Originally posted by PBE6
If we bring this problem into the "real world", we get a very surprising result - this policy will affect the ratio of boys to girls, but in the opposite manner to which the government intended!
The above statement is true given the following assumptions:
1. The chance of giving birth to either a boy or a girl is 50%.
2. The number of families is inf ...[text shortened]... ially the government is shooting itself in the face with a policy like this. Lousy misogynists.
Actually...no. Like Dr. Scribbles said, banning people for flipping heads won't change the probability of the next toss of the person who flipped tails.
Example:
0.4 = P(boy)
0.6 = P(girl)
- Families who want 1 kid.
E(boys) = 0.4
E(girls) = 0.6
E(total of kids) = 1
Proportion of boys = 0.4
Proportion of girls = 0.6
- Families who want 2 kids
E(boys) = 2*0.4^2 + 0.4*0.6 = 0.56
E(girls) = 0.6+0.4*0.6 = 0.84
E(total) = 1.4
Proportion of boys = 0.56 / 1.4 = 0.4
Proportion of girls = 0.84 / 1.4 = 0.6
- Families who want 3 kids
E(boys) = 3*0.4^3+2*0.4^2*0.6+0.4*0.6 = 0.624
E(girls) = 0.6 + 0.4*0.6+0.4^2+0.6 = 0.936
E(total) = 1.56
Proportion of boys = 0.624 / 1.56 = 0.4
Proportion of girls = 0.936 / 1.56 = 0.6
And so on.
Families who want n kids.
E(boys) = n*0.4^n + 0.6*(n-1)*0.4^(n-1) + 0.6*(n-2)*0.4^(n-2) ... 0*0.6
#Note: E(boys) = 0.4* [n*0.4^(n-1)+0.6*(n-1)*0.4^(n-2)+...
E(girls) = 0.4^(n-1)*0.6+0.4^(n-2)*0.6+...+ 0.4^(n-n)*0.6
E(total) = n*0.4^(n-1)+0.6*((n-1)*0.4^(n-2)+(n-2)*0.4^(n-3)+...)
Proportion of boys = E(boys)/E(total)=0.4*E(total)/E(total) - see #Note
= 0.4
Proportion of girls = 1 - Proportion of Boys = 0.6
Hope I haven't made any typo in the equations there...
For any probability P(boy) = x, substitute 0.4 for x and 0.6 for (1-x)