programming and human factors
by Jeff Atwood
Remember The Problem of the Unfinished Game? And the almost 2,500 comments those two posts generated? I know, I like to pretend it didn't happen, either. Some objected to the way I asked the question, but it was a simple question asked in simple language. I think what they're really objecting to is how unintuitive the answer is.
Which reminds me of another question that you've probably heard of:
Suppose the contestants on a game show are given the choice of three doors: behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, opens one of the unchosen doors, which reveals a goat. He then asks the contestant, "Do you want to switch doors?"
Should the contestant switch doors?
This is, of course, the Monty Hall problem. It's been covered to death, and quite well I might add, by dozens of writers who are far more talented than I.
What's interesting about this problem, to me at least, is not the solution, but the vehemence with which people react to the solution – as described in The Drunkard's Walk: How Randomness Rules Our Lives.
It appears to be a pretty silly question. Two doors are available – open one and you win; open the other and you lose – so it seems self-evident that whether you change your choice or not, your chances of winning are 50/50. What could be simpler? The thing is, Marilyn said in her column that it is better to switch.Despite the public's much-heralded lethargy when it comes to mathematical issues, Marilyn's readers reacted as if she'd advocated ceding California back to Mexico. Her denial of the obvious brought her an avalanche of mail, 10,000 letters by her estimate. If you ask the American people whether they agree that plants create the oxygen in the air, light travels faster than sound, or you cannot make radioactive milk by boiling it, you will get double-digit disagreement in each case (13 percent, 24 percent, and 35 percent, respectively). But on this issue, Americans were united: Ninety-two percent agreed Marilyn was wrong.
Perhaps the public can be forgiven their ignorance, but what of the experts? Surprisingly, the mathematicians fare little better.
Almost 1,000 Ph.D.s wrote in, many of them math professors, who seemed especially irate. "You blew it," wrote a mathematician from George Mason University. From Dickinson State University came this: "I am in shock that after being corrected by at least three mathematicians, you still do not see your mistake." From Georgetown: "How many irate mathematicians are needed to change your mind?" And someone from the U.S. Army Research Institute remarked, "If all those Ph.D.s are wrong the country would be in serious trouble." Responses continued in such great numbers and for such a long time that after devoting quite a bit of column space to the issue, Marilyn decided she whould no longer address it.The army PhD who wrote in may have been correct that if all those PhDs were wrong, it would be a sign of trouble. But Marilyn was correct. When told of this, Paul Erdos, one of the leading mathematicians of the 20th century, said, "That's impossible." Then, when presented with a formal mathematical proof of the correct answer, he still didn't believe it and grew angry. Only after a colleague arranged for a computer simulation in which Erdos watched hundreds of trials that came out 2-to-1 in favor of switching did Erdos concede that he was wrong.
You may recognize Paul Erdos from a particularly obscure XKCD cartoon last week. So if you feel like an idiot because you couldn't figure out the Monty Hall problem, take heart. The problem is so unintuitive one of the most notable mathematicians of the last century couldn't wrap his head around it. That's ... well, that's amazing.
How can something that seems so obvious be so wrong? Apparently our brains are not wired to do these sorts of probability problems very well. Personally, I found the text of Jeffrey Rosenthal's Monty Hall, Monty Fall, Monty Crawl (pdf) to be the most illuminating, because it asks us to consider some related possibilities, and how they might affect the outcome:
Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win, either by sticking with your original door, or switching doors?Monty Crawl Problem: Once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open, then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) Now what are the probabilities that you will win the car if you stick versus if you switch?
Paul Erdos was brilliant, but even he realized his own limits when presented with the highly unintuitive Monty Hall problem. For his epitaph, he suggested, in his native Hungarian, "Végre nem butulok tovább". This translates into English as "I've finally stopped getting dumber."
If only the rest of us could be so lucky.
I reckon the PDF's position on Monty Fall is wrong.
When the host opens the door, he is providing you with new information. It doesn't matter how he came by that information, whether he knew in the first place, or whether he's just found out by accident.
It bothers me, because the PDF appears quite learned, but it seems to reveal the author hasn't understood the Monty Hall problem in the first place.
John H on June 23, 2009 8:14 AMPeople who don't get this have "tunnel vision" and can't get past thinking "2 doors" => "50/50 chance". The thing is this ignores important extra information. You need to think of the probability of the outcome of a sequence of events: event 1 is picking 1 door out of three. event 2 is switching (or not). events 1 and 2 are not independent events. I think many similar problems try to trick you by giving extra info that is not necessary. These types of problems train you to ignore the extra stuff. But in this case the extra stuff matters.
DMR on June 23, 2009 8:21 AMI *already knew* about the 2/3 probability thing. I never disputed that. What I'm disputing is people are saying those are only the odds if Monty *knows* what's behind the doors. I've now run a simulation in which he *doesn't* know, and, if he opens a door without the prize, you're still better off switching. So, Monty Fall: yes, you'd still better switch.
As for Monty Crawl - well, if you knew he was going to crawl, then you could always pick door 3. If he then opens door 2, you know it was behind door 1. If you pick 2, then he picks 3, the same thing applies, and you know it's behind 1. There's not really enough info to do a simulation of this. Do you know he's tired?
Zelda on June 23, 2009 8:23 AM@zelda
Sorry, too many posts to keep track of,that I lost track of discussions, though it seems that the Monty Hall & Fall have been resolved.
@moya
I wrote in my post that Monty Hall was not the same as Deal or No Deal (DoND). What I tried to say was the *only* common thread between Monty Hall and DoND is the idea that your initial choice will be wrong the majority of the time.
By the way, you really think DoND is 50/50 at the end (assuming $1 million is still in play)? Run a simulation. Or, just watch the show. I use to watch that show a lot. Howie Mandell always reveals the person's suitcase. It seemed to me about 1 in 30 actually initially picked the $1 million dollar suitcase. I know, not a precise measurement. But hopefully you get the idea.
Nicholas on June 23, 2009 8:48 AM@ those who still thinks it is still 50/50 for Monty Hall problem:
Just ran a test myself using 3 identical name cards: one with my marking, others w/o. I shuffled them around and picked one to my fancy, and here's the result:
Total times tried:70
Wrong:48
Right:22
This says something about the accuracy of your first choice.
@zelda: that's because you are not counting the cases where the host accidentally reveals the car.
Generally, the reason the outcome changes is because the host inserts knowledge into the problem. People who think that the outcome is 50/50 think that the act of the host doesn't change anything, but it gives you extra knowledge.
Frederik on June 23, 2009 9:03 AM@frederik - the cases where the host accidentally reveals the car /should/ be discarded, because the problem states that this does not happen.
"In this variant, once you have selected one of the three doors,
the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car."
This makes sense if you shift out of theory and into practicalities - presumably if this were to happen during a TV show, either the whole game would be abandoned, or you'd be allowed to choose the door which you now know to contain the car!
John H on June 23, 2009 9:11 AMZelda - in the original Monty Hall, the usually unstated assumptions are that:
1) the host *knows* where the car really is
2) will *always* reveal a goat behind an unchosen box
3) will *always* allow you to switch.
If you change these assumptions (such as for "Monty Fall"), the resulting odds can/will change.
The critical (but unintuitive) piece that separates "Monty Hall" from "Monty Fall" is that in "Hall", the host is acting on known information, not random chance like he is in "Fall".
In both Hall/Fall, your original chance of guessing correctly was 1/3, with a 2/3 chance it was in "one of the other two boxes".
But in "Hall", the host *always* eliminates a bad choice from "one of the other two boxes", which since there were only 2 boxes to begin with, now means that there is a 2/3 chance it is in "one of the other *ONE* box". Before the host acted, for you, each box had a 1/3 chance, and there were 2 boxes you didn't pick (so odds are 2/3 that the unpicked *set* contains the car). Then the host uses his special knowledge to reduce the number of choices in the unpicked set, but the probablity the unpicket *set* contains the car remains the same: 2/3.
On the other hand, with "Fall", the host randomly reveals a box. This is equivalent to if a second player was allowed to pick a box, and then his box was shown to have a goat: you have no *new* knowledge special to the unpicked box. When the second player is picking, there are 2 boxes left, and as before, there is a 2/3 chance one of them has the car. Then the second player chooses one of the two, so his odds of getting the car are 50% of 2/3, or 1/3. And the odds the remaining unselected box has the car are 1/3. Now, when the second player's box is opened and the goat is revealed, we know that one of the remaining two much have the car. The new odds remain proportional to the old odds: 1/3 each for player 1 and the unselected box, so the new odds are 50% each. You still don't know anything about the unpicked box.
----------------------
As an aside to all this, I once read a brief interview with the real Monty Hall asking *him* about this problem. His response was interesting: basically, he said he wasn't required to offer the switch, and he usually offered it more when the player *did* pick the correct box, to trick them into losing.
LintMan on June 23, 2009 9:11 AM@Confuse-us Listen to http://www.ted.com/talks/lang/eng/peter_donnelly_shows_how_stats_fool_juries.html . The Monty Hall problem is just a great example of how bad humans are at statistics. Our faults in statistics aren't limited to pretend questions about cars.
@Zelda The different between Monty Hall and Monty Fall is in the events.
(In the below I'm always assuming you pick door 1, it doesn't change the odds just makes the explanation easier).
In Monty Hall [Prize 1, Open 2] and [Prize 1, Open 3] are each 1/2 as likely as [Prize 2, Open 3] and [Prize 3, Open 2]. People have shown why before but basically it comes down to [Prize 2, Open 2] and [Prize 3, Open 3] being consciously turned into [Prize 2, Open 3] and [Prize 3, Open 3].
In the Monty Fall problem this transformation is not happening. [Prize 2, Open 2] is equally as likely as [Prize 2, Open 3]. Just because it didn't happen doesn't mean that it couldn't have happened.
So our potential states, all of equal odds are:
[Prize 1, Open 2]
[Prize 1, Open 3]
[Prize 2, Open 2] X
[Prize 2, Open 3]
[Prize 3, Open 2]
[Prize 3, Open 3] X
Now we know that the two states marked with X didn't happen. The remaining four states gives us a 50/50 shot at having the prize stay or switch.
Now compare to Monty Hall...
[Prize 1, Open 2]
[Prize 1, Open 3]
[Prize 2, Open 2] -> [Prize 2, Open 3]
[Prize 2, Open 3]
[Prize 3, Open 2]
[Prize 3, Open 3] -> [Prize 3, Open 2]
We're not dealing with an even distribution of probability. [Prize 2, Open 3] is twice as likely because Monty is forced to Open 3 if the prize is behind 2. I've represented that here by showing the transformation arrow from the case when monty would have made the 50/50 chance to open door 2 but then has to open door 3 instead in order to not reveal the prize.
Logo on June 23, 2009 9:15 AMZelda -- I question the terms of your simulation of the Monty Fall scenario. You need to completely discard the cases where he falls and opens the door with the car; if those cases are being considered as a probability anywhere, then you're doing it wrong.
These six scenarios are equally likely (assume you pick door #1, and the host does not knock it open:
a: You picked the car, the host knocks open door #2
b: You picked the car, the host knocks open door #3
c: The car is behind door #2, the host knocks open door #2
d: The car is behind door #2, the host knocks open door #3
e: The car is behind door #3, the host knocks open door #2
f: The car is behind door #3, the host knocks open door #3
Eliminating possibilities c and f (because they did not happen!) leaves four choices, each equally likely.
Similarly, consider this problem: I rolled a normal six sided die. I did not get a 5 or a 6. What is the probability that I got a 4?
(In the Monty Hall problem, d and e are NOT equally probable with a and b, but d=e=(a+b) -- hence the advice to switch doors.)
-- Simulationwise, roll 3 3-sided dice. (One for the placement of the Car (C), one for your pick(P), one for the door that Monty opens (M).)
if M = C or M = P, abort and do not add to ANY totals. (Monty opened your door or the door with the car, which did not happen)... Then compare the totals of C=P to C!=P. They should be roughly even.
If we want to run the 1-million door scenario that someone mentioned, I'd invert the M roll to mean the door that Monty DIDN'T open that isn't yours. Roll 3 1-million sided dice. If M=P or M!=C, discard. Again compare totals for C=P to C!=P. Again, they should be roughly even. (Although the count of discarded rolls will be enormous.)
These do NOT model the Monty Hall problem, which as mentioned before will give you a n-1/n chance of winning while switching, where n is the number of doors.
Trevel on June 23, 2009 9:18 AM@SanDiego - The Let's Make a Deal game show was fairly entertaining, as Monty Hall was a top-notch con artist. His mission was, of course, to limit the amount of stuff people would win, while still making it seem possible to win, and people would occasionally win the big prizes, but more often than not, they would make the wrong decisions. A lot of times, Monty would offer the person $100 to walk away instead of opening the door to reveal their prize, and he was an expert in reading people and predicting how they would react to the offer. There are interviews you can read where he describes how he treated people. Contestants on the show would wear costumes, sometimes quite outlandish, and they were in a very artificial and "Monty-controlled" environment from the start, so it was pretty easy for him to manipulate people. Sometimes people would win the cars and boats and stuff, but not very often. BTW, Howie Mandel uses similar techniques on Deal or No Deal, in order to get people to continue playing, even when they have been given a "bank offer" that exceeds the "current expected value" of the game - this fairly often leads them to eliminate high value boxes and reduce the expected value and the subsequent bank offers. It is all very interesting, but more from a psychology standpoint than a mathematical one.
Jasmine on June 23, 2009 9:33 AM@Jasime and it seems like they pick people who don't really understand odds to help with the manipulation. Though maybe the people do understand odds but crack under the pressure and manipulation. Either way it'd be fun to act all bubbly and personable before getting on the show then showing up and turning on the poker face and the statistical mind that should accompany it.
Logo on June 23, 2009 9:37 AM@John H - If you eliminate the possibility that the host can pick the box with the car, you have basically changed Monty Fall into Monty Hall. There is no difference - either way the host can *never* select the box with the car! In Monty Fall as described, it was a possibilty that just didn't happen. That matters.
LintMan on June 23, 2009 9:37 AMI started to look through Pickle's code to see where he went wrong, but Celejar's post convinced me writing my own simple Monty Hall simulator would be more fun. So here's Yet-Another-Monty Hall Sim, this time in Python!
-----
# A simple simulator for the Monty Hall problem
from random import randrange
def MontyHallTest( rounds ):
winsIfSwitch = winsIfKeep = 0
for i in range( rounds ):
prizeDoor = randrange(3) # pick an integer from [0,1,2]
pickedDoor = randrange(3) # pick another integer from [0,1,2]
randomOffset = randrange(-1,1,2) # pick +1 or -1 at random
# modulo arithmetic makes sure our door stays in range
excludedDoor = (pickedDoor + randomOffset) % 3
# can't exclude the prizeDoor, exclude the other one
if excludedDoor == prizeDoor:
excludedDoor = (pickedDoor - randomOffset) % 3
# we had the winner, so keeping it wins
if pickedDoor == prizeDoor:
winsIfKeep += 1
# we didn't have the winner, so switching wins
else:
winsIfSwitch += 1
return [100 * winsIfKeep / rounds, 100 * winsIfSwitch / rounds ]
-----
What I thought was really interesting about doing the exercise of writing the code for the pseudo-empirical (since the number generator is only pseudo random) experiment is how clearly the excluded door falls out of the problem. As you can see in the code, I figure out which door Monty can exclude... but it doesn't matter. I tried my darndest to make it matter, but it just wouldn't! You'd get the same results if you took out all the lines involving the excluded door.
If I picked a winner before, I still have a winner. If I picked loser before, I still have a loser. P(Picked a loser the first time) = 2/3 and P(X=Picked a winner the first time) = 1/3
The only way you can get to 50/50 is throw away your choice and chose again randomly. But why take 50/50 when you can exploit the information you have to get a 2:3 chance of winning?
DM on June 23, 2009 9:47 AM@Logo, that is what I would do... play the game based on expected value, but they would never pick me for the show. I have tried to get on a few game shows, and my personality isn't "fun" enough - that's what they told me at Wheel of Fortune... they want people who will jump up and down and giggle a lot.
Jasmine on June 23, 2009 9:49 AM@Trevel - there was another bug in the simulation. I was allowing the host to sometimes pick the same door as the contestant! Now I get 50/50 again.
What I don't understand is this. If I am playing the game, and I am asked "do you want to switch?", then what we're saying is I haven't got enough information, until I find out whether he *meant* to open a non-prize door. Is that right? Seems weird. I mean, in the 1000-door scenario, if he were to slip and accidentally open 998 doors, leaving mine and one other, is it really still 50/50? Surely after all those doors are opening my original door still has a 1/1000 chance of being corrent?
Zelda on June 23, 2009 9:49 AM...and the comment system ate my indents. Sorry about that, you'll have to add them back in if you want to try my simulator (you win this round, Perl, with your not-getting-eaten by the comment system... you win this round... :P)
DM on June 23, 2009 9:49 AMActually after thinking about it for a while, my own 1/1000 example has clarified it for me. 50/50 it is.
Zelda on June 23, 2009 9:53 AMActually thinking about it, 50/50 does make sense to me in the Fall scenario now. Thanks for helping me hash it out!
Zelda on June 23, 2009 9:54 AM@Celejar
You need to discard the cases where he opens the door with the prize in it, as well.
Trevel on June 23, 2009 9:54 AM@Zelda
Your original door has a 1/1000 chance of being correct, yes. But the other door ALSO has a 1/1000 chance of being correct -- you're swapping 0.1% odds for 0.1% odds, which are even. In the Hall scenario, you're essentially swapping for whichever of the other n-1 doors is best, which means you're going from 1/n to n-1/n, which is an advantage; in the Monty Fall scenario, you're going from 1/n to 1/n, which are equal.
That said, as long as you know he's not offering it maliciously (as he did in actuality, specifically offering when they already picked the car), switching is always the better choice. After all, if the worst case is an even swap -- why not?
Trevel on June 23, 2009 10:00 AMMy initial analysis didn't take into probabilities at all. Not understanding that this was the rules of the gameshow, I thought the guy already knew I had picked the car and so was trying to trick me.
Anonymous on June 23, 2009 10:01 AMThe way that I finally came to understand this is the following game:
Suppose there are 100 doors - 99 hide a goat, 1 hides a fancy pants car. The game show contestant chooses the game, and then the game show host opens 98 doors, revealing goats. Now, the host picks you, the humble reader, and asks you to guess who has the right door. Who do you guess? One of the two doors is right. 1 out of 100 times, the game show contestant will have picked the right door. If the game show contestant doesn't pick the right door, however, that means that the game show host has picked the correct door - which happens 1 - 1/100 = 99/100 times.
Once you wrap your head around this guy, the three door case makes sense.
Martin on June 23, 2009 10:42 AM@LintMan:
"If you eliminate the possibility that the host can pick the box with the car, you have basically changed Monty Fall into Monty Hall. There is no difference - either way the host can *never* select the box with the car! "
This is true, except that I didn't change anything. In Monty Fall as described, the host *does* reveal a goat. The question posed is - host stumbles, accidentally reveals a goat, now do you switch or not?
It *is* exactly equivalent to Monty Hall.
"In Monty Fall as described, it was a possibilty that just didn't happen. That matters."
That would be true if the switch/no switch decision has to be made before he stumbles. But that's not what was described.
John H on June 23, 2009 10:42 AMThe real answer is that there is not car, only goats. It's all a scam. Rigged!
Anonymous on June 23, 2009 10:44 AMQ: If I were to flip a coin 999 times and have it land heads every time, what are the chances of heads the next flip?
I always liked to jokingly answer 100% it's either a heavily biased coin or it has heads on both sides.
Now that I understand the Monty Hall problem, this looks more correct than I first imagined. When the other doors are revealed, the numerator part of the odds goes up for each remaining door instead of what I first assumed, the denominator going down.
Can I now put the coin toss argument to rest by saying that the odds of the next flip are biased by the previous results which I personally just witnessed? I mean, did I kill Schrödinger's cat after all?
Just that one niggling issue which I can't quite shake is this... If I build a Monty simulation just like everyone else, I accidently bias the outcome just like everyone else by doing something like the following:
while (reveal == choice || reveal == prize)
{
reveal = rand() % door_count;
}
Or I could just rename those variables to something like…
while ( my_dice_roll_gives_undesirable_results(dice_roll) )
{ dice_roll = perfectly_random_result(); }
Now that computer simulation does indeed mimic the (real) story behind the problem. However, I can just as easily tell my computer to just pick a single row from my possible outcome table. No simulation required here, I'm rolling my n-sided dice only one time.
I don't know where I'm going with this to be honest, I'm obviously not comprehending one little piece of the puzzle. I know with more coin tosses / dice rolls, my statistics form a lovely bell curve just like any self respecting polynomial should.
Somehow I get this feeling that we all just divided by zero metaphorically speaking and we all agree that that's the correct thing to do.
Note to self. Time for a beer. Lets go shopping. Monty Burns is my hero.
Frogger on June 23, 2009 10:54 AMYou have one prize and two doors. Clearly 50/50 chance to win.
I will now add another door which does not have prize.
Are you chances of winning 1 in 3?
I now give you the opportunity to change your choice, but NOT to the new door. You have a 1 in 2 choice, but I will measure your probability as if it were 1 in 3.
-------
If I change the problem by adding information why should I expect the probability to remain the same as the original problem?
No seriously. We're done here. Go home.
Frogger on June 23, 2009 10:56 AMIn the election of signal to noise, I voted noise.
Frogger on June 23, 2009 10:57 AMIf he had accidentally fallen through the door with the car, am I allowed to now choose that one? Does it matter whether the door with the car he fell through was already the one I had chosen first?
Gary on June 23, 2009 10:58 AMThink about it like this:
You pick door 1. The chance is 1/3 that the car is behind door 1, and chance is 2/3 the car is behind door 2 or 3. The host removes door 3. The chance that the car is behind door 2 or 3 is still 2/3. But door 3 is removed. So the chance that the car is behind door 2 is 2/3.
alwin on June 23, 2009 10:59 AM@alwin
That explanation makes Monty Hall and Monty Fall the same, which everyone is disputing.
Gary on June 23, 2009 11:10 AMJeff,
So let me get this straight, all I gotta do is quote other people and that allows me to make money out of useless blog posts?
Joe on June 23, 2009 11:27 AMMonty Hall is only counter intuitive to people who haven't had basic college level math -- normal people, I guess. It's a super easy Bayes problem otherwise. I don't know why Erdos had a problem with it, but to most mathematicians it is indeed intuitive. And to normal people, simply imagining the example with a million doors usually sets them on the right path.
And Jeff's original Boy/Girl problem is indeed too vaguely specified to answer. The reason we can answer Monty Hall is because we know how the show worked -- ie, we know that the host never reveals the prize.
Anonymous on June 23, 2009 11:28 AMMonty Hall is only counter intuitive to people who haven't had basic college level math -- normal people, I guess. It's a super easy Bayes problem otherwise. I don't know why Erdos had a problem with it, but to most mathematicians it is indeed intuitive. And to normal people, simply imagining the example with a million doors usually sets them on the right path.
And Jeff's original Boy/Girl problem is indeed too vaguely specified to answer. The reason we can answer Monty Hall is because we know how the show worked -- ie, we know that the host never reveals the prize.
Finn on June 23, 2009 11:28 AMHint: Monty Hall never revealed a prize on the show. They did not edit out (throw out) the episodes where this happened.
Anonymous on June 23, 2009 11:40 AM@Gary
My explanation is only for understanding Monty Hall, which is difficult enough for many people... It wasn't a reply to you.
For a long time, the only thing I *did* understand about Monty Hall was that the host knows where the car is, and that he intentionally removes a door without the car.
alwin on June 23, 2009 11:45 AMJC - thanks, I get it now.
ronnie on June 23, 2009 11:58 AMHere is my Monty Fall simulator in Ruby. It may help make sense why the probability is 50%.
def monty_fall(guess, switch)
srand()
car = 1 + rand(3) # put the car behind a random door
srand()
opened_door = 1 + rand(3) # door is opened randomly
if car == opened_door || guess == opened_door
return 0.5
elsif car == guess && switch
return 0
elsif car != guess && switch
return 1
elsif car == guess && !switch
return 1
else # car != guess && !switch
return 0
end
end
You can run the simulation like this:
good_guesses = 0
1.upto(10000) do |i|
srand()
good_guesses += monty_fall(1 + rand(3), ARGV[0] == 'true' ? true : false)
end
puts good_guesses
Mischa on June 23, 2009 11:58 AMThese are ill-posed questions, and the answers are unintuitive only because they give you information but don't indicate whether you're supposed to use it or not.
If I get back a positive cancer test result, I'm immediately put into a different group than someone who hasn't had a test yet. I'm going to get my affairs in order. It means that now I have an 80% chance of having cancer. Ignoring what the test result tells you is nonsense.
Monty could be evil, angelic, stupid, smart, tired, completely random, whatever. It's not always 2/3rds in favor of switching. You have one example of the sample space and try to make a generalization.
Enough already! Talk about Swoopo some more.
Fine. Make it a well defined question:
Suppose the contestants on a game show are given the choice of three doors: behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, *must always* open one of the unchosen doors, which will always contain a goat. He then asks the contestant, "Do you want to switch doors?"
Tim on June 23, 2009 12:02 PMCan we have another post about something interesting? Like stack overflow statistics or something?
Practicality on June 23, 2009 12:18 PMIt really doesn't fail Monty Fall. The problem there is that the host had an equal chance of opening the car door, whether or not it was your door, too. If you picked the car the first time, he has a 2/3 chance of doing that. If you didn't pick the car, then he has a 1/3 chance of doing that. Thus, the fact that he accidentally managed to open a goat door and therefore did NOT ruin the game adds even more information to the player, telling him that he's more likely to have the right door picked in exactly the same proportion as he was unlikely to get it right in the first place.
The two probability events cancel out perfectly, and you end up with a 50/50 chance. The "switch chooses all other doors" rationale does not break down here.
@Gary on June 23, 2009 1:05 PMIt's really a mathematical sleight of hand. If you have 3 choices, and choose one that is wrong, are then offered a new chance, it is really BOTH answers that are correct.
case 1. you have a 2/3rds chance of winning because the 1/3 "losing" choice has been eliminated. switching makes sense.
case 2. you NOW have a 50/50 chance of winning. It is physically and mathematically impossible for it not to be 50/50. however....
if your first choice lost, of course you have better odds of winning by switching. why would you keep the losing choice? here is where the 2/3rds comes in.
Most things in probability or statistics are "provable bullshit". It works, but does not make sense to the brain because the rules are ambiguous, chaotic, or random, and often have several answers depending on context. No wonder most people avoid maths. They suck.
El Guapo on June 23, 2009 1:13 PMI wrote up a little Monty Hall computer simulator in C# back in September 2008, which empirically shows the benefits of switching doors:
http://scottonwriting.net/sowblog/posts/13534.aspx
Although, to me, the most clear explanation is drawing out the probability tree and summing up the probabilities:
http://en.wikipedia.org/wiki/File:Monty_tree_door1.svg
@Mischa
"if car == opened_door || guess == opened_door
return 0.5"
Can you explain to me why car == opened_door would return 0.5?
Gary on June 24, 2009 2:44 AM@Mischa
Yes, what Gary said.
I would have thought:
if car == opened_door
return 1
-- if Monty reveals the car, and I'm now allowed to switch, I'm going to pick the door that I know contains the car!
HOWEVER, the original description of Monty Fall specifically states that car != opened_door: "the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car"
John H on June 24, 2009 3:14 AMOkay, I've got a new question for you Monty Fall lovers:
A man rolls a 3 on a die. What's the probability that he rolled a 6?
@confuse-us
"while ( my_dice_roll_gives_undesirable_results(dice_roll) )
{ dice_roll = perfectly_random_result(); }"
this actually sums up my response to a number of the Monty Fall responses here.
"the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car" -- yet some contend that you can't discard results where he pushes open a different door.
I just rolled a 6. What's the probability that I rolled a 3? ZERO!
All the possible other numbers I might have rolled become irrelevant as soon as the result is announced.
Ugh, in case it's not obvious, Gary and I are sat next to each other having a conversation, but independantly posting comments here. Sorry about the dupes!
John H on June 24, 2009 4:33 AMI didn't believe it. I had to write a program to see that it actually is so. Indeed, the odds of winning go from 33.(3)% to 66.(6)% simply by switching! The answer to why it is so begins to be simple once you understand how it really works.
Here's a short piece of the output of the program; Legend:
The "100" is the initial choice of the guest.
The "1" is the choice where the prize is.
A value of "101" is where the guest managed to guess the prize in the first try.
A value of "-100" depicts the door that is shown to the guest after the choice.
Arrays' indices begin at 0.
"-I" at the end shows an Initially correct choice. too bad we switch.
"---W" at the end means we switched into a correct choice.
Arr:[1,-100,100] switching from [2] to [0]---W
Arr:[100,1,-100] switching from [0] to [1]---W
Arr:[-100,0,101] switching from [2] to [1]-I
Arr:[-100,100,1] switching from [1] to [2]---W
Arr:[1,-100,100] switching from [2] to [0]---W
Arr:[-100,1,100] switching from [2] to [1]---W
Arr:[101,-100,0] switching from [0] to [2]-I
Arr:[100,-100,1] switching from [0] to [2]---W
Arr:[-100,101,0] switching from [1] to [2]-I
Arr:[100,-100,1] switching from [0] to [2]---W
Arr:[100,-100,1] switching from [0] to [2]---W
Arr:[0,101,-100] switching from [1] to [0]-I
Arr:[-100,101,0] switching from [1] to [2]-I
Arr:[100,1,-100] switching from [0] to [1]---W
Arr:[1,-100,100] switching from [2] to [0]---W
Arr:[100,-100,1] switching from [0] to [2]---W
Arr:[-100,1,100] switching from [2] to [1]---W
Arr:[0,101,-100] switching from [1] to [0]-I
Arr:[-100,100,1] switching from [1] to [2]---W
Arr:[-100,101,0] switching from [1] to [2]-I
Arr:[1,100,-100] switching from [1] to [0]---W
Arr:[-100,1,100] switching from [2] to [1]---W
Arr:[100,1,-100] switching from [0] to [1]---W
Arr:[1,100,-100] switching from [1] to [0]---W
Arr:[-100,1,100] switching from [2] to [1]---W
Arr:[100,1,-100] switching from [0] to [1]---W
Arr:[-100,1,100] switching from [2] to [1]---W
Arr:[1,-100,100] switching from [2] to [0]---W
Arr:[0,101,-100] switching from [1] to [0]-I
Arr:[101,-100,0] switching from [0] to [2]-I
Arr:[100,1,-100] switching from [0] to [1]---W
Success rate with changing:0.709677 (form 30 items).
For large groups should tend towards 0.66
What does this show? That we have 33% chance to guess initially.
The other 2 choices represent 66%. One of them is eliminated, it means that the other choice still has the value of 66%.
Or simpler put. We aren't shown another door really.
What the host is actually asking us is this:
Do you want to keep your one choice of 33%, or take the two other choices of 66%? (Since he will always show you one).
Monty Fall state table:
http://spreadsheets.google.com/ccc?key=r7A7ACY3f2G5ZbNon8UjUDA
Takes into account various interpretations of the rules. Turns out you get 50/50 if you declare games where the host's stumble reveals a car, void.
John H on June 24, 2009 5:30 AM@Secure
"If you stick to this door, then it DOESN'T MATTER what happens next, it will remain on 1/3."
Err, so I pick a door, Monty opens both doors shoing a goat and a car, and I still have a 1/3 chance of winning - cool.
Steve on June 24, 2009 6:09 AMIt took me about a minute of reading vos Savants initial column post (before she explained it) to see the validity of her statement and how it was correct. The key absolutely is the foreknowledge on the host's part.
The host must pick the door without the car, with no restrictions if neither door contains the car. You pick a door, with a 1/3 chance. There is then a 2/3 chance that the host HAS to pick a certain door, meaning there is a 2/3 chance that the door the host doesn't pick has the car behind it. The foreknowledge is an additional constraint to the problem that affects the outcome.
workmad3 on June 24, 2009 6:29 AM@Trevel: Yes, after a good night's sleep I realized what I had done wrong. Replacing
next if ($open == $door);
with
next if ($open == $door || $open == $prize);
does indeed result in parity.
Celejar on June 24, 2009 6:39 AM@Steve,
run the test 1000 times, with Monty opening both doors. Do you think you will win any other number than 333 cars (on average)? If yes, which number -- and why?
After you've made the choice, any following action will only show if you've made the right or the wrong choice, i.e. if you are in the right 1/3 or the wrong 2/3 part of the probability -- but it won't change the probability IN THE EXACT MOMENT you've made the choice.
Secure on June 24, 2009 7:28 AMHere is a chart, with ALL the possible options for 3 doors.
http://spreadsheets.google.com/ccc?key=rdo_jMEwKLSVAtYmwajgIyw
Augustin on June 24, 2009 7:40 AMMonty's law: every year or so an A-list blogger will mention my problem, causing an avelanche of millions of stupid replies.
Hope you learned a lesson, Jeff.
Wim on June 24, 2009 7:53 AM"Hope you learned a lesson, Jeff."
Yes. He learned a lesson a long time ago. If you want to increase traffic to your site and increase your potential revenue for ads, post about the Monty Hall problem.
I think I'll create a site specifically dedicated to the Monty Hall problem. And then I'll rake in the moola from all of the idiots who keep arguing the problem into the ground when it has already been done 100 times before.
Matt on June 24, 2009 8:04 AMThat's all very well, but Monty Fall and Monty Crawl are new to most of us.
John H on June 24, 2009 8:06 AM@max: That was hilarious.
@Jeff (Atwood): If nothing else this entry demonstrates the need for a better comments system. A programming blog needs a good way to put codes in the comments. Always pastebin I guess.
Bernard on June 24, 2009 8:06 AM@secure
"run the test 1000 times, with Monty opening both doors. Do you think you will win any other number than 333 cars (on average)? If yes, which number -- and why?"
Before selecting a door, the probability you will chose the car is 1/3.
Once you've selected a door the probability is either 1 or 0, and this will not change.
But, as you don't know if you have coden the car or not there is an inferred probability based upon your knowledge, and this will change as dooors are opened.
Your post is trying to have it both ways, in talking about a 1/3 chance that does not change regardless of the what new information you are provided with. The inferred probability, which is what all these Monty problems is about will change, the actual probability will not change, but is 1 or 0, not 1/3.
Steve on June 24, 2009 8:18 AM"Yes. He learned a lesson a long time ago. If you want to increase traffic to your site and increase your potential revenue for ads, post about the Monty Hall problem"
My guess is that the next post will be on Zeno's paradox, or that one about a plane on a conveyor belt.
Steve on June 24, 2009 8:25 AMHe, I never heard about the plane on the conveyor belt. Thanks! :-) But it is a silly problem, only misleading people who don't grasp the elementary physics of plane lift.
Zeno's paradox is of course nonsense. Apply it to TCP packets, and you could not be reading this!
Anonymous on June 24, 2009 8:40 AM@Gary
Because according to the problem statement those two cases (accidentally opening the chosen door or the door with the car) have to be discarded. Be returning a value of 0.5 those cases are discarded.
The code below is essentially the same, only it makes it clearer that those cases are discarded:
def monty_fall(guess, switch)
srand()
car = 1 + rand(3) # put the car behind a random door
srand()
opened_door = 1 + rand(3) # door is opened randomly
discard = 0
success = 0
if car == opened_door || guess == opened_door
discard = 1
elsif car != guess && switch
success = 1
elsif car == guess && !switch
success = 1
end
return [discard, success]
end
And then you can do:
discard = 0
success = 0
1.upto(10000) do |i|
srand()
result = monty_fall(1, switch)
discard += result[0]
success += result[1]
end
puts success.to_f / (runs - discard).to_f
Mischa on June 24, 2009 9:17 AM@Steve
"Before selecting a door, the probability you will chose the car is 1/3.
Once you've selected a door the probability is either 1 or 0, and this will not change."
You're talking about two different probabilities here. The second doesn't substitute the first, as you seem to argue, but it is correlated to the first: 1/3 of the times you get a 1, 2/3 of the times you get a 0.
"But, as you don't know if you have coden the car or not there is an inferred probability based upon your knowledge, and this will change as dooors are opened."
So you are argueing that when running 1000 tests with
a) Chose a door, both other doors kept close, open the selected door
b) Chose a door, both other doors are opened, open the selected door
then the number of cars you win in a) are different from the number of cars you win in b)? Unfortunately you did avoid my question: How many cars will you win in a), how many will you win in b)? Please give hard absolute numbers, anything else is theoretical blahblah. Here are my numbers: 333 for both.
Hey forget the math problem...I wanna know why there are goats on the show.
Kit on June 24, 2009 9:34 AMWhen I first heard about plane on a conveyor belt (on Mythbusters), I initially thought that it wouldn't move. But then they did something that made me realize that I had neglected something in my initial reasoning: They held the plane still (they used a model plane on a treadmill for a small-scale demonstration).
This made me realize that a plane's wheels are connected to any sort of axle or gearshaft and are actually allowed to freely rotate. So any motion of the conveyor belt would just cause the wheels to spin faster and not have any affect on the plane at all. Once you realize that the reason a plane has wheels isn't locomotion but because overcoming the static friction of a plane resting on the ground is a lot harder than the rolling friction of a wheel.
BA on June 24, 2009 9:40 AMI meant to say that "a plane's wheels are _not_ connected"
BA on June 24, 2009 9:41 AMIt seems to me that the *only* reason one should switch doors based on the host's selection is that the host's selection is *not random*. If it was (if he didn't know what was behind each door) and yet he opened a door to reveal a goat, it still wouldn't make any difference to switch.
Brad on June 24, 2009 10:52 AMThis answer by xebecs on June 22, 2009 8:01 AM
EXPLAINS IT FULLY AND COMPLETELY. I can't see how Charles or anyone else missed this explanation. Simpe and succint.
-------------- REPOST -----------------
It took me a long time to figure it out (after being told I was wrong the first time) but I finally decided this was a good way to think about it:
The key is that the host CAN NOT open the same door you chose. His or her hand is forced if you chose wrong.
2/3 of the time, your initial choice will be wrong, so 2/3 of the time the host's choice is forced.
In those cases, the host MUST open the other wrong door, leaving the third door as the right one. So, 2/3 of the time, the switch will be beneficial.
-------------- THANKS XEBECS -----------------
Nari on June 24, 2009 11:09 AM@Steve,
"Oh sorry, the answer to your question about how many cars I'd win in 1000 guesses is obviously 333 1/3."
"For example in the Monty Fall problem there is only two doors, one of which contains a car, but there is now a probability of 1/2 that your door has the car."
Beside that we are not talking about Monty Fall, and beside that Monty Fall is identical to Monty Hall (read and understand the descriptions -- one of the main problems here in this thread), don't you see the contraction in your own statements? For a probability of 1/2, I would win 500 cars in 1000 guesses.
Secure on June 24, 2009 12:00 PMs/contraction/contradiction/
Secure on June 24, 2009 12:03 PMLookit, when you first picked the door, you had 1/3 chance of getting the car. Agreed? Now close your eyes. While you are not looking, maybe somebody opens another door, maybe they don't. Still 1 in 3, because nobody changed what was behind your door.
If you have the chance to change your selection to, in effect, *ALL* of the doors you didn't initially pick, that's a 2/3 chance of containing the car. Regardless of whether any of those doors opened or closed while you weren't looking.
So when Monte opens a door and reveals a goat, all he's doing is presenting you with the opportunity to switch your selection from the ONE door you already chose, to instead selecting BOTH of the unchosen doors. Which is a no-brainer, since the not-originally-chosen two doors has a 2/3 chance of having the car.
Duh.
Larry on June 24, 2009 12:47 PM@secure
"Unfortunately you did avoid my question:"
Oh sorry, the answer to your question about how many cars I'd win in 1000 guesses is obviously 333 1/3. But that wasn't what your first post was saying. You were saying that having made the choice, the odds wouldn't change however many doors were opened, and so the probability of there being a car behind the door would always be 1/3. From this you made the eronious claim that,
"So, if there is one other door left, and behind one of the two doors (your original choice and the other) there definitely is a car, the total probability for a car MUST add up to 1. Now do the math, and no, 1/3 + 1/2 is NOT 1."
For example in the Monty Fall problem there is only two doors, one of which contains a car, but there is now a probability of 1/2 that your door has the car.
Steve on June 24, 2009 1:09 PMSuch a nice thread, I really want to add to the total confusion. The key to understanding is to not concentrate on the other door, but to the original choice.
When chosing 1 out of 3 doors, and behind 1 of these is a car, I think everybody with a basic understanding of math agrees that you have a chance of 1/3 to get the car. If you stick to this door, then it DOESN'T MATTER what happens next, it will remain on 1/3. Except for the trivial case that the thing behind the door is removed, it doesn't matter if
- No other door is opened.
- Both other doors are opened.
- A million doors with goats are added.
- A million doors with cars are added.
- Monty slips on a banana peel, causing a microexplosion of matter and antimatter when hitting the floor.
- Jeff finds a prove for P=NP, but writes such a confusing blog post about it that nobody believes him.
No matter what, if you stick to your initial door of choice, you still have a chance of 1/3 to get the car. How should it change as long as this door is not affected by the following actions?
So, if there is one other door left, and behind one of the two doors (your original choice and the other) there definitely is a car, the total probability for a car MUST add up to 1. Now do the math, and no, 1/3 + 1/2 is NOT 1.
Let me give some pseudocode for it:
Case 1: immediately open the door
Value = Random(1..3)
if Value = 2 then print "You won!"
Case 2: Wait for a looooong time before opening the door
Value = Random(1..3)
...
// Many actions here that DON'T change Value
...
if Value = 2 then print "You won!"
When you run this 99,999 times, how many times (on average) will you get "You won!" in case 1 and in case 2 (modification of the random seed does not count)?
Secure on June 24, 2009 1:58 PM@secure
"Beside that we are not talking about Monty Fall, and beside that Monty Fall is identical to Monty Hall"
Yes you were, and no it isn't.
"(read and understand the descriptions -- one of the main problems here in this thread)"
Have _you_ read and understood the pdf.
"don't you see the contraction in your own statements? For a probability of 1/2, I would win 500 cars in 1000 guesses."
No. You don't win 500 cars in a 1000, you win 333 1/3 cars in 666 2/3 guesses.
Steve on June 25, 2009 2:12 AM@Steve,
now at least I know where all the confusion comes from. You, and the document, are talking about two completely different probabilities. I'm talking about the probability for FULL RUNS of the game:
Select a door -- some action happens -- open the selected door (NO CHANGE)
A probability of 1/2 means 500 cars in 1000 runs, 1/3 means 333 cars in 1000 runs.
"No. You don't win 500 cars in a 1000, you win 333 1/3 cars in 666 2/3 guesses."
And I don't even want to know the confused math behind this statement. What's with the remaining 333 1/3 guesses, where are they in the probability calculation? Schroedinger guesses, car and no car at the same time? Well, you can't know if you win the car until you open the door, better keep it closed and you have a probability of 1 for a win.
Secure on June 25, 2009 4:00 AM@Secure
"I'm talking about the probability for FULL RUNS of the game:
Select a door -- some action happens -- open the selected door (NO CHANGE)"
That's a trivial problem - always 1/3. The whole point of the problem is how the odds change given certain information after you've mad your selection - and hence if you should switch.
"And I don't even want to know the confused math behind this statement. What's with the remaining 333 1/3 guesses, where are they in the probability calculation?"
OK, then I won't tell you, Rosenthal document explains it better than I will in any event.
However, for the benefit of anyone else mad enough to still be reading this - the remaining third of the guesses are those where the random door that is opened reveals the car. This possibility is not part of the Monty Fall problem, so has to be excluded from the set of possible outcomes.
Jeff - "I think what they're really objecting to is how unintuitive the answer is."
That's is a bizarre interpretation given the discussions in the comments. The question was roundly condemned as being unclear by mathematicians and academics. Maybe you only skimmed the first couple of comments???
Reading the comments makes it entirely clear how the sentence could have been phrased to avoid any ambiguity.
Philip on June 25, 2009 6:16 AM@Steve,
last try, I promise. ;)
"That's a trivial problem - always 1/3. The whole point of the problem is how the odds change given certain information after you've mad your selection - and hence if you should switch."
Whoa, you did it again -- contradicting yourself. You say the odds for my originally chosen door are trivially always 1/3. Then you tell me that the odds will change. This includes a change for the odds of the originally chosen door, because the probability always adds up to 1. So the odds for my door are ALWAYS 1/3 -- except when they change. Or what?
"where the random door that is opened reveals the car. This possibility is not part of the Monty Fall problem, so has to be excluded from the set of possible outcomes."
So Rosenthal is just as confused, which comes at no surprise given the controverse discussion about this problem. He gives a definition of a problem (Monty Fall: "...accidentally pushes open ANOTHER door, which just happens NOT to contain the car..."), and then calculates the probability for a complete different problem, which you state as the solution for the original Monty Fall problem -- which it isn't, as you said yourself.
"Whoa, you did it again -- contradicting yourself. You say the odds for my originally chosen door are trivially always 1/3. Then you tell me that the odds will change."
No contradiction. The odds of chosing the door with the car are 1/3. Once you have chosen the door the odds change in the light of new knowledge. (As I pointed out before these are the inferred odds, not the actual odds that are always 1 or 0).
You claim that the inferred odds do not change regardless of which doors are opened. This is wrong.
For example, if one of the two remaining doors is chosen at random and opened - if it shows a goat the chances that your door now contains the car has increased to 1/2. This seems confusing, but it is more clear if you consider the case when the random door is opened to show a car. Now do you think you still have a 1/3 chance of a win?
And if both doors are opened, to show two goats do you think there is still a 1/3 chance of you having the car? What if Monty now offers you half the value of the car rather than whatever is behind your door. Would that be a good deal?
Steve on June 25, 2009 7:00 AMPeople have trouble with this because in grade school they make a mistake in chain probabilities (two coinflips should be independent) and overgeneralize. Now they treat the door choice as the event - coinflip. No. The car and goats are not reshuffled after Monty opens a door. If they were, that would be the 50% independent event people confuse this for.
Serious on June 25, 2009 7:31 AM@Steve,
OK, promise broken, but now I have a very last question to clarify:
"if one of the two remaining doors is chosen at random and opened - if it shows a goat the chances that your door now contains the car has increased to 1/2"
You really want to tell me that the INTENTION makes the difference? For the very SAME ACTION (opening one of the two other doors and revealing a goat), the probability for your selected door is 1/3 when the action happened by will and 1/2 when the action happened by accident?
Do you REALLY want to tell me THIS?
How does it happen? Psychokinetic influence of mathmatic rules?
Secure on June 25, 2009 8:15 AMThis probably too late for anybody to see but here's another take.
Let's say there are 26 doors. Chances of you picking the right one is 1 in 26. Monty opens 24 of the doors showing goats. He always does this. It will always be your door and one other. Do the odds still say 50/50. I would switch instantly since Monty narrowed it down for me.
I would lose only once in 26 times on average. That would be when I picked the right door. The other 25 times I would win, as an average.
So whether it is 3 doors or 300, it makes sense to switch.
Rich S. on June 25, 2009 8:49 AM@Steve,
Finally, I've simulated it -- it was about time. 1/2, I have to say. Intention makes a difference, it's psychokinetic influence. ;)
I was right in the sense that when sticking to the selected door, even when another door with a car was opened, the probability is 1/3. But when these cases are removed, then of course it will change the overall statistics, as a simple truth table will show (Player always choses Door1 and sticks to it).
Door1 - Door2 - Door3 - Opened
Car - Goat - Goat - Door2 - Win
Car - Goat - Goat - Door3 - Win
Goat - Car - Goat - Door2 - Removed
Goat - Car - Goat - Door3 - Lose
Goat - Goat - Car - Door2 - Lose
Goat - Goat - Car - Door3 - Removed
The trick is that the random selection of the door is part of the game; it influences the distribution, because all 6 combinations are equally likely with 1/6.
In the normal game, a door with a goat is always opened, thus there is no randomness and the door opening does not influence the distribution of states.
Door1 - Door2 - Door3 - Opened
Car - Goat - Goat - Door2 or 3 - Win
Goat - Car - Goat - Door3 - Lose
Goat - Goat - Car - Door2 - Lose
Only three different states with a probability of 1/3 are possible, the first state is not partitioned into two states. It can be done, but then the probability of the state is partitioned, too, given that Monty opens one of both doors at random:
Door1 - Door2 - Door3 - Opened
Car - Goat - Goat - Door2 - Win (1/6)
Car - Goat - Goat - Door3 - Win (1/6)
Goat - Car - Goat - Door3 - Lose (1/3)
Goat - Goat - Car - Door2 - Lose (1/3)
Now everything is clear and easy. Thanks for a very inspiring discussion with a happy end. ;)
Secure on June 25, 2009 9:38 AMForget about the montey hall problem, that's a peice of piss. What I'm angry about is that Jeff still won't admit he fucked up when he asked the boy/girl problem. His explanation for why people are pointing this out is bullshit. I'd respect him more if he was willing to admit a mistake, but this is just annoying and smug.
Breton on June 25, 2009 11:51 AMI love listening to the whining of the many so called "intelligent" people who were simply wrong.
It's very easy to attack and criticise the semantics of the question. How about being honest with yourself and realising that you can actually (shock! horror!) be wrong.
It's painfully obvious that most of the angry, dismissive responses here are just face saving, pure and simple.
Also, can I Just point, again, to the link Jason M posted to the High Priest and Primary Advocate, the go to guy for anything Bayesian Statistics, demonstrating how the wording that jeff used is mangled and wrong. Are you going to tell him that he's only saying that because he doesn't want to admit that beyesian statistics is too unintuitive for him to understand? I mean, how much more do you need to understand what happened here?
Breton on June 26, 2009 2:11 AM"It's very easy to attack and criticise the semantics of the question. How about being honest with yourself and realising that you can actually (shock! horror!) be wrong."
I'm not sure who you're actually directing this at, but if it's regard with the boy/girl problem- Properly phrased, as in the original source material that he derived the original question, it is easy to arrive at the same answer that Jeff did. When the question is mutilated, and presented in the way that jeff ACTUALLY presented the question, you arrive at a different solution than that which was intended. Jeff screwed up the wording of the problem, simple as that. We didn't get the solution wrong, he got the question wrong and won't admit it.
I don't care a wit about face saving. I just hate smug idiocy.
Not that any of it actually matters. It is just the internet after all.
Breton on June 26, 2009 1:54 PMI find it hard to believe that so many mathematicians would so vehemently protect a stance they haven't proved. Any intuition about conditional probability even would direct one in the right direction.
The reasoning is simple. You have a 2/3 chance of getting it wrong in the first place, so changing will result in a correct choice two times versus one.
Andre on June 27, 2009 11:02 AMHard to believe
sexy costumes on June 28, 2009 5:28 AMI created a learning object to help my statistics students (Year 2 university) to better understand the probabilities underlying the Monty Hall Dilemma. The learning object gives you a chance to play both the classic three-door game and a modified 20-door version. It also provides a set-based explanation of the probabilities for each version. In the three-door game, you really are twice as like to win by switching, and in the 20-door game, you are 19 times as likely to win by switching. The learning object is quite easy to use, and my students tell me it's actually rather fun. Give it a try!
David on June 28, 2009 6:11 AMThe way in which I understood the Monty Hall problem easiest was to examine starting conditions.
You have a 33% chance of choosing the car at the outset. If this is the case and you swap after the host reveals a goat, you choose a goat.
You have a 66% chance of choosing a goat at the outset. If this is the case and you swap after the host reveals a goat, you choose the car.
Andrew Care on June 28, 2009 9:53 AMI was thinking for a while on the same puzzle, but now, I am happy that so many great minds have been tricked by this puzzle.
OFF:
"They just debond it from carbon"
Nope, the oxygen is released from the water.
E.g.:
http://en.wikipedia.org/wiki/Water_splitting
http://en.wikipedia.org/wiki/Photodissociation
I was thinking for a while on the same puzzle, but now, I am happy that so many great minds have been tricked by this puzzle.
OFF:
"They just debond it from carbon"
Nope, the oxygen is released from the water.
E.g.:
http://en.wikipedia.org/wiki/Water_splitting
http://en.wikipedia.org/wiki/Photodissociation
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