Dear Uncle Colin,

In a culture where people have children until they have a daughter and then stop, what proportion will be daughters?

* Families Retain Extra Girls Evidently

Hi, FREGE, and thanks for your message ((For the purposes of this question, I shall assume that daughters show up with a probability of 0.5.)) !

Spoiler alert: it’s a half. There are several ways to see this, and I’ll show you two.


Half of all the families will consist of a single daughter.

Half of the remaining families will consist of two children, including only one daughter

Half of the remaining families will consist of three children, including only one daughter.

… and so on.

If we have $n$ families, we clearly have $n$ daughters. How many children altogether? It’s $n\left(\frac{1}{2}\times 1 + \frac{1}{4} \times 2 + \frac{1}{8} \times 3 + \frac{1}{16}\times 4 + \dots\right)$.

Alternatively, that’s $\sum_1^\infty n 2^{-n}$.

We can use a generating function there: we know that $(1-x)^{-2} = 1 + 2x + 3x^2 +\dots$. Evaluating that at $x= \frac{1}{2}$ gives $1 + 2\times \frac{1}{2} + 3 \times \frac{1}{4} + \dots$, which is double our sequence; we need $\frac{n}{2} \left( 1 - \frac{1}{2} \right)^{-2}$.

That works out to be $\frac{n}{2}\times 4 = 2n$ children altogether.

So half of them are daughters.


Let’s do an experiment. Keeping track of every result, flip a coin until you get a head. Then repeat. Repeat. Keep repeating.

Your results don’t look any different to just flipping coin after coin after coin - how you divide them up doesn’t change anything. So, on average, half of them will be heads.

Same goes for the children: half will be daughters.

Hope that helps!

- Uncle Colin