One Boy Per Family, One Girl Per Family, and Why ln 2 Is Not a Half

Imagine a town with one rule about family size. Every couple keeps having children until a boy is born, and then stops. Nobody uses any other criterion, boys and girls arrive with equal chance at every birth, and no family gives up. After a generation, what fraction of the children in the town are boys?

Half. The rule cannot move it. That answer is easy to state and easy to state wrongly, because two different quantities are both called "the fraction of boys" and only one of them is exactly a half. Getting that difference right is most of the mathematics here.

Why the rule looks like it should do something

The rule is aimed at boys. It ends a family the instant a boy appears and keeps a family going as long as girls keep coming. Every family finishes with exactly one boy and possibly a long row of girls in front of him, so the intuition that the town should fill up with girls is at least as popular as the intuition that it fills up with boys. Both are wrong, and the reason is the same in each case: the rule decides when to stop asking, and it never touches the answer to any individual question.

One family, exactly

Let $G$ be the number of girls in a family. Having $G = k$ means $k$ girls followed by a boy, so

Those probabilities sum to 1, which is the formal way of saying every family does eventually stop. Half of all families are one boy and nothing else. A quarter are one girl then a boy. An eighth are two girls then a boy, and so on down.

The average number of girls is the corresponding weighted sum:

Every family also contains exactly one boy, with no averaging needed. So the expected number of boys and the expected number of girls per family are both 1, and the average family has two children.

The ratio of the counts

Now add up a whole town. With $m$ families the expected number of boys is $m$ and the expected number of girls is $m$ as well, so

By the law of large numbers the two totals each concentrate around $m$ as the town grows, so the actual observed share converges to a half as well. In a real town of any size worth calling a town, you would measure something indistinguishable from fifty-fifty.

The expected share is not one half

Equation (3) is a ratio of two averages. It is not the average of the ratio, and here the two genuinely differ. Take a single family and ask for the expected value of its boy share. The family has $G + 1$ children and exactly one boy, so the share is $1/(G+1)$, and averaging that gives a logarithm:

Well above a half, and for a reason you can see without the series: half of all families are a single boy, and their share is 100 per cent. No family can ever have a share of 0 per cent, because the boy is guaranteed. The distribution is lopsided, and averaging a lopsided distribution of ratios does not give you the ratio of the averages.

Ten families still average 0.5249. A hundred families average 0.5025, and a thousand average 0.50025, so the excess shrinks like $1/(4m)$. It never turns into a bias in favour of girls and it never quite disappears, which is why the honest phrasing of the answer names the whole town rather than a household.

No stopping rule can do better

The result is not special to this rule. Number the births in a family $1, 2, 3, \dots$ and let $D_n$ be the number of boys minus the number of girls after $n$ of them. That is a simple symmetric random walk. A stopping rule is a decision, made at each point, to continue or stop, using only what has already happened. If the rule stops after finitely many births on average, then

which is the optional stopping theorem, and it says exactly that the expected number of boys equals the expected number of girls. Our rule has $T = G + 1$ with $\mathbb{E}[T] = 2$, comfortably finite, so equation (5) applies and equation (3) follows with no computation at all.

Try to be cleverer and the theorem stops you. Suppose families stop at the first boy or after $n$ children, whichever comes first, which is what a real population with a family-size limit would look like. Then

for every $n$, still dead level. The only way to bias the counts would be a rule that reads a birth before it happens, and that is not a stopping rule.

When the coin is not fair

Suppose each birth is a boy with probability $p$. A family then averages $(1-p)/p$ girls and still exactly one boy, so the boy share of the expected counts is

The stopping rule hands back whatever ratio the births had, untouched. Real human births run at roughly 1.05 boys per girl, so $p \approx 0.512$, and a town following this rule would settle at 0.512 rather than 0.500. The expected share for a single family follows the same logarithm as before, $-p\ln p / (1-p)$, which returns $\ln 2$ at $p = 1/2$.

Sources and further reading

• The distribution in equation (1) — Geometric distribution
• The theorem doing the real work in equation (5) — Optional stopping theorem
• Why the observed share and not only the expected counts settles down — Law of large numbers
• The figure the model idealises away — Human sex ratio

The two numbers that matter here were computed both ways before publication: the series in equations (2) and (4) in exact rational arithmetic, and a simulated town of two million families, which returned a boy share of 0.5000 with one boy and 0.9999 girls per family. The small-town bias was measured separately and matches the curve in figure 2.