# Suppose there is a population with the tradition that the women bear children until they have one boy. What would be the ratio of boys to girls in this population?

## Suppose that the probability to give birth to a boy is the same as to give birth to a girl. Also, the women in this population keep bearing children until they have exactly one boy.

Jun 16, 2018

$50 : 50$

#### Explanation:

There are two ways of explaining:

1. (Logic)

By introducing the rule that women keep bearing children until they have a boy, the number of children is effected ($n$) and not the probability ($p$) and therefore not the ratio of the gender of the children. The probability is still $\frac{1}{2}$ and therefore the ratio is still $50 : 50$.

2 (Mathematic :) )
To determinate the number of boys or girls in one family, we need to find the expacted value of the number of childrens within one family.
$E \left(x\right) = 1 \cdot P \left(1\right) + 2 \cdot P \left(2\right) + \ldots + n \cdot P \left(n\right)$
$E \left(x\right) = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + \ldots + n \cdot \frac{1}{{2}^{n}}$
$E \left(x\right) = {\lim}_{n \to \infty} \left({\sum}_{i = 0}^{n} \left(\frac{i}{2} ^ i\right)\right)$

sum_(i=0)^(n)(i/2^i)\stackrel{!}{=}(2^(n+1)-n-2)/2^n
To prove this by induction, we need to show that this is true for $n = 0$ and that it is true for $n + 1$

Base case
sum_(i=0)^(0)(i/2^i)\stackrel{!}{=}(2^(0+1)-0-2)/2^0
0\stackrel{!}{=}(2-2)/1
$0 = 0$

Inductive step

sum_(i=0)^(n+1)(i/2^i)\stackrel{!}{=}(2^(n+1+1)-(n+1)-2)/2^(n+1)
(n+1)/2^(n+1)+sum_(i=0)^(n)(i/2^i)\stackrel{!}{=}(2^(n+2)-n-3)/2^(n+1)
Sub in sum_(i=0)^(n)(i/2^i)\stackrel{!}{=}(2^(n+1)-n-2)/2^n
(n+1)/2^(n+1)+(2^(n+1)-n-2)/2^n\stackrel{!}{=}(2^(n+2)-n-3)/2^(n+1)| * 2^(n+1)
(n+1)+2 * (2^(n+1)-n-2)\stackrel{!}{=}2^(n+2)-n-3
n+1+2^(n+2)-2n-4\stackrel{!}{=}2^(n+2)-n-3
2^(n+2)-n-3\stackrel{!}{=}2^(n+2)-n-3

Q.E.D.

Therefore,
$E \left(x\right) = {\lim}_{n \to \infty} \left(\frac{{2}^{n + 1} - n - 2}{2} ^ n\right)$
$E \left(x\right) = {\lim}_{n \to \infty} \left(2 + \frac{- n - 2}{2} ^ n\right)$
$E \left(x\right) = 2 - {\lim}_{n \to \infty} \left(\frac{\textcolor{b l u e}{n + 2}}{\textcolor{red}{{2}^{n}}}\right)$

Since ${\lim}_{n \to \infty} \left(\textcolor{red}{{2}^{n}} > \textcolor{b l u e}{n + 2}\right)$
${\lim}_{n \to \infty} \left(\frac{\textcolor{b l u e}{n + 2}}{\textcolor{red}{{2}^{n}}}\right) = 0$

$E \left(x\right) = 2 - 0 = 2$

Knowing that the ratio is
$\textcolor{red}{E \left(x\right) - 1} : \textcolor{b l u e}{1}$
$\left(\textcolor{red}{red} = g i r l s , \textcolor{b l u e}{b l u e} = b o y s\right)$
$\left(2 - 1\right) : 1$
$\iff 1 : 1$
The ratio of boys to girls is $1 : 1$