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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?
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Assume Prob(boy) = Prob(girl) = 1/2.
Prob(1 kid) = Prob(boy) = 1/2, Prob(2 kids) = Prob(girl)*Prob(boy) = (1/2)*(1/2) = 1/4, …, Prob(N kids) = (1/2)^N.
The proportion of boys to girls (b/g) for the country is the inverse of the expectation of the girls to boys (g/b). For 1 kid, we have 1 boy (b=1) and 0 girls (g=0) and an expected g/b of (g/b)*Prob(1 kid) = 0*1/2. For 2 kids, the expected g/b is g/b*Prob(2) = (1/1)*(1/4). For 3 kids, (2/1)*(1/8). For N kids, (N-1)*(1/2^N)
Thus, expectation of g/b = 0/2 + 1/4 + 2/8 + 3/16 + 4/32 = sum((N-1)*(1/2^N)) for all N=1,2,… = 1.00.
Hence, country’s b/g = 1.
Why not higher? Because we have a 1/2 chance of terminating with 1 boy and 0 girls.
An alternate explanation:
The expected number of children a family will have is 1/P(b) where P(b) is the probability of having a boy. If P(b)=0.5, then each family is expected to have two kids. For a family to have two kids, the first must be a girl and the second a boy.
e(g)=0*0.5+1*.25+. . .=1
e(b)=0.5+0.25+. . .=1
e(b)/e(g)=1
ratio is 1/1
explanation
boys will be=1/2*1(first one is boy)+1/2*1/2*1(second one isboy)+(1/2)^3*1(third one is boy)……=(1/2)/(1-1/2)=1
girls will be=1/2*0(first one is boy)+(1/2)^2*1(second one is boy)+(1/2)^2*2(third one is boy)…..=1
proportion is 1/1
note: p*n sum of products of probablity of having n (boys or girls) in a family is p
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