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)

  • $\begingroup$ Hint: For every child born, what is the probability that it is a boy? So if $n$ babies are born in total, how many do you expect them to be boys? (Does it matter which family is having a baby?) $\endgroup$ – TMM Oct 22 '12 at 12:51
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    $\begingroup$ There is an extremely thorough discussion of this question on MO: mathoverflow.net/questions/17960/… $\endgroup$ – Qiaochu Yuan Oct 22 '12 at 12:54

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.

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    $\begingroup$ As Douglas Zare points out in his answer to the MO question above, the question is ambiguous; there is no definite proportion of girls to boys, only 1) an expected number of girls, 2) an expected number of boys, 3) an expected proportion of girls, and if you interpret the question as asking for 3) then you can't divide 1) by 1) + 2) to compute it. $\endgroup$ – Qiaochu Yuan Oct 22 '12 at 14:21

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 $\ln(2)$.

  • $\begingroup$ Andreas: I do not think one can reasonably interpret this question as asking for a ratio of expectations; it asks about the proportion of boys to girls, not of expected boys to expected girls. So it's asking for B/G, or perhaps one of the related ratios G/B or G/G+B, etc. Each of these is a random variable, and ordinarily one of the most interesting properties of a random variable is its expectation. So I don't think the value of E(B)/E(G) is responsive to the question, but if someone else does, then, unlike Monty, I won't object to someone computing something other than I would have. $\endgroup$ – WillO Jun 5 '13 at 19:02
  • $\begingroup$ @WillO: It's debatable. The wording in the question here is "After some time, what is the proportion of boys to girls in the country"? If $N$ is the number of families in the country, and $B_i$ and $G_i$ are the number of boys and girls in the $i$th family (obviously $B_i=1$, for instance) and $\mathcal{B}=\sum_{i=1}^{N} B_i$ and $\mathcal{G}=\sum_{i=1}^N G_i$, then I'd say the question is asking for "the proportion of" $\mathcal{B}$ to $\mathcal{G}$. You could interpret this as the random variable $\mathcal{B}/\mathcal{G}$ I guess, but note that $E[\mathcal{B}/\mathcal{G}] \neq E[B_i/G_i]$. $\endgroup$ – ShreevatsaR Jun 5 '13 at 19:10
  • $\begingroup$ Shreevatsa: Yes, it seems clear to me that the problem is asking for the properties of the random variable ${\cal B}/{\cal G}$, which of course is not the same as $B_i/G_i$, unless $n=1$. (Slight correction to your comment, though: The expectations are equal, both being infinite.) Again, I accept that what seems clear to me might not seem clear to others, and one can calculate anything at all as long as one indicates what one's calculating. Monty's problem, of course, is that he keeps saying ridiculous things like $G/B=1$ (without taking expectations), which anyone can see is wrong. $\endgroup$ – WillO Jun 5 '13 at 19:29

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).

  • $\begingroup$ WillO: The ratio of B/G = 1 = G/B. Both with the statistical errors (which in case that only boys are born extends to infinity). $\endgroup$ – user81104 Jun 5 '13 at 17:40
  • $\begingroup$ monty: I'm sorry that I have no idea what it means for two things to be "equal with statistical errors"; could you please provide the definition? $\endgroup$ – WillO Jun 5 '13 at 18:58
  • $\begingroup$ This sequence 01|01|001|1|001|0001|1|01|1|01|01|1|1| does not supply G/O = 1. That means , there is a deviation. Another sequence would show another deviation. Infinitely many sequences woulkd show 1 - also infinitely many trials of a single family! Not 31% G/(G+B). $\endgroup$ – user81104 Jun 5 '13 at 19:07
  • $\begingroup$ Monty: Perhaps my question wasn't clear. Can you please define the phrase "equal with statistical errors"? $\endgroup$ – WillO Jun 5 '13 at 19:25

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.

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    $\begingroup$ The ratio of two expected values is not usually the expected value of their ratios; see Douglas Zare's answer to the MO question above. $\endgroup$ – Qiaochu Yuan Oct 22 '12 at 14:20
  • $\begingroup$ Monty: Actually the question concerns the ratio of boys to girls, not girls to boys. Do you also believe the ratio of boys to girls is "1/1 in every case"? $\endgroup$ – WillO Jun 5 '13 at 15:50

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.


protected by Asaf Karagila Jun 5 '13 at 19:11

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