[twelvechairs]

If we are determined to look at order of birth (ie. have a BG and GB) then we should consider all cases by order of birth, so where 'B' represents the boy we know and 'b' or 'g' represents a child we dont, and the first character represents the first child, and the second the second childe - we have:

Bb, bB, Bg, gB

50%.

[NickPollard]

You need to look at it in a Bayesian sense.

Consider all the possibilities and then look at the outcomes they produce.

BB BG GB GG

These are all equally likely, right? They each have P = 1/4. If we took 1000 fathers who had two children each, and lined them up, we would expect 1/4 of them to have children matching each of the above pairings. That means 250 of each. With me so far?

Now, we know that we are dealing with a father who has at least one son. If we went along the line and asked each father 'Do you have at least one son: Yes/No?', then 750 would answer yes, and 250 (the GG fathers) would answer no.

If a random father comes up to us and says 'I have at least one son', we know that he is from those 750 - we have selected a subset to deal with. Of those 750, only 250 have a second son, so the probably is 250/750 or 1/3.

[Deestan]

You're assuming a different selection model, in which a father of two boys is twice as likely to volunteer information if he has two boys: "Pick a child at random, then if it's a boy - say: I have a boy [blablabla] what is the other?"

However, you're doing this on an already selected sample by discounting all the girl-girl pairs. The selection rule in your logic then becomes a two-step "If you have at least one boy, then pick a child at random, then if it's a boy, say: [blablabla]"

Given that, your analysis is correct.

But, seeing as it is a wildly different selection model than what everyone else seem to work with, you should be explicit about it.

[incremental]

Imagine a similar problem but with red and blue poker chips. Say, for example, that I have a bag and I pull out two chips, one at a time.

In this problem, would you still try to distinguish the two identical red poker chips using your logic?

[twelvechairs]

Okay. To respond to both of you - I have put this in programming just to check, and I think you are taking a different inference from the question than I am.

Under your inference, the man wouldnt have mentioned anything unless he had at least one male child. (in which case you can say the GG scenario is gone, but GB BG and BB are equally probable)

Under my inference, the man just told me the sex of one child at random.... (in which case BB is twice as probable as GB or BG - where he could have equally said 'i have at least one girl')

That sound reasonable to you? I have it in ruby form if you are interested :)

The blogpost linked to by someone above (http://blog.tanyakhovanova.com/?p=221) uses this explanation, which is very different (to me) than the one in the main link, and I can see why this gets to 1/3:

"A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son."

[incremental]

"Under your inference, the man wouldnt have mentioned anything unless he had at least one male child. (in which case you can say the GG scenario is gone, but GB BG and BB are equally probable)"

No. I'm just assuming he has two children, and randomly mentions something about one of them. The GG scenario is only eliminated after he makes his statement, because we then know he has at least one boy.

[twelvechairs]

Ah, then I do think you have an error of logic.

Put it this way. before he says it, we have GG, GB, BG, BB

after he says "I have a child that is [MALE OR FEMALE]" we have (where the capital letter is the child whose sex has been mentioned, and the lowercase letter is the other child):

Gg, gG, Gb, gB, Bg, bG, Bb, bB

So if he has said the sex is male, then we have four combinations left:

gB, Bg, Bb, bB.

Understand that we go to more scenarios (8) based on which child is mentioned, before we go to fewer. Actually the order of the children is something you can and should ignore, however as you are holding on to it, I show it this way....

[incremental]

I'm afraid this is wrong - you shouldn't distinguish between Bb and bB. In this problem, they are not different states, so counting them messes up your probability calculation.

Someone had a nice link higher up: http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-proble...