Stopping at the First Boy

Key idea. The stopping rule decides how many children a family has, but not the sex of any single birth.

Consider $n$ families. Each family stops at its first boy, so each has exactly one boy. Let $G_i$ be the number of girls in family $i$ . The family has $G_i=k$ when its first $k$ children are girls and the next is a boy, so

\mathbb{P}(G_i=k)=\left(\tfrac12\right)^{k+1},\qquad k=0,1,2,\dots,

and

\mathbb{E}[G_i]=\sum_{k\ge 0}\frac{k}{2^{k+1}}=1 .

The $G_i$ are i.i.d. In these $n$ families there are $n$ boys and $G_1+\dots+G_n$ girls. The proportion of boys is therefore

\frac{n}{n+\sum_{i=1}^n G_i}=\frac{1}{1+\frac1n\sum_{i=1}^n G_i} \;\longrightarrow\;\frac{1}{1+\mathbb{E}[G_1]}=\frac12

almost surely, by the strong law of large numbers.

Alternative. Every birth is an independent fair coin flip, whatever the reproductive rule. Over the whole population the births form one long sequence of fair flips, so the long-run fraction of boys is $\tfrac12$ .

Remark. What converges to $\tfrac12$ is the population ratio. It is not the average of the per-family proportions. A family with $G$ girls has a fraction $1/(G+1)$ of boys, and

\mathbb{E}\!\left[\frac{1}{G+1}\right]=\sum_{k\ge0}\frac{1}{(k+1)2^{k+1}}=\ln 2\approx 0.693 .

This average is high because the many small all-boy families each count as much as the rarer large families.

\boxed{\tfrac12}