Boys and Girls Revisited

Question

Consider a country with $n$ families, each of which continues having children until they have a boy and then stop. In the end, there are $G$ girls and $B=n$ boys.

Douglas Zare's highly upvoted answer to this question computes the expected fraction of girls in a population formed of complete families and explains why we shouldn't expect it to equal $1/2$. My current question concerns a different statistic, namely the probability that there are more boys than girls (after all families have finished reproducing). This probability turns out to be exactly $1/2$, and I'm looking for an intuitive explanation of why.

Indeed, for fixed $n$, it's not hard to see that
$$Prob(G=k)=\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$ (The binomial coefficient is the number of ways to assign $k$ indistinguishable girls to $n$ distinguishable families.)

Therefore $$Prob(G < B)=\sum_{k=0}^{n-1}\binom{n+k-1}{k}\cdot {1\over 2^{n+k}}$$ It's not hard to check that this sum is exactly equal to $1/2$ (and therefore, in particular, independent of $n$).

That is, regardless of the number of families, we always have the surprisingly (to me) simple formula

$$Prob(G < B )=1/2$$

(Note that this implies $Prob(G>B)$ is strictly less than $1/2$ --- because there's always some probability weight on the event $G=B$ --- though an application of Stirling's formula shows that $Prob(G>B)$ converges to $1/2$ as $n$ gets large.)

My question is:

Is there some simple intuitive reason I should have expected this result?

Answer 1

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.

Answer 2

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.