I hate to beat a dead horse here, but I have not seen a correct answer.

The space of solutions for having two kids is (B, b), (B,G), (G,g). There is no ordering of the kids.

We are told that at least one is a girl which removes the (B,b) case.

But (and this is the tricky bit) we have:

(B,G) with the "named" girl being G --> here the other child will be the boy

(G,g) with the "named" girl being G --> here the other child will be the girl "g"

(G,g) with the "named" girl being g --> here the other child will be the girl "G"

Out of possible three choices there two possible ways for the other child to be a girl. This makes the probably 2/3!

Not quite right. There is a similar problem for which your solution will work: suppose we have two bowls. One contains a white and a black marble, the other one contains two white marbles. You pick a marble at random, turns out you picked a white marble, what is the probability of the other marble in the same bowl also being white? The answer is 2/3 for this problem as shown by your solution.

But in this case of children, it is in fact twice as likely to have a boy and a girl as it is to have two girls. Consider that no matter what sex your first child is, you have a 1/2 chance to get a boy and a girl (your second child has to be of the other sex). But to get two girls, you must first have a first child girl (probability 1/2), then a second child girl (probability 1/2). The probability for this is 1/2 x 1/2 = 1/4. So if we translate to the marble case, for this to be properly translated, you need two bowls with white+black marbles, and one bowl with white+white. The probability will now be 1/2 for the other marble to be white. Which is the same answer as you get to the boy girl problem with a "named" girl.

So to keep it all straight: 1/3 for the "given that both aren't boys" case, 1/2 for the "given one child who's definitely a girl" case, 2/3 for the marble case which is different but similar.

**Edited by shakingdavid, 15 June 2011 - 11:07 AM.**