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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?

The Solution

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There is always one boy in each family. Determining the number of girls in each family will allow us to calculate the ratio. The number of girls in each family is equal to the expected value, which is the probability multiplied by the value (number of girls). Let's start with the probability of just one girl. Here is the sequence of events:
$$\left(\dfrac{1}{2} \text{ chance it's a girl}\right) \to \left(\dfrac{1}{2} \text{ chance it's a boy}\right) = \dfrac{1}{4} \text{ chance}$$
We can do this each time.
$$\left(\text{1 girl}\right) \cdot \dfrac{1}{4} + \left(\text{2 girls}\right) \cdot \dfrac{1}{8} + \left(\text{3 girls}\right) \cdot \dfrac{1}{16} + \dots$$
The series is
$$\sum_{n = 1}^{\infty} \dfrac{n}{2^{n+1}}$$
We now have to calculate this sum. Let's write it out
$$S = \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{4}{32} + \dfrac{5}{64} + \dots$$
Before simplifying, we rewrite this sum in a different way.
$$S = \dfrac{1}{4} + \dfrac{1 + 1}{8} + \dfrac{1 + 2}{16} + \dfrac{1 + 3}{32} + \dfrac{1 + 4}{64}$$
We take all the fractions that have a 1 in the numerator.
$$S = \left(\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots\right) + \dfrac{1}{2}\left(\dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16} + \dots\right) $$
The sum of the left parentheses is $\dfrac{\frac{1}{4}}{1 - \frac{1}{2}} = \dfrac{1}{2}$ and the right parentheses just equal S.
We now have this equation:
$$S = \dfrac{1}{2} + \dfrac{S}{2} \to S = 1$$
So, we conclude that the expected number of girls in each family is 1, and the expected number of guys is 1. Therefore, the ratio is 50/50.