is a video of Russ and David on youtube...

http://www.youtube.com/watch?v=MfgX0fyNeLc

I hope it is obvious who is who :-)

@David

You're right, we can rule out BB.

But I believe that it still matters where in the game you project the final outcome. Let's take it a step further.

(A) A couple has one child, it is a girl. What are the chances that they will have a boy and a girl after their second child is born?
(B) A couple has two children and one of them is a girl. What are the chances that they have both a boy and a girl.

In (A) all we need is the probability of the second child being either a girl or a boy.

In (B) we need to do a bit of permutation. The first birth can be either a boy or a girl. The odds are 50% that it will be either. If the first is a boy, we know that the second will be a girl, because we know final result. But if the first is a girl, the second can still be either a boy or a girl, so we have another 50% hurdle to overcome before we reach the desired result. Another way of stating it would be like this:

(B1) First child: Boy, Second child: Girl
(B2) First child: Boy, Second child: Boy
(B3) First child: Girl, Second child: Boy
(B4) First child: Girl, Second child: Girl

We can rule out B2, which leaves three possibilities. We have a 1/3 chance of ending up with two girls, and a 2/3 chance of a boy and a girl.

I believe that my original question is still relevant. Where in the game are we projecting the final outcome? At the beginning, or halfway through?

@Russ,

Still, you have no reason to believe all Mixed Sibling parents are always going to say "I have at least one girl", since they could say "boy" just as likely, and we have no information that they are biased in that way. So the B/G population is reduced by 50%, making it equal to the G/G population, making the answer 50%.

The question Jeff posed was NOT about tabulating the total population of people with two kids and at least one girl; it WAS about a specific person who has kids, and in light of that you logically must question why EVERY Mixed Sibling parent would say they had at least one girl, instead of at least one boy. They wouldn't, thus, 50%.

The reason for all the confusion is that the question is ambiguous in how the selection to make the reveal was made. If the woman COULD have said she had a boy, ie, she was just picking a random child's sex to reveal, then things are different and the odds are 50%. If the rule was something like, if you have a daughter reveal it, then it's 2/3.

I flip two coins. I look at them both, pick one randomly to reveal, and tell you it's a H. The odds on tails are now 50%. I may have revealed a T for all you know, we are only playing once.

I flip two coins. I look at them both and only reveal H. If I had flipped TT I would have thrown away this flip and redone it. Now the odds on the other one being T are 2/3.

Giggle ... Jeff is going to love the bullshit we have done to the place ... errr, why do I use his first name? Not like I know him or anything! From now on I shall refer to him as Lord Overflow.

Pardon my garbage ... my old reliable computer just started whining like a banshee (can't afford to retire/fix due to CPP/UI back on in Jan.), so I am using my work-supplied one without due shackles ... my hot water heater blew up on my birthday and I wrecked my shop vac on the cement dust, and my son tells me about his fireworks stash in his bedroom. FAIL.

Giggle ... Jeff is going to love the bullshit we have done to the place ... errr, why do I use his first name? Not like I know him or anything! From now on I shall refer to him as Lord Overflow.

Pardon my garbage ... my old reliable computer just started whining like a banshee (can't afford to retire/fix due to CPP/UI back on in Jan.), so I am using my work-supplied one without due shackles ... my hot water heater blew up on my birthday and I wrecked my shop vac on the cement dust, and my son tells me about his fireworks stash in his bedroom. FAIL.

Giggle ... Jeff is going to love the bullshit we have done to the place ... errr, why do I use his first name? Not like I know him or anything! From now on I shall refer to him as Lord Overflow.

Pardon my garbage ... my old reliable computer just started whining like a banshee (can't afford to retire/fix due to CPP/UI back on in Jan.), so I am using my work-supplied one without due shackles ... my hot water heater blew up on my birthday and I wrecked my shop vac on the cement dust, and my son tells me about his fireworks stash in his bedroom. FAIL.

Now I AM AN IDIOT ... some error message about renaming the template! Honest!

@jon

The two scenarios you described are the same.

1) there are two coins and at least one of them is a H
2) there are two coins and both of them aren't T

@bglick
"In (B) we need to do a bit of permutation. The first birth can be either a boy or a girl." You're introducing order, in which case, you are right, the probability is 2/3.

But again, nothing in the original question indicates that order must be taken into consideration.

@David

"My question was absolutely not a probability question beyond the simple fact that I asked you what was the odds that MY second child would be a boy."

No, your question was "what are the odds I also have a girl?", and that most certainly IS a probability question. The two questions are not the same. One is related to gender probabilities at birth. The other is not.

I'm really sorry that you don't get this. The chances of any given child being born a particular gender under normal circumstances are 50%, no-one's debating that. That is NOT the question, though. The children have already been born. Birth probabilities are ONLY relevant in sizing the categories.

You are part of the population of people with two children and at least one boy. Of that population 67% have a mixed-gender pair. Therefore, the odds of your other child being a girl is 67%. You asked for these odds, and the only way to calculate them is as a probability calculation against a statistically-valid population. The probabilities in play at the time your children were conceived or born is utterly irrelevant to the question you asked.

@Justin

The difference in the 50% scenario is that if repeated, it might not always be a G that is revealed. I'm not throwing away BBs.

It's not clear in the original question how the woman chose to reveal she has a girl. If she was randomly revealing the sex of one child, for example, then GG is twice as likely as BG and GB, so it's back to 50%. If she could NOT have revealed she had a boy, it is 2/3. So if the repeatably example it's always a daughter, it's 2/3.

@Jason

"Still, you have no reason to believe all Mixed Sibling parents are always going to say "I have at least one girl", since they could say "boy" just as likely, and we have no information that they are biased in that way."

It doesn't matter! If they said boy, there would be a 67% chance the other sibling was a girl. If they said girl, there's a 67% chance the other is a boy. In this case, they said girl, and we were asked the probability of a boy. The answer is 67%.

The actual gender is irrelevant, it's simply a question of opposites. That's why the MixedGender category has the most influence - it's the largest, and it's also the one we're effectively calculating membership probability for.

Ed, your breakdown of my odds was incorrect. The breakdown is:

original probability (25%) * chance that it was identified as a girl. (50% or 0%). I listed both combinations to show that, which you ignored.

With your numbers:
----
A GIRL | B girl - 25% * 50%
A BOY | B girl - 25% * 0%
A GIRL | B boy - 25% * 100% -- different
A BOY | B boy - 25% * 0%

A girl | B GIRL - 25% * 50%
A boy | B GIRL - 25% * 100% -- different
A girl | B BOY - 25% * 0%
A boy | B BOY - 25% * 0%
----

You can't pick 100% for anything! The 50% chance is "is the child identified A, or is it B". The top half shows the combinations if A is picked, the bottom half shows the combinations if B is picked. Hence, the relevant lines are "25% * 50%" or "25% * 0%" (in that there is a 50% chance that A's gender is revealed or that B's gender is revealed). The fact that you pick 100% for anything is revealing you don't understand what the combinations actually are.

So, for the last time, here's the break down, with absolutely nothing skipped:
Chance that it was A's gender revealed: 50%
A GIRL | B girl - 25% * 50%
A BOY | B girl - 25% * 50%
A GIRL | B boy - 25% * 50%
A BOY | B boy - 25% * 50%

Chance that it was B's gender revealed: 50%
A girl | B GIRL - 25% * 50%
A boy | B GIRL - 25% * 50%
A girl | B BOY - 25% * 50%
A boy | B BOY - 25% * 50%

All probabilities are still equal.

Now we rule that the revealed gender (wether it was A or B) is a girl, and we get:
A GIRL | B girl - 25%
A GIRL | B boy - 25%
A girl | B GIRL - 25%
A boy | B GIRL - 25%

Guess what? They are still all equal possibilities! Which means that there is a fifty/fifty chance that the other child is a boy, assuming we wait until this late stage to make the prediction, which we are.

To get to the 2/3rds chance, you do it like this:

"I predict this woman, who I already know has two children, has a girl and a boy".

There are four equal options at this stage (at 25% each), and when one of the genders is revealed, one of the options has to be removed. This means that, if you are still in the running, you have a 2/3 chance of being right. But you also have a 1/3 chance of being wrong, and a 25% chance of getting blown away in the first round. So that makes the total probability break down:
right: 2/3 * 3/4 = 9/12
wrong: 1/3 * 3/4 = 3/12
wrong: 1/4 = 3/12

By the way: 2/3 * 3/4 = 1/2.

If you eliminate the possibility of getting the prediction wrong at the start, then you are looking at a different sample set.

What makes this different to the Monty Hall problem is that, in the MH problem, you can't get blown away in the first round. If you pick the door without the prize, Monty shows the contents of the OTHER door without the prize. Your initial pick had a chance of 1/3 - the fact that Monty eliminates a door means your initial pick still only has a chance of 1/3, as there was no new information about that door. The OTHER door, however, is the only survivor of the [set of doors you did not pick], and thus has the cumulative probability of that set: 2/3rds. But - if you come into the game late, after the doors are eliminated - you only have a 50/50 chance of picking the winner.

Anyway, I've wasted enough time on this, and I'm getting over my SIWOTI syndrome now. If this doesn't explain it for you, spend some time at Wikipedia reading up the difference between COMBINATIONS and PERMUTATIONS - right now, you are collapsing a combinatorial problem into a permutation problem, and wondering why you get the answer wrong.

Will the last person who still thinks the answer is 50% please turn out the lights.

"It doesn't matter! If they said boy, there would be a 67% chance the other sibling was a girl. If they said girl, there's a 67% chance the other is a boy. In this case, they said girl, and we were asked the probability of a boy. The answer is 67%."

It does matter, and if you scroll up and get the two versions of my javascript code, and try them out. If you don't think my javascript code proves that it DOES matter, then please point out where I've misstepped in the code.

For all the 2/3ers trying to understand the 50%ers, think about this scenario:

You catch the eye of a woman across the room. You know she has two kids for some arbitrary reason (let's say it's party where you can only enter if you have two children). She has one child with her, who happens to be a girl. Other people at the party have boys with them, so she COULD have had a boy with her for all you know. But in this problem, you see her right now with a little girl tagging along. What are the odds the other child is a boy? This is 50%

The 2/3 scenario would be the same party, but there's a sign on the front which says you can only bring one child, and it has to be a girl. Now you see someone coming out of that building, the odds of the other child being boy are 66%.

@Jon

"I flip two coins. I look at them both, pick one randomly to reveal, and tell you it's a H. The odds on tails are now 50%"

No, this is NOT right. This is exactly the issue Jeff is talking about.

After flipping the two coins, you have the following possibilities:

HH
HT
TH
TT

If you tell me one is H, TT is eliminated, and we have three remaining possibilities:

HH
HT
TH

T is present in 2 of those 3 outcomes, so the probability is 2/3, NOT 50%.

"I may have revealed a T for all you know, we are only playing once."

OK, so you revealed a T. That leaves:

HT
TH
TT

H is present in 2 of those 3 outcomes, so the probability is 2/3.

WHATEVER result you choose to reveal, the chance of the opposite result also coming up is 2/3. It doesn't matter which one you reveal, or whether you reveal a different one each time. Whenever you say H, there's a 2/3 chance of a T. Whenver you say T, there's a 2/3 chance of an H. If you say T 5000 times in a row, the chances of there also being an H will be 2/3. Every. Single. Time.

@Russ

You'd agree that if I only revealed at the first coin, it'd be 50% right? That's equivalent to a random reveal (a reveal where you could either reveal H OR T). But if the rule is I only reveal Ts, (not revealing anything on HH), of course it is 66%.

New Years dinner interrupted my entry .

if the problem is described in a matrix like this:

GG BB
GB BG

where each quadrant of the matrix represents a real statistical probability(25%). And the BB quadrant is eliminated. The remaining subset shows a distribution of 2/3 for BG/GB vs 1/3 for GG

GG
GB BG

the real issue is how the original question is interpreted.

"What are the odds that person has a boy and a girl?"

And the answer to that is still 50%.

statistically it is likely (66%) that it is a GB/BG pair. But in fact, the odds the second child is a boy or a girl are not causally affected by that fact. The chance the second child is a boy is still 50%. (for all children born, half (a little more than half actually) are male and half are female).

knowing the gender of a first child does improve YOUR ability to guess (statistically), but it does not change the actual chances of male or female gender for the second child. That is, the 66% results from a sampling statistic, and not a statistic which shows a predictive correlation.

Otherwise, parents with one child could expect a greater probability of their second child being the opposite gender. But no one I know expects this. Because the gender of a first child does not predict the likelihood the second child is the opposite gender. Exactly for the reasons I posted earlier GB and BG constitute one set, not two (as shown in the matrix).

The children are either "a girl and a boy" or a "girl and a girl". They are not "a girl and a boy", "a boy and girl", and "a girl and a girl".

how you interpret the notion of "odds" is how you arrive at your conclusion. You can think of "odds" as being your chance at guessing, or you can think of "odds" as the likelihood the second child is a boy. Each will lead you to different conclusions. Both conclusions appear valid, they just mean different things.

@Jon

the woman randomly revealing the sex of one child does not in any way change the probability of the sex of the other child. those are still each 50%.

as has been said a number of times, the 4 and only 4 combinations of 2 kids are

BG
GB
GG
BB

if the woman randomly reveals a boy, the choices left are

BG
GB
BB

therefor the chance of the other child being a girl is 2/3

if the woman randomly reveals a girl, the choices left are

BG
GB
GG

therefor the chance of the other child being a boy is 2/3.

@Russ,

It doesn't matter which gender they tell you. Which ever gender they HAPPEN to tell you, you're only encountering that information from a pool of people, 50% of which have GG, and 50% which have BG, because half the BG parents gave the opposite gender initially.

@weevil

I'm not addressing the 67% scenario, which a quick search through this thread and the other you'll see that I entirely agree with.

Let's look at the three possibilities:

BG
GB
GG

Two things to note: 67% of the possibilities contain one boy and one girl (67% is the correct answer for "You meet a random couple with two children and ask if they have any girls; they say yes")

HOWEVER, 50% of the _girls_ are in the "two girls" scenario, and 50% of the girls are in the girl-boy scenario; on meeting a random girl who tells me she has only one sibling, there is a 50% chance that she'll have a sister and a 50% chance that she'll have a brother.

To clarify, let's give everyone a name and assign random pairs, not necessarily siblings:

Book and Jayne (Male)
River and Simon
Mal and Zoe
Inara and Kaylee

River and Zoe are paired with men, and Inara and Kaylee are paired with women. If I meet one of those four at random, there's a 50% chance that they're paired with a man.

If I were, however, to put the pairs in separate rooms, and tell you to peek in and see if there's a girl in a certain room, selected at random, if you said Yes (of which there is an unrelated 75% chance) then there's a 67% chance that it's a girl and a boy, and a 33% chance that it's two girls.

Two different questions, two different answers.

Which one did Jeff ask? -- the 67% one. We know that because, as good software designers, we went back and clarified a vague requirement -- and that's probably the best lesson to be learned from this whole exercise.

@Russ

Ahhh... I think I see where you are coming unstuck.

You can't look at the original population and assume that those proabilities persist.

That is - you can't just say "okay - we have our original options, now remove one... there we go".

No - new information has been provided and that changes the ORIGINAL probabilities. At least one of the siblings is a girl. That means the chance of GG and BB occuring have now changed.

In the original data they were the same (25% each) for BB and GG, but they are DIFFERENT after the new information!!! BB reduced to 0%, because we know there must be at least one girl, and GG doubles because we KNOW there is at least one girl, which could be either of the "G's" in GG, which many of us have taken to be first or the second born.

BG and GB don't increase EVEN THOUGH BB is eliminated because GG doubled. So the chance of BG is still 25%, and the chance of either mix is 50%.

Russ - if you still don't agree then I strongly suggest we agree to disagree.

Trevel has it right.

There are three states, yes.

BG
GB
GG

But those states may be equally likely, or may not, it depends how you read the problem. Given a certain reading of the problem, we're twice as likely to be GG as either GB or BG alone.

I understand well the 2/3 case. If I go around asking people if they have at least one girl, for everyone who says yes, the odds of the other child being a boy are 2/3. But it's not clear enough whether this problem is that case -- we do need to consider, in a Bayesian way, the probability of the GG case.

@Trevel
"Which one did Jeff ask? -- the 67% one. We know that because, as good software designers, we went back and clarified a vague requirement -- and that's probably the best lesson to be learned from this whole exercise."

precisely where did jeff clarify this point?

Possible states:

BB
BG
GB
GG

Each state is 0.25

We observe a girl. The odds of this happening were 50% (we could have observed a boy).

That's the 50% case.

@Russ
There is no reason why you'd need to first calculate the probability of me being in the "two children including one boy" category since I have told you that this was my situation. Therefore the probability for me to be in that category is 100%, not 66%. And therefore the probability of my second child being of the other gender is 50%.

I too cannot understand why, be it for one second, you do not want/cannot see the problem in a simple way.

having read through the threads and been thoroughly confused as to how people keep getting 50%, attaching extra bits of info that aren't in the question, and how convoluted the responses to such a simple question are.

I've come to one question, and one idle wondering...

Question: Jeff, what is the tool you use for building all your diagrams?

Idle wondering: do bloggers use these questions as a trick to drive comment rates nuts, you don't even have to say anything controversial - just say: monty haul - discuss, or 'will the airplane fly from the conveyor belt' ;)

@Philip

"BB reduced to 0%, because we know there must be at least one girl, and GG doubles because we KNOW there is at least one girl"

I agree that BB reduces to 0. GG, however, does NOT double. That's ridiculous, since it wrecks the ratio. Apply the percentage changes to a concrete example and you'll see how wrong that is. 1000 people, 250 BBs, 250 GGs, 500 mixed, right? You tell me you have a girl, so we remove the BBs. If you're seriously saying it's now 50%/50%, that means the number of mixed has DROPPED from 500 to 375. That's impossible. There's still 500 mixed and 250 GGs. The percentage expands to 100, giving us 67/33, but the ratio remains the same.

@Jon

"You'd agree that if I only revealed at the first coin, it'd be 50% right? That's equivalent to a random reveal (a reveal where you could either reveal H OR T)"

I agree that if you reveal the first coin, it's 50%. That's NOT the same as a random reveal, however - it's the same as saying "my FIRST child is male". That is not the question we are addressing. Jeff's question was not dependent on whether the older or younger child was identified.

cag: <a href="http://brooksmoses.livejournal.com/137327.html?nc=19">this post I wrote</a> explains why I think it's 50%. And, yes, I'd like to turn out the lights on this comment thread. :)

Short form? We have UTTERLY NO CLUE. And "50%" is the standard answer that humans give in conversation when they have no clue. (Yes, I am being somewhat flippant about giving it a number.)

Although there is additional information in the question, there is a lot of assumption inherent in Jeff's analysis that wasn't in the question. Let me enumerate some:

1.) The parent has been selected from a population in which the probability of having a girl baby is 50%.

1a.) The ratio for people in general is 50-50.

1b.) The parent has been selected from a population equivalent to "people in general".

2.) The parent has been selected by picking a random parent from the set of all parents in this population, with equal probability of picking any of the parents.

2a.) This could be generalized to "the probability of picking a given parent is uncorrelated with the gender of their children." Jeff has not even gone this far with his analysis.

2b.) The parent has not been selected by a process of picking a random girl child with equal probability, and then picking one of her parents with equal probability. (Real-life example: You have a child on a girl's softball team. You randomly pick another parent at a game, and find out they have two children.)

2c.) The parent has not been selected by a process of picking that is likely to pick one of a small group of parents already known to me. For example, in my case, I know exactly one person who has two children, at least one a girl, who has the mathematical sense of humor necessary for her to say to me, "I have two children. One of them is a girl." As a result, for me in practice the probability is closer to 90% for both girls.

2d.) The parent has not been selected from a subset of the population known to have a specific gender selection. (Quasi-real-world example: I talk to a parent at a support group for parents of mixed-gender children.)

3.) The data communicated is limited to the literal facts given in the description, not to the facts likely implied by the way that the description was worded.

3a.) There is some reason why an analysis of the sort that would take a large corpus of conversations among people, select the ones that would lead to providing those peices of information but not the piece of information of the gender of the other child, and finding the probabilities for the speakers in question, would be inappropriate.

3b.) We are not to read the description as implying that the person has said something literally equivalent to "One of my children is a girl." People almost never say that unless they are implicitly saying, "The other one is not."

And so on. In short, to come up with the analysis Jeff gives -- as a "inarguable right answer", mind you! -- we must use information beyond the information in the question, and specifically we are expected to include (not-quite-true) knowledge about gender distributions among children, an assumption that the selection of this parent has matched the simplest mathematical ideal, and ignore all of our knowledge about when people are likely to say things.

This is, quite frankly, a preposterous set of things to include and exclude for anything that has real-world relevance rather than being a mathematical curiousity. And I think it's actually pretty insulting when a person takes a lack of willingness to play along with the standard (cultural! not universal!) rules of mathematical curiousities as a lack of understanding of the mathematics.

Moreover, I think it is rather dangerous when people ignore the assumptions inherent in their mathematical curiousities, as this leads to ignoring the assumptions inherent in their real-world examples. And that leads to engineering failures.

I think it suppose to be 100%.

If the first statemens is: "if someone has two children and one of the is a girl", then we can argue about the above probability.

But if someone met another person, and said directly that "one of them is a girl", it's not possible if both are girls.
So, we should eliminate that possibility of both are girls.

Cheers.

@Russ
There is no reason why you'd need to first calculate the probability of me being in the "two children including one boy" category since I have told you that this was my situation. Therefore the probability for me to be in that category is 100%, not 66%. And therefore the probability of my second child being of the other gender is 50%.

I too cannot understand why, be it for one second, you do not want/cannot see the problem in a simple way.

I stopped reading the comments after 1/3.
as a former programmer, both side made sense to me. it's all depends on how you first "read and interpret" the question.

really, the question didn't have anything suggest that the order would be a factor, so without thinking too much when answering the question, 50% seems to be the most straightforward answer.

but if you really put it in a program and do "random", that's when you get the 2/3 chances of have a boy and a girl.

why so serious?

@Breton

This post, where he explained the answer he was looking for was 67%. You'll find it at the top of the page, along with his explanation as to why.

@David

You're right that nothing in the original question indicates that we have to consider order, but simply because we are not given that information, we have to enter it as a variable.

In A we know the order. The first child was a girl.

In B we have no idea which came first, so we must account for either possibility.

@canchi

Maybe all the people arguing for 50% are people that can't code?

I agree with David. If the question was stated this way...

"Let's say, hypothetically speaking, you met someone who told you they had two children, and THEY AREN'T BOTH BOYS. What are the odds that person has a boy and a girl?"

Then the answer is 2/3. But it wasn't stated that way. Given that we know there's at least one girl, then we only need to solve for the other child. That's why the answer to Jeff's question is 50%.

@David

"Therefore the probability for me to be in that category is 100%, not 66%. And therefore the probability of my second child being of the other gender is 50%."

You can take your membership of the "two children, one boy" category to be a given rather than calculate it, if you prefer. That's fine.

The fact remains, however, that within that group - which we have established you are part of - 67% of people have mixed-gender children. Therefore, as a flag-waving card-carrying member of that category, there is a 67% chance that your other child is a girl.

I fail to see how, if you accept you are indeed in the two-child-one-boy category, you can then go on to claim that the probabilities defining that category don't apply to you.

*sigh* I can't stay away. Here's another way to look at it:

We start with the four-way split:
B1/G2 - 25%
G1/B2 - 25%
B1/B2 - 25%
G1/G2 - 25%

The woman says "one is a girl", which means that we now have three possibilities:
B1/G2
G1/B2
G1/G2

But are they equal? No. She either identified the first child or the last child. That's a 50% split. So, the break down is:

B1/G2 - 1/3 * 50%
G1/B2 - 1/3 * 50%
G1/G2 - 1/3 * 50% * 2 (as there are two combinations that work)

So, as we can see: she is twice as likely to identify the G/G combination. So our final break down is:
B1/G2 - 25%
G1/B2 - 25%
G1/G2 - 50%

for a 50/50 chance that the other child is a boy.

Remember: it's a combinatorial problem, not a permutation problem.

@bglick
"but simply because we are not given that information, we have to enter it as a variable"

See, that's what's beyond me. So simply because the question is ambiguous you choose to make it more complicated instead of simpler? Sorry, but I choose the simpler version. Nobody can say I'm wrong, just like I cannot say anyone is wrong to choose the complicated version.

@Robert

So, you agree that there's a 33% chance of it being a family with 2 female children, but think that since there's two potential children for her to identify it's twice that?

That doesn't make much sense, does it?

If you're selecting *families* at random, it's 67%; if you're selecting *girls* at random, it's 50%.

Or do you really think that if you have three mothers standing in front of you, each with two kids as given in the combinations listed, the one with two girls gets to answer twice? That's the only way 50% works in your example.

@Robert

"So, as we can see: she is twice as likely to identify the G/G combination."

This is the key point on which we disagree. It may or may not be true that the likelihood of a parent saying "I have a girl" is doubled for GG. It could equally be argued that the probability is lowered in that case since they would be more likely to say "I have A girl" in the mixed gender case. It's highly, highly subjective and I don't accept that you can attach a probability to it with enough confidence to allow it to affect the result.

More to the point, it's irrelevant. It's outside the scope of the problem. If you replace children with coin tosses, the underlying 2/3 math is identical and all of this frippery is thrown out. We are given the premise that the couple have two children and one is a girl. Arguing about the premise is missing the point - it's a PREMISE. If you don't accept the premise, you're answering a different question.

@robert:

"""she is twice as likely to identify the G/G combination."""

That doesn't make the G/G combination more likely.

All you proved was that if a woman has two kids and at least one of them is a girl, then it is more likely for one of them picked at random to be a girl(because the BB case has been excluded). Then you made an illogical jump to a conclusion about the sex of the other child.

Trevel said "This post, where he explained the answer he was looking for was 67%. You'll find it at the top of the page, along with his explanation as to why."

Then I am forced to conclude that Jeff picked a faulty analogy in his original question. He should have asked a question about gambling, where the order things occur in is significant. He obviously doesn't give a shit about babies.

As to the rest of you, I have posted TWO javascript programs. One results in 50%, the other results in ~67%, and they both look reasonable to me. Anyone with firebug can run these programs in the firebug console. I could even make bookmarklet versions if that's what it'll take to get you to run them. These programs show definatively, at the very least, that the assumption about how the parent reveals the gender DOES in fact make a difference, by experiment.

The confusion comes from the fact that given two coin tosses, the results are seemingly different if the coins are numbered than if they are not. For example, HH will give a 1/4 possibility with numbered coins, but by picking random coins, the first one (picked at random) has 1/2 possibility to be H and now that we know the first H, the second H must be 1/3 (since the possibilities are HT TH HH). So HH has 1/2*1/3 = 1/6 possibility.

HH is HH no matter what order they are given. So there is confusion as to which is correct. 1/4 or 1/6.

the question is lack some informations. people would ignore the sequence (BG = GB). that's why people answer 50% (the possibilities are 50% boy, 50% girl).

An update to the php simulation...

Simulations without at least one girl are thrown out.

<?php
define('GIRL', 1);
define('BOY', 2);

$iterations = 100000;

$has_two_girls = 0;
$one_girl_one_boy=0;

$counted_simulations=0;

for ($x=0; $x < $iterations; $x++) {
$children = array();

$children[] = rand(GIRL,BOY); // first child is born
$children[] = rand(GIRL,BOY); // second child is born

/*
Throw out simulations that don't have at least one girl
*/
if (in_array(GIRL, $children)) {
$counted_simulations++;
if (in_array(BOY, $children)) {
$one_girl_one_boy++;
}
else{
$has_two_girls++;
}

}

}

print "Has two girls: ".round(($has_two_girls / $counted_simulations)*100)."%<br/>\n";
print "Has a girl and a boy: ".round(($one_girl_one_boy / $counted_simulations)*100)."%<br/>\n";
?>
Produces:
Has two girls: 33%
Has a girl and a boy: 67%

Justin - no, I pointed out that if one of the children is identified as a girl at random, then the G/G combination is twice as likely as either of the B/G or G/B combinations. That is because there are two combinations of G/G, and only one each of B/G and G/B.

So - there are two G/G combinations, 1 B/G and one G/B. Each combination is equally likely. Thus there is a 50% chance of a girl and a boy.

Remember: G/B, B/G, and G/G are three permutations and four combinations. Combinations are what counts for probabilities.

Russ - I never said that there was a 33% chance of picking the G/G. I said that there were three permutations, but the G/G represented two combinations.

If you don't understand the difference between permutations and combinations, then I suggest you stop trying to defend your answer. I've shown not one, but several different ways of coming up with the 50% chance.

@David

You're right. The point I was trying to make is simply that the question _is_ at least somewhat ambiguous, and that how we understand it makes a difference in our final answer.

I certainly wasn't trying to tell you that you're wrong. And I'm wondering how I came across as picking one answer over the other.

Here's how I interpreted it. The person has two children, a and b. The person has told you that one is a girl, I.E.

girl(a) || girl(b)

(Notice that this is an inclusive OR)

There are three situations which satisfy that statement.
girl(a), girl(b)
girl(a), !girl(b)
!girl(a), girl(b)

Of those three situation, two involve one girl and one boy.
So the probability is 2/3.

However, I believe some people were interpreting it this way:
The person has two children, a and b. The person has told you that one (child a) is a girl, I.E.

girl(a)

There are two situations which satisfy that statement.
girl(a), girl(b)
girl(a), !girl(b)

So 1/2 of the possible situations involve one girl and one boy.

So yes, the question was ambiguous, specifically the meaning of "one". It could be interpreted as "this one is a girl" or "at least one is a girl". Hopefully the formatting of the question into common programming syntax will help some people understand. I had to formalize it myself before I got it. =)

@Equitastic
This simulation would be perfect if the question was "among all possible combinations of two children, taking the order in which the children were born into consideration, what are the odds that the combinations that have at least a girl also have a boy."

That is, at least to me, nothing like the question that was originally asked, which I could rephrase like this: "among ONE couple that has two children including one boy, what are the odds that the other child is a girl". And since the question doesn't mention order as being an important factor, I choose the simplification of order being irrelevant. And the result to this is 50%.

Here's how I interpreted it. The person has two children, a and b. The person has told you that one is a girl, I.E.

girl(a) || girl(b)

(Notice that this is an inclusive OR)

There are three situations which satisfy that statement.
girl(a), girl(b)
girl(a), !girl(b)
!girl(a), girl(b)

Of those three situation, two involve one girl and one boy.
So the probability is 2/3.

However, I believe some people were interpreting it this way:
The person has two children, a and b. The person has told you that one (child a) is a girl, I.E.

girl(a)

There are two situations which satisfy that statement.
girl(a), girl(b)
girl(a), !girl(b)

So 1/2 of the possible situations involve one girl and one boy.

So yes, the question was ambiguous, specifically the meaning of "one". It could be interpreted as "this one is a girl" or "at least one is a girl". Hopefully the formatting of the question into common programming syntax will help some people understand. I had to formalize it myself before I got it. =)

@Breton

I've been playing with your 2nd javascript example. I believe it produces a 50% result because it is dismissing too many outcomes. In particular, it counts no results if there is a mixed pair and the parent identifies the boy. You are treating a specific case as a general one.

The general case here is that when you have two (births, coin tosses, tokens from a hat) and identify one of them, there is a 67% chance that the OTHER (birth, coin toss, token) will be the opposite and only a 33% chance it'll be the same. That is, BG (unordered) is more likely than BB.

Your code, however, is basically discarding any scenario in which the parent identifies a boy. This is too specific. To solve the general case, the if statement should have TWO branches - if a girl is identified, check for a boy sibling; if a boy is identified, check for a girl sibling.

By ONLY checking for an identified girl, you are constraining the entire problem domain to the single case that Jeff used as an example.

Interestingly, by playing around with which numbers get checked etc, I can make any outcome become the most likely, but they all congregate around 50/50 and 67/33. You may have proved something here, I'm just not sure what it is :-)

Aren't two different problems being confused?

If I've already got a daughter and my wife is currently in the delivery room, the odds of the new baby being a boy or a girl are surely 50/50.

Those odds don't change two minutes later, but ...

If I have two children the question is what are the odds of me having a boy and a girl when the boy/boy scenario is eliminated, thus the 66%/67%/two-thirds argument.

This to me seems like the tossing a coin thing. If I toss a coin 999 times and get heads each and every one of those 999 times, what's the odds of the 1000th toss being heads. 50/50 of course. But what's the odds of tossing 1000 heads in a row. Somewhat longer.

The 66% question:

Let's say, hypothetically speaking, you met someone who told you they had two children. You ask them if they have a girl, and they say yes. What are the odds that person has a boy and a girl?

The 50% question:

Let's say, hypothetically speaking, you met someone who told you they had two children. They mention that Sarah, they're daughter, plays soccer. What are the odds that person has a boy and a girl?

You should bring up the Monty Hall problem and see how readers react =)
The NYT has a decent explanation: http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A967958260&sec=&spon=&pagewanted=all and simulation http://www.nytimes.com/2008/04/08/science/08monty.html

Like all great debates, I think we are debating different things.

I think all of this comes down to a badly worded question leading to different interpretations. Reading through the comments I can see this as the central problem.

Can we all at least agree on this:

"What are the odds of a couple who have two children having both a boy and a girl when at least one child is known to be a girl?”

Answer 2/3

"What are the odds of a pair of children being boy and girl when at least one child is known to be a girl?”

Answer 1/2

"Someone told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?"

Answer 1/2 or 2/3 (depending on interpretation it could match either of the two questions above).

The children have a 50/50 chance of having a same sexed or opposite sexed sibling.

The parent has a 2/3 chance of having mixed sex children.

@Jon
The questions are the same. What you know after reading either of the two questions is exactly the same: the couple has two children and one of these children is a girl. Whether you asked or you were told doesn't change anything. If it does, you'll have to show me how. Given that, what's the correct answer, 50 or 66?

@Robert

"If you don't understand the difference between permutations and combinations, then I suggest you stop trying to defend your answer. I've shown not one, but several different ways of coming up with the 50% chance."

Yes, and none of them hold. It's been shown over and over that it doesn't matter which child is identified in this problem, so there is no benefit in identifying the two combinations gG and Gg. Identifying two combinations for the third permutation DOES NOT change the result. It DOESN'T MATTER how likely the parent is to say "I have a girl" instead of "I have a boy".

If they say they have a girl, there's a 67% chance the other child is a boy. If they say they have a boy, there's a 67% chance the other child is a girl. It is completely irrelevant which one they choose to say, since we are calculating the probability that the sibling is the opposite and that number is always the same.

Chris, you correctly identify the three scenarios, but you've assumed that they are all equally likely. They aren't.

If order is not important, you end up with G/x, where X is fifty fifty girl/boy. If order is important, you end up with G/x * 50% and x/G * 50%, where x is still fifty fifty girl/boy.

"I've been playing with your 2nd javascript example. I believe it produces a 50% result because it is dismissing too many outcomes. In particular, it counts no results if there is a mixed pair and the parent identifies the boy"

Why would I count them? In the question, the parent hasn't identified a boy. Knowing that, I discard all the possibilities where parent has first identified a boy. From there, I can determine the odds of the gender of the remaining child. I'm not sure how to interpret the question in a way where I count parents that have first identified a boy. In the question, this parent is clearly not one of them, so their odds have no bearing on the solution.

@Russ

>>The general case here is that when you have two (births, coin tosses, tokens from a hat) and identify one of them, there is a 67% chance that the OTHER (birth, coin toss, token) will be the opposite and only a 33% chance it'll be the same.

Flip two coins. Identify ONE, randomly, without even looking at the other. Then count the number of times the other one is the opposite. This will get you 50%.

I finally figured out the problem. After choosing a specific girl, asking if the other person is a girl is 50%, but asking if the family is BG without specifying any particular girl is 66%.

Suppose you have:

B1/G1
G2/B2
G3/G4

If they have someone picked specifically, you can end up with any of G1, G2, G3 or G4. So whichever is picked, the other sibling is:

B1, B2, G4 and G3 respectively.

So when there is a specific identity associated with the "girl", the answer is 50%. But if it's not specifically chosen, then G3 and G4 point to the SAME family and you end up with three cases where G3 and G4 are half as likely as G1 and G2.

So this is why if you meet a family and they SHOW you the girl and tell you if she's the oldest or not, then the odds are that the other one is also a girl is 50%.

But if you DON'T know if she's the oldest or not, then it's 66%.

Consider this problem, is it the same?

Let's say, hypothetically speaking, you met someone who told you they had two children. They reveal the sex of one of the children. What are the odds the other child is the opposite sex?

@robert:

"""Remember: G/B, B/G, and G/G are three permutations and four combinations"""

Uhhh, actually there are 4 permutations and 3 combinations.

Permutations:
BG 1/4
GB 1/4
BB 1/4
GG

Combinations
BG 1/2
GG 1/4
BB 1/4

out of the 4 PERMUTATIONS of 2 kids, there is only ONE way a woman can have 2 kids who are both girls, and that if both of her kids are girls. The fact that the GG result makes a child picked at random more likely to be a girl does not make the GG result more likely!

@Philip

Interesting. I agree with this statement:

"The children have a 50/50 chance of having a same sexed or opposite sexed sibling."

If this is the perspective you have been driving from, your position becomes more understandable. Could you elaborate slightly on how your first two statements are semantically different, however? I think if you take the perspective of one of the siblings, then 50% makes sense, but if you take an 'external' perspective 2/3 is right. I think your first two statements are both 'external' and therefore both 2/3.

@Justin

>The fact that the GG result makes a child picked at random more likely to be a girl does not make the GG result more likely!

It may or it may not. The way Jeff is reading the question it doesn't. I don't particularly lean either way. The 'told you' part t could mean a random sample.

I'll try to explain why I changed my mind (search for "Olav" on previous comments if you're interested).

Let's put it this way. You get 10k parents with two kids. Statistically (supposing the chance of being a boy/girl is 50/50), you'll get:

2,500 bb
2,500 bg
2,500 gb
2,500 gg

That's:

2,500 bb
5,000 bg (order doesn't matter)
2,500 gg

Now, the father tells you he has, at least, a girl. That eliminates the first 2,500 bb from the equation, so there are 5,000 bg and 2,500 gg. That's the 2/3 Jeff is talking about. At least that's how I understood it.

My mistake on the first place was that I read it too fast and I was too excited to write an answer. Too much testosterone, as I said before.

@Jon

"Flip two coins. Identify ONE, randomly, without even looking at the other. Then count the number of times the other one is the opposite. This will get you 50%."

Assuming you mean that you identify a coin at random with every pair of tosses, and then note whether or not the other is the same or opposite, and tally the results, I suggest you try it. You will not get 50%. There are a number of simple programs on this page that will simulate it for you. 67% of the time, the other coin will be the opposite of the one you randomly identify.

"Consider this problem, is it the same?

Let's say, hypothetically speaking, you met someone who told you they had two children. They reveal the sex of one of the children. What are the odds the other child is the opposite sex?"

I believe this is stating the general case, whereas Jeff's example is a specific case of it. I think a lot of the 50% errors have been due to incorrectly expanding the specific case to the general case.

The answer to the general case is 2/3.

@Russ

"I've been playing with your 2nd javascript example. I believe it produces a 50% result because it is dismissing too many outcomes. In particular, it counts no results if there is a mixed pair and the parent identifies the boy. You are treating a specific case as a general one."

This is simulating the "parent picks the child to identify" (random) selection, which naturally gives a 1/2 result.

If the program only discarded BB pairs, that would simulate the "parent is asked whether either child is a girl" scenario, and give 2/3.

It absolutely matters how the clarifying information is revealed.

"5,000 bg (order doesn't matter)"

the order that the parent tells you, does kind of matter if you're going to interpret it like that.

bb
bg
gg
gb

now the parent has told you it's a girl, you're only left with

gg or
gb

bg does not count, because the parent did not tell you that they had a boy.

@Jon Your question is the same. The answer is 2/3

@Russ

"Assuming you mean that you identify a coin at random with every pair of tosses, and then note whether or not the other is the same or opposite, and tally the results, I suggest you try it. You will not get 50%. There are a number of simple programs on this page that will simulate it for you. 67% of the time, the other coin will be the opposite of the one you randomly identify."

This is clearly not true. The other coin will be the opposite only in the HT or TH case, which happens 50% of the time. The other two cases (HH and TT) make up the other 50%.

@Derek

It is omitting every case where the identified child is male. This means that it does not account for identified males with female siblings, which is a valid interpretation of the problem in its general form. You cannot draw valid conclusions from such a distorted population.

The correct way to do it (as was the case with Breton's first simulation) is to generate an unbiased population, THEN ask questions about it. If you don't arbitrarily exclude data from the initial population (as is the case with real birth rates) and THEN ask the questions, you will find that it doesn't matter if you randomly identify a boy (67% chance of girl sibling) or a girl (67% chance of boy sibling).

@Russ

"Assuming you mean that you identify a coin at random with every pair of tosses, and then note whether or not the other is the same or opposite, and tally the results, I suggest you try it. You will not get 50%. There are a number of simple programs on this page that will simulate it for you. 67% of the time, the other coin will be the opposite of the one you randomly identify."

Ok, so we have a whole table is full of pairs of flipped coins.

We aren't throwing away any pairs in this example, every flip one was just revealed randomly. But nothing was removed from the table.

We know the general distribution for a table full of flipped coin pairs:

HH .25
HT .25
TH .25
TT .25

So how can the table, the GENERAL CASE, have 67% opposites? By randomly revealing a coin we didn't change anything on the table.

Now, if we put a sign on the table that says 'every pair must have at least one H' and throw away all the TTs, we get 67% pairs.

@Derek

"This is clearly not true. The other coin will be the opposite only in the HT or TH case, which happens 50% of the time. The other two cases (HH and TT) make up the other 50%."

For pity's sake, run the numbers instead of running your mouth. When you identify the first coin, you ELIMINATE one of the identical pairs, and therefore they no longer add up to 50%. That is the WHOLE POINT of this blog post! If you ran one of the simulations, or even did it yourself with a coin, you would SEE the results and you wouldn't keep making unsubstantiated claims.

I'm not answering anything else from anybody that hasn't actually tried this out instead of just posting the first thing that pops into their head.

Vorlath,

That last statement is incorrect. The age of the girl relative to her sibling is irrelevant. If your knowledge is that the family has two children, and you see a girl, the probability is 50% that the other one is a girl.

"It is omitting every case where the identified child is male. This means that it does not account for identified males with female siblings, which is a valid interpretation of the problem in its general form."

The original question did not ask us to expand the question to its general form. I discard bg AND bb in my second example, because the parent did not identify a boy. Since the question is asking me to assume that the parent has identified a girl, that "distorts" the population in precisely the way I simulate. Why would I count parents who have identified a boy, in the odds of a case where the parent has identified a girl?

I believe the mistake for some is in thinking BG and GB are the same. They are not. There are 2 children and to get the correct number of combinations you must allow for either child to be a boy or girl.

ex.
I have 2 children. What are the odds one of them is a boy?
There are 4 possible combinations
BB
GG
BG
GB
Three of which include a boy. So there is a 3 out of 4 (75%) chance I have at least 1 boy.

Now add the other piece of info provided, one of my children is a girl.

ex.
I have 2 children. One is a girl. What are the odds one of them is a boy.
There are 3 possible combinations (BB is excluded since one is a girl)
GG
BG
GB
Two of the combinations include a boy. Thus there is a 2 out of 3 (approx. 66.67%) chance that my other child is a boy.

Note this has nothing to do with which was born first, their age, if they are twins, if the mother is a stone face liar, or any other information. Its a word problem, don't add information that isn't there.

If you come up with a different answer than what I did above, you are either trying to solve a different problem or are wrong.

Am I correct in thinking the confusion, other than the original question being a criminal use of the English language, is based on what part of the set is being compared to the possible elements?

The breakdown of the 2-child couples is as follows:

25% - 2 boys
25% - 2 girls
50% - 1 boy and 1 girl

It's 50% to have a boy and a girl if all 2-child couples are the set.

It's (2/3)*100% to have a boy and a girl if only couples that have at least 1 girl are the set.

It's (2/3)*100% to have a boy and a girl if only couples that have at least 1 boy are the set.

Am I missing something?

@Russ

"If you don't arbitrarily exclude data from the initial population (as is the case with real birth rates) and THEN ask the questions..."

That's the whole problem with the original puzzle. You (the solver) isn't asking questions, but receiving data that may or may not be consistently generated. If you can ask "are there any girls?", then yes, the 67% result is correct. If not, you're back to 50%.

xooorx has it absolutely right with the discussion on "preferential revealers".

@Russ

The semantic difference is in the focus of couple or the of child.

Focus on the child:
Children 1: bB
Children 2: Bb
Children 3: bG
Children 4: Bg
Children 5: gG
Children 6: Gg

How many combinations of children are there? 6
How many combinations are there (excluding birth preference)? 3 (all boy, all girl and mixed).

Combinations are:
BB – two boys
BG – a boy and a girl
GG – two girls

Take away BB and you get 50/50

Focus on the couple:
Couple1: BB
Couple2: BG
Couple3: GB
Couple4: GG

How many combinations of couples are there? 4.

Take away the option of two boys and you get 2/3 mixed and 1/3 all girls.

Restated - Child focus:
If there are two children and one is a girl, what is the chance the other will be a girl? 50%. This is, of all children who are girls, 25% have older brothers, 25% have older sisters, 25% have younger brothers 25% have younger sisters.

Restated - Couple focus:
If a couple have two children, and one is a girl, what is the chance the other will be a girl? 1/3. This is because, of all couples with at least one girl, 1/3 of couples have all girls.

So one focuses on the child and the other the couple.

Ok, let me take a stab at explaining this:

There's a large group of mothers in the world: The ones who have two children. Of those:

25% have had a boy and then another boy;
25% have had a boy and then a girl;
25% have had a girl and then a boy;
25% have had a girl and then another girl;

So 75% of those mothers have at least one girl, right?
Of THOSE 75%:

1/3 have girls only;
2/3 have one boy and one girl;

So, if the mother in the problem has said to you she has a girl, she is on the 75% group.
Thus, she has a probability of 2/3 of being in the group that has a girl and a boy.

Hope that makes sense.

The key ambiguity is the question of whether you are considering the odds of you being told the sex of the child.

Go up to random people, ask if they have two children, ask if they have a girl, then it's clearly 66%.

Go up to random people, ask if they have two children, ask them to pick one child and tell you the sex of it, they happen to tell you girl, it's 50%.

One is a random sample we need to consider the odds of (the odds that we could have gotten a boy as a result instead) because with a random sample, we are more likely to have stumbled across the GG case.

Which question Jeff's corresponds to is a matter of interpretation.

@Jon

"So how can the table, the GENERAL CASE, have 67% opposites? By randomly revealing a coin we didn't change anything on the table."

Just. Try. It. I'm sick of repeating this. Go and do it before commenting further.

Flip two coins 100 times and record the results. Now flip the coins again and 'reveal' one coin. Draw a line through every result that does not contain the result (i.e. all the HHs or all the TTs). Now look at the results that are not struck through. There will be less identical pairs than mixed pairs. The ratio will be around 67/33. The larger your result set, the nearer to 67/33 the ratio will be.

@Russ

Let me get this straight. The following events occur:

1. Two coins are flipped
2. One of the coins is randomly revealed to me
3. I'm supposed to guess the other coin.

Your proposal is that if I guess the opposite value of the revealed coin after step 2, I'll win 67% of the time?

Another way of saying it:

The answer depends on how you know that one child is a girl.

If you know that one was a girl because you took a random sample from the space of children and drew a girl, then the answer is 50%.

If you know that one was a girl because you sampled randomly from the space of households and discarded households with two boys, then the answer is 67%.

Wow, over 600 comments. Reminds me of the "fizz buzz" or "Can a plane take off from a conveyor belt" type of question. It's an endless binomial thread of true-false, yes-no, zero-one, you're an ass and I'm not.

For the definitive answer to Jeff's question, search this page for "Yvon". For my take, soldier on below.

In my one year of "Statistics for the Social Sciences" (ie, stats for artsy farts who struggle with their checkbook) I mainly remember the admonition to always define the "relevant population".

If 100 families have sex and each produces 2 rugrats, then according to Father Nature, the distribution will be as follows:

1. 25 families have 2 girls.
2. 25 families have a boy followed by a girl.
3. 25 families have a girl followed by a boy.
4. 25 families have 2 boys.

Now Jeff has kindly told me ahead of time that the families we are to consider already have at least 1 girl. Praise the Lord, it is a Bayesian gift from heaven, a prior probability! Now the relevant population is not the original 100 families, but the 75 families with at least 1 girl. Do the math yourself (clue: it ain't 50/50).

@Jon

You have presented the same information: 2 children, one is a girl.

How the data was acquired means nothing. At least for the original problem anyway.

@Russ

And this is the ambiguity Jon is talking about.

Couple/parent focus - this is about flipping coins and crossing out ones that don't match. You end up with your values.

Child focus - this is about one flip. Look Reveal 1 coin. What are the chances you will get another just like it, and what is the chance you will get a different one.

Are we asking about the proablity of the sex of a specific child, or are we asking about the probability of the sex of the set of children?

I initially took this to mean what is the sex of the other child (50/50) but the question is too vagure to be sure.

@Derek

"receiving data that may or may not be consistently generated"

The initial population IS consistent - it's 25/25/50. It doesn't matter in the least how you arrive at the information that one of the children is a particular gender.

Like I said, try it yourself. I explained the steps a couple of posts up. Generate an initial population. Then 'reveal' your additional info - you can do this with a coin toss, a magic 8 ball, or just say 'girl' over and over again to simulate parental bias. It DOESN'T MATTER. Whatever your revealed information is, remove impossible pairs from your initial population, and from the remaining outcomes compare the indentical pairs to the mixed pairs. They will NOT be 50/50 with a decent size initial population. No matter how you choose your additional information, the ratio will approach 67/33. Try it.

@Jon

Absolutely. How you obtain the ">= 1 girl" info is of utmost importance.

@Philip

"Child focus - this is about one flip. Look Reveal 1 coin. What are the chances you will get another just like it, and what is the chance you will get a different one."

I can't see how the original problem could be interpreted like this. I thought it was pretty clear that the children were ALREADY born, and this wasn't a question about "what's the NEXT child/coin/token going to be".

If that's how people are reading it, though, then it explains at least some of the 50%s.

Jon is right. It's about the sampling. Here's another version of the problem to show this.

You meet someone who tells you they have two children.

You ask, "Is at least one a girl?"

"Yes."

You then ask something even stranger. "Then, if only one is a girl, tell me whether it's the older child or the younger child. If you have two girls, just say 'older' or 'younger' at random."

You hear either "older" or "younger." What is the probability that the other child is a boy?

To convert it into coins:

You have table of 100 flipped coin pairs. I think we can all agree that at this point, there is an equal number of matched pairs as there are opposites, right? That is, there are 50 matched pairs, and 50 opposites. So if you were to randomly select a coin, you'd be equally likely to have picked from a matched pair as a pair of opposites?

OK. Now start our test. You pick a coin randomly from the table. If it's a T, pick again until you get an H. H is the girl in our example. What are the odds the other coin in the pair you picked is a T? 50%. Why? Because of the random nature of your pick, you were twice as likely to pick from a HH pair as you were HT or TH individually. So it's .5 for HH, and .25 each for HT and TH.

OK. Now the 67% solution.

You're looking at the same table with 100 pairs of flipped coins. This time, you pick a random PAIR from the table. If that PAIR doesn't have a T, pick another pair. If it does have an H, what are the odds the other coin in that pair is a T? This is 50%.

How we know the information matters.

@Russ the question did does not include the stipulation that you're asking whether the parent has at least one girl, while your coin flipping scenario does require that you ask this when you're eliminating/including results.

You're asking, is either of the children a girl?
the parent, is not being asked this question. They are instead volunteering information about the gender of one of their children.

@Derek

"Your proposal is that if I guess the opposite value of the revealed coin after step 2, I'll win 67% of the time?"

NO. This is NOT about future probabilities. In the original question, the couple have two children already.

After step 1, eliminate the impossible combinations and see what the results tell you about the historical pairs containing that result have been. There will be more mixed pairs than identical pairs. When you toss the coin, you are deciding the 'reveal', i.e. telling yourself that one of the children is a girl. It is NOT predicting the probability that the NEXT child will be a girl. That's a totally different question and not what the original puzzle was about.

@Russ

There's a difference between asking "is there a girl" and being told "there is a girl". When asking, you can certainly eliminate the BB pairs. However, if you're being told, you will hear "there is a boy" from a BB pair, and you may be told either "there is a boy" or "there is a girl" from a GB pair.

So eliminating invalid pairs never happens in the "told" case, since you never get any false (impossible) information.

Brian G,

"That last statement is incorrect. The age of the girl relative to her sibling is irrelevant. If your knowledge is that the family has two children, and you see a girl, the probability is 50% that the other one is a girl."

Only if you know that the girl is the oldest or not.

You can only have ONE representative for each family. In the case of a family with children Elaine and Cindy, you can only include ONE of them to be the representative girl. Either one will do, but ONE it must be.

So say we name all three kinds of families:

Family #1: Bob,Michelle
Family #2: Nicole,John
Family #3: Elaine,Cindy

If you create a set from which to pick the family, you can only include ONE representative girl from each family. Family #3 can use either one, but only ONE can be picked to represent the family at any given time.

{Michelle, Nicole, Elaine}

Pick one of these three representatives. Two of them have brothers.

You could likewise choose from a different set of representatives.

{Michelle, Nicole, Cindy}

And again two of them have brothers.

However, if you pick a particular person and ask if the other sibling is a boy, then it's 50/50. It doesn't have to be a name, it could be if they are the oldest girl or the youngest girl. If you know which, then the other sibling will be 50% a boy.

>>You're looking at the same table with 100 pairs of flipped coins. This time, you pick a random PAIR from the table. If that PAIR doesn't have a T, pick another pair. If it does have an H, what are the odds the other coin in that pair is a T? This is 50%.

I obviously meant 67% here like the intro to that paragraph. Frak I'm going to sleep.

@Russ

""Your proposal is that if I guess the opposite value of the revealed coin after step 2, I'll win 67% of the time?"

NO. This is NOT about future probabilities. In the original question, the couple have two children already."

Both coins are already on the table. Their values are set. I'm just revealing one of them to you. Same difference.

Vorlath,

"Only if you know that the girl is the oldest or not."

What if I don't know that, but I do know whether she is the tallest or not?

@Breton

"Russ the question did does not include the stipulation that you're asking whether the parent has at least one girl, while your coin flipping scenario does require that you ask this when you're eliminating/including results."

No, I'm saying you generate the initial population first, without asking ANY questions. Once you have a decent sized population, THEN you ask questions. "Flip two coins 100 times and record the results. Now flip the coins again and 'reveal' one coin."

I think one of my earlier descriptions may have been unclear, as this keeps coming up.

@Breton

"Since the question is asking me to assume that the parent has identified a girl, that "distorts" the population in precisely the way I simulate"

No, you're distorting the population as you generate it. That's not what happens in the real world. The initial population is 25/25/50, and it is THEN distorted by adding information and distorting the available outcomes. When generating your population, you should have no exclusions. Either add another branch to the if, or have a preliminary loops that generates your population without any conditionals and run the analysis afterwards (ie. by discarding all the MM pairs - 'choosing' female - and looking at what's left).

>>You're looking at the same table with 100 pairs of flipped coins. This time, you pick a random PAIR from the table. If that PAIR doesn't have a T, pick another pair. If it does have an H, what are the odds the other coin in that pair is a T? This is 50%.

>>I obviously meant 67% here like the intro to that paragraph. Frak I'm going to sleep.

And "If that PAIR doesn't have a T" should be "doesn't have an H".

@Russ

I also thought it was clear, but will allow for confusion due to ambiguous language. I believe others are looking at the provided info and seeing a different problem or are trolling.

@Derek
@Jon
@Brian G
You are adding information that has nothing to do with the original problem (my apologies if you are talking about a different problem).

A person has two children, one is a girl. You can ask, be told, or get it from a stone tablet, it makes no difference.

@statistical anomaly

Thank you :-) I was starting to crack up here. Since you are evidently solving the same problem as me, I'm making a long-awaited move in the direction of my bed since it's about eek!'o'clock in the morning, and you can take over :-)

@statistical anomaly

>>A person has two children, one is a girl. You can ask, be told, or get it from a stone tablet, it makes no difference.

It does matter. A peron has two children. You happen to be walking by their house and them with one of their children, a girl. That's a random sample, so we need to consider the odds that of GG differently. Basically, we got information about ONE child, and it happened to be girl. This is picking a random coin from the table.

A person has two children. You ask them if they have a girl. They say yes. Now we've basically asked information where the answer depended on BOTH children. This is picking a whole coin pair from the table.

@statistical anomaly

>>A person has two children, one is a girl. You can ask, be told, or get it from a stone tablet, it makes no difference.

It does matter. A peron has two children. You happen to be walking by their house and them with one of their children, a girl. That's a random sample, so we need to consider the odds that of GG differently. Basically, we got information about ONE child, and it happened to be girl. This is picking a random coin from the table.

A person has two children. You ask them if they have a girl. They say yes. Now we've basically asked information where the answer depended on BOTH children. This is picking a whole coin pair from the table.

"No, I'm saying you generate the initial population first, without asking ANY questions. Once you have a decent sized population, THEN you ask questions. "Flip two coins 100 times and record the results. Now flip the coins again and 'reveal' one coin."

@Russ
but I don't get to ask questions in the original problem proposed. Why have you added this ability now?

The parent has either told me that they have a boy, or they have told me that they have a girl. In my code, I eliminate the population of parents who would have told me that they have a boy. What remains is the relevant population.

statistical anomaly,

"A person has two children, one is a girl. You can ask, be told, or get it from a stone tablet, it makes no difference."

As has been demonstrated many many times, it makes all of the difference. I've seen several false statements of 2/3 where that number really should not apply.

If the way you learn that one of children is a girl is by a method where each individual child is equally likely to have been mentioned to you, then the probability is 1/2. This is due to the fact that in a GG family, girls are twice as likely to be mentioned as in a GB or a BG family. These situations DO happen and they are NOT imaginary.

I have two coins

one is penny, look, here it is
-penny-

the other coin is in a bag and is either a penny or a nickel
-bag-

what is the probability that I have a penny and a nickel ?
well, the choices are:
bag contains a penny and I have
penny, penny
or
bag contains a nickel and I have
penny, nickel

the answer is 50%

I have two coins

one is penny, look, here it is
-penny-

the other coin is in a bag and is either a penny or a nickel
-bag-

what is the probability that I have a penny and a nickel ?
well, the choices are:
bag contains a penny and I have
penny, penny
or
bag contains a nickel and I have
penny, nickel

the answer is 50%

@statistical anomaly

How you get the info *does* matter. If the person self-identifies by saying "one is a girl", that cuts the BG group in half, since the person could have just as easily said "one is a boy".

The other benefit to asking "is one a girl" is 100% certainty of the two siblings if the answer is "no".

@Russ, statistical anomaly,

Being asked is different from being told, because by being told, certain statistical truths about the encounter are forced to be predetermined, based on what is being told. When you ask, they are not, since the question is entirely up to you and could be anything.

@Russ

"I can't see how the original problem could be interpreted like this. I thought it was pretty clear that the children were ALREADY born, and this wasn't a question about "what's the NEXT child/coin/token going to be".

If that's how people are reading it, though, then it explains at least some of the 50%s."

Even if both children have been born, knowing that someone has a child that is female, there is a 50% chance that the other is female. Regardless of present, past, anything. That is, there was 50% chance that the other child was born female, 50% chance the other child IS female and 50% chance the next child will be female.

That is - on your list of coin flips, take out all the ones that don't start with at T.

You will find there is a 50/50 chance of TH or HH.

Why take away all that start with a T? Because in the example with have a "Known" girl. So we start with a girl. The chance of the girl having a sister or brother are the same regardless of any other factor.

Even though the coin flip has happened, we still end up with 50/50.

That's what I mean by focus. Are we asking "what is the chance that this person has mixed siblings knowing they have 1 girl" or are we asking "what is the chance that a person with at least 1 girl has mixed siblings"? The first is 50% the second is 2/3.

@Jon

You are seeing a different problem than I am. The one I see involves nothing more than working out the four possible combinations of boys and girls when there are two children. Then removing the one invalid combination since one of the children is a girl. It has nothing to do with the sample size etc.

@statistical anomaly

The entire debate is: which problem does the question correspond to?

Is 'told you they have a girl' a random sample or not?

If the phrasing was instead:

"You asked them if they had a girl and they said yes." then it's not, and we're clearly in the 67% case.

If it was instead:

"You walked by someone's house, you knew they had two children, you saw a girl in the window." (girls and boys being equally likely to hang out in windows, etc.) then we're in the random sample case, the 50% case.

statistical anomaly,

A strict frequentist approach to this problem is bound to collapse under the weight of its own implicit assumptions as we more closely inspect the different permutations of the problem.

You meet someone who has two children. The older child is a girl. What is the probability that the younger child is a boy?

We know the answer to this, right? It's 50 percent. Yet, if you assert that the way we come across the information is irrelevant, isn't that problem a subset of the one which claim only has one right answer (2/3)?

To put it another way, you say that if there are two children and at least one is a girl, the probability of a boy is always. Yet I have a given an example which fits this premise, and the answer is not 2/3.

Hey Jeff,

Have you ever considered using a nested style comments system that allows replies?

This is insane trying to read this LOL

If BG == GB then the answer is 1:1

if BG != GB then the answer is 2:1

I don't think anyone can refute *that*. However I'm not sure that BG != GB because I can't convince my self that order matters in the problem as stated given that is uses no ordering words.

"Then removing the one invalid combination since one of the children is a girl. It has nothing to do with the sample size etc."

your valid combinations are

bg
gb
gg

if you ask if they have a girl, and the answer is yes, then it's equal odds between these three possibilities. If on the other hand, they've volunteered the information, and there's even odds that they could have revealed the gender of one child or the other.

in that case, the difference between bg and gb must be the order in which they are revealed. You must in this case eliminate bg, leaving only gb and gg.

@Brian G

Stipulation that the older child is a girl isn't about changing how the information is obtained, it is changing the question! We now can rule out BB and BG, hence why the probability changes.

I have to say this argument about true random sampling a total red-herring here.

in other words, if bg and gb are to be two distinct groups, then we can decide that one group is the 50% of mixed pairs that revealed they had a girl, and the other 50% is the group that revealed they had a boy. We eliminate the second group, because that's obviously not who we're talking to.

@Jon
@Derek
@Brian G

I talking about the original problem which boils down to a combinatorial exercise. I really think you guys are reading to much into it, but that's my opinion. The problem I see is a simple as the following:

ex. 1
I have 2 children. What is the chance one of them is a boy?
There are 4 possible combinations
BB
GG
BG
GB
Three of which include a boy. So there is a 3 out of 4 (75%) chance I have at least 1 boy.

Now add the other piece of info provided, one of my children is a girl.

ex. 2
I have 2 children. One is a girl. What are the chances one of them is a boy.
There are still 4 possible combinations, but only 3 are valid(BB is exclude