50% vs 66% -or- B-G vs G-B

People are confusing Events with Outcomes!!!

An Event is a set of related Outcomes. An Outcome is, simply put, all that one can reasonably measure and/or record

about an (ideally instantaneous) observation.

For example, consider the Events on the roll of one standard six sided die:

- Event Even is Outcome of 2, 4, or 6

- Event Odd is Outcome of 1, 3, or 5

- Event Factor-of-3 is Outcome of 3 or 6

- Event One is Outcome of 1

There is also the special Event called the Sample Space that is the set of all possible Outcomes. The Sample Space

for the one die roll is of course 1, 2, 3, 4, 5, or 6.

Note: An Event CANNOT contain an Outcome that is NOT contained in the Sample Space! This is because, by definition,

the Sample Space must contain all possible Outcomes for a given context.

An Event is said to occur when the Outcome one observes matches one of the Outcomes that defines the Event.

(Ideally, the match should be exact, but in real life the match can be close enough.) And, yes, it is possible for

one to define Events such that more than one occurs within a single (instant) observation. If the roll had an

Outcome of 3, then Events Odd and Factor-of-3 have occurred (simultaneously).

Note also that the Outcomes in the Sample Space are (ideally) distinct. In other words, if one is to work with a six

sided die that is labelled with 1, 2, 2, 3, 3, and 3, your theoretical Sample Space might need to look like 1, 2A,

2B, 3A, 3B, and 3C, and you may need Events like Rolled-a-2 (2A or 2B) and Rolled-a-3 (3A, 3B, or 3C). In real

life, you're not likely to tell the 2's and 3's apart, but your theory (of 2A, 2B, 3A, etc.) will aid you in getting

the odds right.

Each Outcome may have a numerical weight of its own. Ideally all Outcomes should have the same weight, but this is

not always the case. Still, for most problems, giving each Outcome the same weight works well enough.

Each Event, in turn, has a weight equal to the sum of the weights of its Outcomes. Assuming the above Outcomes each

have a weight of 1, then Events Even, Odd, Factor-of-3, and One have weights of 3, 3, 2, and 1.

One can normalize these weights by dividing each with the weight of the Sample Space. This will give the weights,

going from 0 to 1, that is conventionally used in Probability. For the one die roll, this is a weight of 6. Hence,

the Probability of Event Even is 3/6 = 50%. In other words, P(Even) = 50%. Also, P(Odd) = 3/6 = 50%, P(Factor-of-3)

= 2/6 = 33.3%, P(One) = 1/6 = 16.7%.

As many have already pointed out, the possible Outcomes for the one is a Girl, odds on a Boy problem are:

(Boy, Girl), (Girl, Boy), (Girl, Girl)

The Sample Space of this problem has a weight of 3 (keeping it simple, only two kids, gender mix at 50/50, and no

twins). The observation being recorded is the gender of the first born and the gender of the second born. Hence the

Outcomes here are ORDERED sets - (Boy, Girl) and (Girl, Boy) are NOT the same! (Mathematically speaking, an Outcome

is really more like a Vector or a Matrix.)

The Event we would like to track is Girl-and-Boy:

- Event Girl-and-Boy is Outcome of (Boy, Girl) or (Girl, Boy)

Events are NOT ordered sets! {(Boy, Girl), (Girl, Boy)} IS the SAME as {(Girl, Boy), (Boy, Girl)}.

An Outcome should be (in principle, anyway) a perfect record of all that can be observed. Dropping information, such

as birth order, often brings misleading results. This is because, ideally, all Outcomes should have equal weight,

and following the mechanics of the situation is your best hope of achieving this. With birth order, Girl-And-Boy

is an Event with a weight of 2. Without birth order, Girl-And-Boy is an Outcome with a (presumed) weight of 1.
If

you insist on dropping birth order from your Outcome record, you may miss out on whether the Girl-And-Boy Outcome

needs a weight of other than 1.

Event Girl-and-Boy has a Probability of 2/3 = 66.7% (weight of Girl-and-Boy divided by weight of Sample

Space).

Since the problem originally asks for the odds of a boy and a girl, that is simply the weight of
Girl-and-Boy to

the weight of Only-Girls:

{(Boy, Girl), (Girl, Boy)} : {(Girl, Girl)}

2 : 1

What makes this problem Bayesian is in the way information is used to remove Outcomes from the Sample Space and

Events. Once this reduction happens, the weights on Events and the Sample Space can change, which in turn changes

the probability.

Let's say in addition to one is a girl we are also told and she is the first born. The Sample Space then

becomes:

(Girl, Boy), (Girl, Girl)

because the information she is the first born now makes (Boy, Girl) impossible.

By definition, Sample Space indicates all possible Outcomes FOR THE GIVEN CONTEXT, and so (Boy, Girl) must be

removed from Event Girl-and-Boy too:

- Event Girl-and-Boy is Outcome of (Girl, Boy)

And so, in the context of one is a girl and she is the first born, the weights of Sample Space and Girl-and-Boy

are 2 and 1.

Thus P(Girl-and-Boy) is now 1/2 = 50%, and the odds change to:

{(Girl, Boy)} : {(Girl, Girl)}

1 : 1