Puzzle
In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country?
My solution (not finished)
If we assume that the probability of having a girl is 50%, the set of possible cases are:
Boy (50%)
Girl, Boy (25%)
Girl, Girl, Boy (12.5%)
...
So, if we call G the number of girls that a family had and B the number of boys that a family had, we have:
$B = 1$
$P(G = x) = (1/2)^{x+1}*x$
So
$G = \Sigma (1/2)^{x+1}*x$
I feel like the sum of this infinite serie is 1 and that the proportion of girls/boys in this country will be 50%, but I don't know how to prove it!
Thanks!
Mike Scott is correct that you don't need to sum the series, but suppose you want to. Each family has 1 boy-that is easy. Each family has 50% chance of no girls, 25% chance of 1, etc. So the average number of girls is $$\sum_{i=0}^\infty \frac{i}{2^{i+1}}$$ The way to sum this is to remember that $$\sum_{i=0}^\infty a^{-i} = \frac{1}{1-1/a}=\frac a{a-1}$$ Now if you take the derivative with respect to $a$: \begin{align} \frac{d}{da}\sum_{i=0}^\infty a^{-i} &= \frac d{da}\frac a{a-1} \\ \sum_{i=0}^\infty{-i}a^{-(i+1)} &= \frac {a-1-a}{(a-1)^2} = \frac{-1}{(a-1)^2} \\ \sum_{i=0}^\infty \frac{i}{a^{i+1}} &= \frac{1}{(a-1)^2} \end{align} For $a=2$, $$\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}=1$$ So there is an average of one girl per family as well.