In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?

I just dipped into a book, The Drunkard's Walk - How Randomness Rules Our Lives, by Leonard Mlodinow, Vintage Books, 2008. On p.107 Mlodinow says the chances are 1 in 3. 

It seems obvious to me that the chances are 1 in 2. Am I correct? Is this not exactly analogous to having a bowl with an infinite number of marbles, half black and half red? Without looking I draw out a black marble. The probability of the second marble I draw being black is 1/2.
In a family with 2 children there are four possibilities:

1) the first child is a boy and the second child is a boy (bb)

2) the first child is a boy and the second child is a girl (bg)

3) the first child is a girl and the second child is a boy (gb)

4) the first child is a girl and the second child is a girl (gg)

Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg. Out of those three possibilities the only one with two girls is gg. Hence the probability is $\frac{1}{3}$.
There are many ways to explain this, but let me share with you how I do it in my class. I set $q=P(\mbox{You say that he has at least one daughter} \mid \mbox{He has $\{B,G\}$})$.

If you sample at family level, then you ask the father if he has at least a daughter or you check official records. In that case $q=1$.

If you sample at child level, by for instance meeting the father in the company of one child or otherwise happening to find out about one child, then $q=1/2$. In this case you will half the time miss this family because you happen to see the son first.

Using Bayes' formula, we get
$$
P(\{B,G\}) \mid \mbox{You say that he has at least one daughter})
= \frac{P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\})}
{P(\mbox{You say that he has at least one daughter})}.
$$

Since
$$P(\mbox{You say that he has at least one daughter}) = P(\mbox{You say that he has at least one daughter} \mid \{B,G\}) P(\{B,G\}) + P(\mbox{You say that he has at least one daughter} \mid \{G, G\}) P(\{G, G\}),
$$

we get 
$$
P(\{B,G\} \mid \mbox{You say that he has at least one daughter}) = (q \times 1/2)/(q \times 1/2 + 1 \times 1/4) = 2q/(1 + 2q),
$$

and it follows that 
$$
P(\{G, G\} \mid \mbox{You say that he has at least one daughter})=1/(1 + 2q).
$$

If $q=1/2$, we get $P=1/2$, and $q=1$ gives $P=1/3$.