Boys and Girls Revisited

Symmetry. Put them all together and tell them to multiply forever. The question then becomes whether the $n$-th boy was born at the $2n$-th birth or later, or not, i.e., who is the majority among the first $2n-1$ births: boys or girls.

This is just a variant presentation of fedja's truly wonderful solution. It took me a while to catch on to the idea there, so I'm offering this in case it helps clarify it for anyone else. All credit for the insight, though, properly belongs to fedja.

For the purpose of determing the probability that the $n$ families wind up with more boys than girls altogether, it's convenient to imagine births take place in the following fanciful way. There is a "magic fertility wand" which is passed from family to family and which causes the family that possesses it to produce one child per day until they have a boy, at which point they pass the wand along to the next family. If and when every family has their boy, the wand is given to a pair of rabbits, who use it to produce a male or female offspring each day without cease. All births take place at a hospital/veterinary clinic.

On the $(2n-1)$st day, the hospital reports whether they've delivered a preponderance of males or females. Obviously it's a 50:50 chance for either result. If it's more males, there are at least $n$ of them, which means there are exactly $n$ boys and possibly some male rabbits, but certainly no more than $n-1$ girls -- in any event, the families are done reproducing and there are more boys than girls. If, on the other hand, the hospital reports more female births, then there have been fewer than $n$ males born, which means the wand hasn't yet reached the rabbits, so all the births have been boys and girls, with at least $n$ of them girls, so the boys will never outnumber the girls.