Family Size

Problem 1

Consider a couple beginning a family; in the interest of promoting a diverse family, they decide to have children until they have both sexes. How many children can they expect to have?

Hint

We have already solved a very similar problem estimating the number of (Bernoulli) trials until the first success. The idea there was to evaluate probabilities of a sequence of trials having a certain length and then sum up an infinite series.

Solution of Problem 1

Assume $p$ is the probability of having a boy, $1-p$ that of having a girl. The sample space for our problem consists of just two kinds of sequences:

$BB\ldots BG$ and $GG\ldots GB.$

For a sequence of length $n,$ the probabilities are $(1-p^{n-1}p$ and $p^{n-1}(1-p),$ respectively, $n\ge 2.$ The mathematical expectation is then

$\begin{align}\displaystyle \sum_{n=2}^{\infty}&n((1-p^{n-1}p + p^{n-1}(1-p)) \\ &= p(1-p)\sum_{n=2}^{\infty}n((1-p)^{n-2}+p^{n-2}) \\ &= p(1-p)\bigg(\frac{2-p}{(1-p)^2}+\frac{2-(1-p)}{(1-(1-p))^2}\bigg) \\ &= \frac{1-p+p^2}{p(1-p)}. \end{align}$

(For away to obtain such sums in a closed form see a separate discussion.)

This result sits well with the formula $E=\frac{1}{p}$ for the expected length of a sequence of trials until the first success. Since the series in the latter starts with with $p$ which is omitted as meaningless in the present problem, we would get

$\displaystyle\bigg(\frac{1}{p}-p\bigg)+\bigg(\frac{1}{1-p}-(1-p)\bigg)=\frac{1-p+p^2}{p(1-p)},$

conforming our calculations. The graph of $\displaystyle f(p)=\frac{1-p+p^2}{p(1-p)}$ shows that the expectation is at minimum when $p=1/2$ and goes to infinity as $p$ approaches either $0$ or $1.$

graph of the expected value of family size depending of the probabilities of sexes

For $p=1/2,$ the expectation equals $3.$

Problem 2

Consider a couple beginning a family; At the outset they decide they will have children until they have a child of the same sex as the first one. How many children can they expect to have?

Solution of Problem 2

The sample space for this problem consists of sequence

$GBB\ldots BG$ and $BGG\ldots GB$

that, for length $n\ge 2,$ come with probabilities $(1-p)^{2}p^{n-2}$ and $p^{2}(1-p)^{n-2},$ respectively. The expectation then is given by

$\begin{align}\displaystyle E &= \sum_{n=2}^{\infty}n((1-p)^{2}p^{n-2}+p^{2}(1-p)^{n-2}) \\ &= \frac{(1-p)^2}{p^2}\sum_{n=0}^{\infty}np^{n}+\frac{p^2}{(1-p)^2}\sum_{n-2}^{\infty}n(1-p)^{n} \\ &= \frac{(1-p)^2}{p^2}S(p)+\frac{p^2}{(1-p)^2}S(1-p), \end{align}$

where $\displaystyle S(q)=\sum_{n=2}^{\infty}nq^n.$

I leave computing $\displaystyle S(q)=\frac{q^{2}(2-q)}{(1-q)^2}$ so that the expectation in this case is given by

$\displaystyle \begin{align} E &= \frac{(1-p)^2}{p^2}\frac{p^2(2-p)}{(1-p)^{2}}+\frac{p^2}{(1-p)^2}\frac{(1-p)^2(1+p)}{p^{2}}\\ &= (2-p) + (1+p) = 3! \end{align}$

Independent of $p!$ Which is a great surprise in its own right. But there is more. Note that the formula has been derived under an implicit assumption that $p$ is neither $0$ nor $1.$ But what if, say, $p=0!$ What if, by a fluke of a fate, women stopped producing boys. Then obviously the only possible children component in a family would be two girls. In another extreme $(p=1)$, all families would consist of two boys. Although neither case is plausible, I think that the circumstance merits being mentioned as a naturally occurring discontinuity.

A diversion

Moscow street matress advertisement: increase nativity rate

Moscow street board promoting beds and sofas under the slogan "Foster increase of the nativity rate."

References

  1. P. Nahin, Duelling Idiots and Other Probability Puzzlers, Princeton University Press, 2000, pp 55-57

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