See here: https://en.wikipedia.org/wiki/Geometric_distribution. There are two parametrization of the geometric distribution:

1. You compute the number of failures before success, here the mean is $1/p$;

2. You compute the number of trials before success, here the mean is $(1-p)/p$.

In the first case, you have $P(X = 0) = 1/2$ and in the second case $P(X = 0) = 0$. All this to say you simply use the wrong formula for the mean.

You should have $E(X) + 1 = (1/.5) + 1 = 3$.

See here: https://en.wikipedia.org/wiki/Geometric_distribution. There are two parametrization of the geometric distribution:

1. You compute the number of trials before success, here the mean is $1/p$;

2. You compute the number of failures before success, here the mean is $(1-p)/p$.

In the first case, you have $P(X = 0) = 0$ and in the second case $P(X = 0) = 1/2$. All this to say you simply use the wrong formula for the mean.

Once you have set the gender of the first child, you then look for the number of trials you need to get the other, letting that be $X$. Hence you should have $E(X) + 1 = (1/.5) + 1 = 3$.