A family has two children. You know that one of the two children is a girl. What is the probability that the other child is a girl as well? A classical problem for a probability course exam. Unfortunately, the problem is undefined, because the problem can be intrepid in multiple ways which lead to different results.

Let $G$ be the case that we know that one of the children is a girl. Let $GG$ be the case that both children in the family are a girl. By Bayes rule

$$

P(GG|G) = \frac{P(G|GG) P(GG)}{P(G)} = \frac{1 \times 1/4}{3/4} = \frac{1}{3}.

$$

Because $P(G|GG) = 1$, $P(GG) = 1/4$, and $P(G) = 1/2$. Assuming that the ratio boy girl is exactly fifty/fifty.

Let’s go back to the child of which we know that she’s a girl. She must have a name (we assume that every child has a name). Let her name be $N$ and let $G_N$ be the case that a girl is named $N$. The probability that a girl is called $N$ is set to be $P(G_N) = p$. $p$ is close to zero but not equal to zero, because there are many girl names and none of these names are used too often. Note that if a family has two girls the chance that at least one of the girls is called $N$ is

$$

P(G_N|GG) = 1 – P(\textrm{not }G_N|GG) = 1 – (1 – p)(1 – p) = 2p – p^2.

$$

By Bayes rule

$$

P(GG|G) = P(GG|G_N) = \frac{P(G_N|GG) P(GG)}{P(G_N)} = \frac{(2p – p^2) \times 1/4}{p} = \frac{1}{4} (2 – p).

$$

Because $p$ positive but close to zero we find that $P(GG|G) \approx \frac{1}{2}$. A different results than the result in the previous paragraph.