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](https://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys-closed) 
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?
> 



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.


