[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...