I am coming back to a "primary school" probability problem I heard quite a while ago. It reads as follows. A couple have given birth to two children. At least one of those children is a boy. What is the probability that the other child is a girl? (Assume priors are 1/2).
The solution I read considered four cases, (BB, GB, BG, GG) and judged that one of those cases is impossible given the data. All the other possibilities are equally likely. Therefore the answer is 2/3.
Why can't we go to the formal Bayesian model:
$$P(G|D) = \frac{P(D|G)P(G)}{P(D)} $$ Where D are the data. $P(D|G)$ and $P(G|D)$ are not the same kind of thing, right? Because the data conditioned on is different. G is that a girl is born, D is that a boy is born and another child is born. $P(D|G) = \frac{2}{3}$ makes sense to me. 2 children are born, one of them is a girl, the same argument as for the above solution applies. But I can't provide further justification than that. $P(D) = \frac{3}{4} $ is true, by the same reasoning as the argument for the above solution. $P(G) =\frac{1}{2}$ under the model's approximation. Then $P(G|D) = \frac{4}{9} $.
The assumption that is perhaps the most contentious is $P(D|G) = \frac{2}{3}$, but how is $P(G|D) = \frac{2}{3}$ any better? Is there any background theory that can deem the one right and the other wrong? Or does this approach simply not work, and we need to nod our heads and not think too much (surely that's not right!).
Edit: Answer: G is not the event of 1 girl being born. G is the event of at least 1 girl being born after two birth-events. Therefore $P(G) = \frac{3}{4} = P(D)$. It follows that $P(G|D) = \frac{2}{3}$.