Ratio of Expected values of Boys to Girls

Question

In a country where everyone wants a boy, each family continues having babies till they have a boy. After some time, what is the proportion of boys to girls in the country? (Assuming probability of having a boy or a girl is the same)

Answer 1

Let the expected number of girls in any given family is $g$. If a boy is born first (probability =$\frac{1}{2}$) then $g=0$, but if a girl is born (probability =$\frac{1}{2}$) then $g=1 + g$ ($1$ for the girl already born to the couple and $g$ for the fact that state is reset to original state where couple have to keep breeding until a boy is born)

$\implies$ $g = 0 (\frac{1}{2}) + (g + 1) \frac{1}{2}$

$\implies$ $g = 1$

Similarly, lets say the expected number of boys in any given family is $b$. If a boy is born (probability =$\frac{1}{2}$) then $b=1$, but if a girl is born (probability =$\frac{1}{2}$) then $b=0 + b$ ($0$ because now the newborn is girl and $b$ for the fact that couple are back to where they started but with an extra girl and couple keep breeding)

$\implies$ $b = 1 (\frac{1}{2}) + (b + 0) \frac{1}{2}$

$\implies$ $b = 1$

Lets say town has $n$ couples. Therefore, expected number of girls in town is $E[g_1] + E[g_2] + ... +E[g_n]$

But, $E[g_1] = E[g_2] = E[g_n] = g = 1$

$\implies$ Expected number of girls in town $= n$

Similarly, Expected number of boys in town $= n$

This implies ratio of expected values is $n:n = 1:1$

Answer 2

Let's talk about the expected fraction of boys in the population, and let's just consider one family. The fraction of boys is $F = \frac{B}{B + G}$. If we follow your rule, this family will stop having children precisely when they get a boy, so $B = 1$, implying $F = 1/(1 + G)$. Now, let's note that $$P(G = 0) = \frac{1}{2},\quad P(G=1) = \left(\frac{1}{2}\right)^2$$ and that in general $P(G = k) = \left(\frac{1}{2}\right)^{k+1}$. This then means that \begin{align} E[F] &= E\left[\frac{1}{1 + G}\right] \\ &= \sum\limits_{k = 0}^\infty \frac{1}{1 + k}P(G = k) \\ &= \sum\limits_{k = 0}^\infty \frac{1}{1 + k}\left(\frac{1}{2}\right)^{k+1} \\ &= \log(2). \end{align}

Dealing with more than one family is much more complicated.