# Boys and Girls Revisited

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?
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The families have children until they have boys. We at MO have these questions until what? –  Mark Sapir May 30 '13 at 7:16
Apparently, until we figure out everything we want to know about this setup :). The blue-eyed islanders were discussed on Tao's blog for a year or so and I'm not 100% certain that thread will never be revived again... –  fedja May 30 '13 at 10:57
Well, it does not have to be the blue-eyed islanders. Maybe it will come back as the surprise test. –  Michael Renardy May 30 '13 at 13:38
This question may be beside the point of the original question. But is there a direct combinatorial computation of Prob(G<B) as the sum shown in the post aside from, say, an inductive argument? –  Hansen Mar 2 at 19:57

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.

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You can view it like that too but my mental picture was different: I just put all couples into one big family and told them to multiply. The first few girls and the first boy born would then go to the first family, the second group like that to the second, etc. until every family gets their children. The rest of the offspring gets discarded. –  fedja May 30 '13 at 11:52
Unfortunately, that would obscure the picture quite a bit because this way they can stop well before they reach the magic count of $2n-1$. Let's agree to send the extra children to the orphanage for now and decide what to do with them later when they grow up... –  fedja May 30 '13 at 14:33
fedja's excellent argument answers a question that has been bothering me ever since reading Doug Zare's famous MO answer, which is whether there's some model in which the expected proportion of girls really is 1/2. Imagine Family 1 having a child at t=1, a second child at t=2 if the first is a girl, a third child at t=3 if the second is a girl, etc., until they have a boy. Then Family 2 picks up where Family 1 leaves off, until they have their boy. Now compute "the proportion of girls" by picking a random positive integer T and looking at all the children born up to time T. (cont'd) –  Timothy Chow Jun 1 '13 at 23:45
In this model the expected proportion of girls is 1/2. This doesn't contradict Zare's observations because the number of families is not fixed in this model, and the "observation time" T can easily land in the middle of an "unfinished family." This model, or something like it, may be what some of the people who objected to Zare's MO answer were groping to formulate precisely. –  Timothy Chow Jun 1 '13 at 23:48
Tim: it is also worth noting that after everyone has reproduced, we do have $E(G/G+B-1)=1/2$. –  Steven Landsburg Jun 2 '13 at 14:57

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.

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I too failed at first to grasp Fedja's wonderful idea, but for me it all came clear when I read his response to Emil's comment. Thanks for elaborating it. –  Steven Landsburg May 31 '13 at 0:19
So there remains hope that you will also grasp in finite time the explanation of my colleague Rhett Butler, who had just this wonderful idea some days before Fedja. I don't understand why it appears to be so difficult. I grasped it the first time when he explained it to me. Take a very long random sequence of n bits 0 and 1, he said, subdivide it in as many, say B, short segments as there are ending with 1. Then you will have realized the result of one family with very good statistics. Since the long sequence has to be chosen such that it ends with an 1, there is a slightly greater chance –  user34246 Jun 1 '13 at 9:02
there is a slightly greater chance of having more 1, call that number B, than 0, call that number G. The difference will be about 1/n but will never result in G/(G-B) = 31 %. (The average of percentages will have this limit - but that is of no interest with respect to the official question concerning the ratio of girls to boys being 1). The statement by D. Zare: "for a large population such as a country, the official answer of 1/2 is approximately correct, although the explanation is misleading", is painly wrong. Unfortunately RB cannot comment himself. Yesterday he has been admitted to hos –  user34246 Jun 1 '13 at 9:03
@Mellius, I am sorry to hear about your colleague, and certainly wish him well. I've found many of Rhett Butler's contributions at MathOverflow, particularly on historical matters, to be interesting and informative. But I don't see how his comments at the other question (linked to near the top of the OP's question here) are relevant to fedja's answer to this question or to my elaboration of it. I'll leave it to others to assess the appropriateness of his comments there and yours here. –  Barry Cipra Jun 1 '13 at 14:44
@Mellius, I do like the image of babies on a conveyor belt. It makes me think of I Love Lucy and the candy factory: youtube.com/watch?v=HnbNcQlzV-4 :-) –  Barry Cipra Jun 2 '13 at 14:13