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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?

My solution (not finished)

If we assume that the probability of having a girl is 50%, the set of possible cases are:

Boy (50%)

Girl, Boy (25%)

Girl, Girl, Boy (12.5%)


So, if we call G the number of girls that a family had and B the number of boys that a family had, we have:

$B = 1$

$P(G = x) = (1/2)^{x+1}*x$


$G = \Sigma (1/2)^{x+1}*x$

I feel like the sum of this infinite serie is 1 and that the proportion of girls/boys in this country will be 50%, but I don't know how to prove it!


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In this example, you don't need to sum the infinite series. For each birth, the probability of a boy is 50% and the probability of a girl is 50% (for the purposes of this exercise; the real world is more complicated). Therefore, the ratio for all births must be 50:50. The number of previous births in each family, and the genders of those children, is irrelevant. –  Mike Scott Feb 4 '11 at 17:28
There is a thorough discussion of this question at mathoverflow.net/questions/17960/… for anyone who's interested. –  Qiaochu Yuan Feb 4 '11 at 17:29
As the MathOverflow discussion Qiaochu linked to points out, your approach will get the approximately right answer but for the wrong reasons. The expected ratio of boys to girls is not necessarily equal to ratio of the expected number of boys to the expected number of girls. –  Rahul Feb 4 '11 at 23:21
thebigquestions.com/2010/12/22/a-big-answer-2 why is this not the answer? –  KevinDTimm Oct 7 '14 at 21:03

3 Answers 3

up vote 2 down vote accepted

Mike Scott is correct that you don't need to sum the series, but suppose you want to. Each family has 1 boy-that is easy. Each family has 50% chance of no girls, 25% chance of 1, etc. So the average number of girls is $$\sum_{i=0}^\infty \frac{i}{2^{i+1}}$$ The way to sum this is to remember that $$\sum_{i=0}^\infty a^{-i} = \frac{1}{1-1/a}$$ Now if you take the derivative with respect to a and evaluate it at a=2 So $$\frac{d}{da}\sum_{i=0}^\infty a^{-i} =\frac{d}{da} \frac{1}{1-1/a}=\frac{1}{(a-1)^2}=1$$ for $a=2$. So there is an average of one girl per family as well

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This is a trick question!

This question is very simple if you just learn to accept that most of the information given is completely irrelevant...

It doesnt matter how many families continue to have children and how many stop at 1 or 2...its no more relevant than what car they drive...

None of the information provided alters the statistical probability of a child born being male or female...its still 50%

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If the probability is 50% then we can expect from four families to typically have:

B B, B G, G B, G G

However because families stop when they have a boy, we do not gain the boy from the first row or the girl from the second row: B B, B G,

Therefore we actually get three girls and three boys.

From 8 families we would expect:

B B B, B B G, B G B, G B B, B G G, G B G, G G B, G G G

and we would actually keep: B, B, B, GB, B, GB, GGB, GGG,

for a total of 7 G and 7 B.

I expect proportion to remain 50% for each expected outcome from larger numbers of families but haven't proved anything as such. This is a lot simpler though and doesn't require summation notation or any advanced maths.

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