## Rules for the perfect birthrate

Matthen2 posted this problem on twitter and I thought it was interesting. If families all have children until they have exactly one daughter, on average how many children of either sex will they have? In this post I play with the probabilities involved.

In order to answer this question let us consider what is the probability that a family has $n$ children. Since they have children until they have a daughter, and each time they have a child the probability of it being female is $\frac{1}{2}$, the probability of having $n$ children is $\left(\frac{1}{2}\right)^n$. In order to determine the average number of children per family $\langle~n\rangle$  we must therefore evaluate the following sum:

$\langle~n~\rangle~=\Sigma_{n=1}^{ \infty}\left(\frac{n}{2^n}\right)$

The first terms of the sum are: $\frac{1}{2}~,~~\frac{2}{4}~,~~\frac{3}{4},...$. The answer to this sum is readily found by asking WolframAlpha (the answer is 2). Is it possible to determine the summation with good old pen and paper? The solution is not immediately obvious as the summation is neither a geometric, nor an arithmetic sum. In order to make the sum, let us make a table of the table above:

$\begin{array}{cc}1/2&+\\2/4&+\\3/8&+\\..&=\end{array}$

It is clear that this sum is the same as:

$\begin{array}{cccccc}1/2&+&&&&\\1/4&+&1/4&+&&\\1/8&+&1/8&+&1/8&+\\..&&&&&=\end{array}$

OOOOH! Now the columns are geometric series and they can be easily summed. The resulting values are:

$1+\frac{1}{2}~+~~\frac{2}{4}~+~~\frac{3}{4}+...=2$

So the average number of babies per family is two, since each family has exactly one daughter, this means that on average each family has one son. So to achieve a “perfect” birth rate of 2, all we have to do is have enough babies to have exactly one daughter. But never, ever have two daughters.

PS: The previous argument requires families to  be prepared to have infinite number of babies, if necessary. What if each family decided to have at most $m$ babies? The the average number of girls in each family would b:

$\langle~g~\rangle~=1-1/2^m$

which is the probability of having $m$ boys. The number of boys would be:

$\langle~b~\rangle~=\frac{m}{2^m}~+~\Sigma^m~(i-1)/2^i$

Where the first second represents the average of the number of boys in families with one girl, the first term is the probability of having $m$ boys weighted by the number of boys. With Wolfram‘s help:

$\langle~b~\rangle~=1-1/2^m$

So if people are going to follow the above rules, but only want to have – say – 4 children, the number of children per family will be $latex~2~(1-1/16)=15/8$. And still equality of boys and girls!

I am a researcher in solar energy at the University of Oxford. I am interested in mathematics, programming and trying to understand why things work. I also like the great outdoors and riding my bike.
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### 7 Responses to Rules for the perfect birthrate

1. joe parker says:

It’s an interesting phenomenon well-known amongst population geneticists that the actual replacement rate is around 2.1 – not only this, but the sex ratio (boys:girls) is about 1.05 (ish).

This is because young men tend to kill themselves by doing stupid things like going caving before they get a chance to reproduce: in the actively-reproducing cohort (in natural human populations at least) the sex ratio is back to ~1.0.

What’s less well-known is that fixie riders tend to a sex ratio of infinity. So perhaps nature knows someting we don’t :p

• joe parker says:

Addendum to the above: even more bizarrely, the sex ratio is an evolutionarily plastic trait, meaning that it can alter over time (e.g. human populations under stress may end up producing more males).

2. ambruge says:

It is much easier if you calculate the expectation by seeing how many boys you are going to have. Every couple has a girl for sure; now they have an extra boy w.p. 1/2, another w.p. 1/4, and so on. So the expected number of boys is 1/2 + 1/4 + 1/8 + .. =1 (I have to figure out how to add math formulas). So the expected number of children is 1+1=2. Now if you have at most m-1 boys, then the expectation of the number of boys is 1/2 + … 1/2^(m-1) = 1 – 1/2^m. Notice though that here you don’t have equality between boys and girls: every couple has exactly one girl, and less than one boy on average!

• the expected # of boys is:
0 1/2 + 1 1/4 + 2 1/8 +3 1/16 … = 1

If you decide to have m children at max, then there is also the possibility of having exactly m boys and no girls – the two expectation values are still the same!

• ambruge says:

Of course the value of the two series is the same, it is just easier to calculate the $\sum 1/2^n$. So when you calculate the expectation, you do it by seeing what is the probability that you have at least k boys, and then add up these probabilities. It is just double summation (or Fubini, if you want to be fancy 🙂

Yep you are right, I forgot the possibility of having no girls. Strange that the expected # of boys and girls is the same! There must be some nice symmetry argument behind it.

OK, when you may have infinitely many children, then let $b$ be the number of boys. Then the first kid is either a girl, giving $b=0$ or a boy, in which case from the second child on the whole process is the same as the original one (this is the Markov property, so in a fancy language we have a Markov chain here). So if the expectation on $b$ is $B$, then we get the equation
$latex B = 1/2 ( 1 + B)$
which gives $B=1$ right away.

Now, if you have at most $m$ children, then let $b_m$ be the expectation. The above argument now gives
$b_m = 1/2 (1 + b_{m-1})$,
and taking into account $b_0 = 0$ we arrive at $b_m = 1 - 2^{-m}$.

3. prometejs says:

So this calculation means that no 2 families have the same number of children?

4. prometejs says:

And how do you came up with:
g(average)=1-1/2^m
So if every family has 4 children, it means that on the average they all have 15/16~1 girl(s)? Something doesn’t make sense to me.