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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)

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It stays 50%. As long as the chance for each child is 50%, it won’t change. You will have 1/2 of the families with just one child, a boy, 1/4 with one girl and one boy, 1/8 with two girls and one boy, etc. the average will be one girl and one boy per family.

Obviously the ratio of boys to girls could be any rational number, or infinite. If you either fix the number of families or, more generally, specify a probability distribution over the number of families, then B/G is a random variable with infinite expected value (because there’s always some non-zero chance that G=0).

Both $B/G$ and $G/B$ are random variables. If the “some time” in the question is long enough so that each family has had its one boy, then $B$ is the number $n$ of families, while $G$ is a random variable whose expectation is also $n$. So if one interprets the question as being about the ratio of the expectations of $B$ and $G$, then the answer is $1$. If, on the other hand, one interprets the question as being about the expectation of the ratio (which is not the same as the ratio of expectations!) then things get more interesting. The expectation of $B/G$ is infinite, simply because $\infty$ is one of the possible values of $B/G$, since there is a non-zero probability that every family has a boy on the first try and so $G=0$. As for $G/B$ and related non-obvious matters, see the answers at the MathOverflow question cited in Qiaochu Yuan’s comment. An amusing calculation is that, when there is just a single family (so $n=1$), the expectation of $B/(B+G)$, which could be regarded as the proportion of boys among the children of the unique family, is $\ln(2)$. Specifically, $B/(B+G)$ has the possible values $1/k$ for all positive integers $k$, and the probability of the value $1/k$ is $2^{-k}$; the expectation is thus given by an infinite series, whose sum turns out to be $\ln(2)$.

The expected number of boys per family is clearly 1. The probability that a family will have exactly $k$ girls is $(1/2)^k \cdot (1/2) = (1/2)^{k+1}$, as the first $k$ “draws” need to be a girl, while the $(k+1)$’th draw needs to be a boy (if not for the latter, the family would have more than $k$ girls).

Hence, the expected number of girls in a family is

$$(1/2) \cdot 0 + (1/4) \cdot 1 + (1/8) \cdot 2 + (1/16) \cdot 3 +\dots = \sum_{k=0}^{\infty}(1/2)^{k+1} k$$

Due to a well known property of geometric distributions, this expection is equal to 1. The ratio of boys to girls will thus be 1.

If every *family* continues having children until they have one boy, then every *family* will have one boy; *except* those who haven’t stopped having children *yet*, or those that are not able to continue due to various circumstances. Let’s ignore these, as we don’t have enough information to consider such. (Likewise how many births are twin boys?)

The expected number of girls in *families which have a boy* is $1$, by reason that this count a geometrically distributed random variable (of failures before success), and for $\;G\sim\mathcal{Geo_0}(p)\;$, then $\;\mathsf E(G) = (1-p)/p\;$.

This leads to a *theoretical* expected ratio of $1:1$ boys to girls in the country.

Of course, we will *actually* expect the ratio to be somewhat higher, because there are *families which have no boys* (yet). But we can’t say what *proportion of* families these may be.

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