You may have heard about the one-child policy. Imagine if a country decides to have a one-son policy. The question is: will it lead to a gender imbalanced society? Intuition suggests that there will be more girls in the land because of this policy. Let’s analyse this by estimating the expected values.
The analytical solution
Assume that the probability of either sex is equal, 1/2. We have seen earlier that if p is the probability of something, then the expected value (denoted by E) or the average time we need to wait for the first success is 1/p. In our case, if N is the number of children in a family with 1/2 as the parameter, E(N) = 1/(1/2) = 2. Let Nm be the number of male children and Nf be the number of females, N = Nm + Nf and E(N) = E(Nm) + E(Nf) = 2. E(Nm) for a one-son policy regime has to be 1. That means E(Nf) = 2 – 1 = 1. So the expected number of males and females are equal! Not convinced? We will illustrate with examples.
Testing the concept
1 child scenario: The simplest case is one child in a family. The child can be M (male) or F (female). p(M) = 1/2, p(F) = 1/2. Nm(M) = 1, Nm(F) = 0, Nf(M) = 0, and Nf(F) = 1.
E(Nm) = Nm(M) x p(M) + Nm(F) x p(F) = 1 x (1/2) + 0 x (1/2) = 1/2. E(Nf) = Nf(M) x p(M) + Nf(F) x p(F) = 0 x (1/2) + 1 x (1/2) = 1/2. |
A bit of explanation on the notation:
Nm(M) is the number of boys in an M scenario, and Nm(F) is the number of boys in an F scenario.
Nf(M) is the number of girls in an M scenario, and Nf(F) is the number of girls in an F scenario.
Maximum 2 children in a family: The combinations are M, FM, FF.
p(M) = 1/2, p(FM) = 1/4, p(FF) = 1/4
Nm(M) = 1, Nm(FM) = 1, Nm(FF) = 0
Nf(M) = 0, Nf(FM) = 1, Nf(FF) = 2
E(Nm) = Nm(M) x p(M) + Nm(FM) x p(FM) + Nm(FF) x p(FF) = 1 x (1/2) + 1 x (1/4) + 0 x (1/4) = 3/4. E(Nf) = Nf(M) x p(M) + Nf(FM) x p(FM) + Nf(FF) x p(FF) = 0 x (1/2) + 1 x (1/4) + 2 x (1/4) = 3/4. |
Maximum 3 children in a family: The combinations are M, FM, FFM, FFF.
p(M) = 1/2, p(FM) = 1/4, p(FFM) = 1/8, p(FFF) = 1/8
Nm(M) = 1, Nm(FM) = 1, Nm(FFM) = 1, Nm(FFF) = 0
Nf(M) = 0, Nf(FM) = 1, Nf(FFM) = 2, Nf(FFF) = 3
E(Nm) = Nm(M) x p(M) + Nm(FM) x p(FM) + Nm(FFM) x p(FFM) + Nm(FFF) x p(FFF) = 1 x (1/2) + 1 x (1/4) + 1 x (1/8) + 0 x (1/8) = 7/8. E(Nf) = Nf(M) x p(M) + Nf(FM) x p(FM) + Nf(FFM) x p(FFM) + Nf(FFF) x p(FFF) = 0 x (1/2) + 1 x (1/4) + 2 x (1/8) + 3 x (1/8)= 7/8. |
More Proof?
We will do one final time, where the family waits for a boy but stops at 4. The combinations are M, FM, FFM, FFFM, and FFFF.
p(M) = 1/2, p(FM) = 1/4, p(FFM) = 1/8, p(FFFM) = 1/16, p(FFFF) = 1/16.
Nm(M) = 1, Nm(FM) = 1, Nm(FFM) = 1, Nm(FFFM) = 1, Nm(FFFF) = 0
Nf(M) = 0, Nf(FM) = 1, Nf(FFM) = 2, Nf(FFFM) = 3, Nf(FFFF) = 4
E(Nm) = Nm(M) x p(M) + Nm(FM) x p(FM) + Nm(FFM) x p(FFM) + Nm(FFFM) x p(FFFM) + Nm(FFFF) x p(FFFF) = 1 x (1/2) + 1 x (1/4) + 1 x (1/8) + 1 x (1/16) + 0 x (1/16) = 15/16. E(Nf) = Nf(M) x p(M) + Nf(FM) x p(FM) + Nf(FFM) x p(FFM) + Nf(FFFM) x p(FFFM) + Nf(FFFF) x p(FFFF) = 0 x (1/2) + 1 x (1/4) + 2 x (1/8) + 3 x (1/16) + 4 x (1/16) = 15/16. |