Failed to debunk an argument against a simple conditional probabilty question During family festivities last Christmas, someone read out the following question from a puzzle book:

I have two children, who aren't both boys. How probable is it that they are both girls?

Using some very shaky stats, I worked out that:
P(Both Girls | Not Both Boys)
= P(Both Girls) / [1 - P(Both Boys)]
= (1/2 x 1/2) / (1 - 1/2 x 1/2)
= 1/3
(Which was also the answer in the book)
However, my brother in-law contested this and said (effectively):

Well, if they're not both boys, then one of them is guaranteed to be a
girl. In which case, it's just 50/50.

Frankly, I couldn't argue against his reasoning (then or still).
How could I have responded?
Thanks!
 A: Four possibilities to start with, one is excluded
boy,boy
boy,girl
girl,boy
girl,girl
Since you know that option boy,boy is not one of them, you are left with 3 options. There is now only one event of interest "girl,girl" out of three, so the probability is 1/3.
In this case, it appears that the brother-in-law is considering the combinations, not permutations. Hence, the 50/50 suggestion.
A: I think https://stats.stackexchange.com/users/176202/frans-rodenburg answer is correct, in that there are two ways of having a boy and a girl: BG and GB.
Your stats look sound to me.  Assuming the probability of having a boy is equal to that of having a girl, $P(G)=P(B)=1/2$, then the probability of having two boys is equal to that of having two girls $P(GG)=P(BB)=1/4$, and the probability of having one of each is the complement of the sum: $P(BG or GB)=1-1/4-1/4=1/2$.  The proba of having no boys is then $P(GG or GB or BG)=1/2+1/4=3/4$.  Using Bayes' theorem, the probability of having two girls given there are not two boys is: $$P(GG|not BB) = P(not BB|GG) \frac{P(GG)}{P(not BB)}= \frac{1/4}{3/4} = 1/3$$
since the probability of not having two boys given there are two girls is 1.
