If population keep trying till they have girl child, what will be the probability of population having more girls than boys and vice versa?

I was solving this problem:

In a world where everyone wants a girl child, each family continues having babies till they have a girl. What do you think will the boy to girl ratio be eventually? (Assuming probability of having a boy or a girl is the same)

The solution given was:

Suppose there are N couples. First time, N/2 girls and N/2 boys are born (ignoring aberrations). N/2 couples retire, and rest half try another child. Next time, N/4 couples give birth to N/4 girls and rest N/4 boys. Thus, even in second iteration, ratio is 1:1. It can now be seen that this ratio always remain same, no matter how many times people try to give birth to a favored gender.

My doubt is that will following be the case:

P(population will have more girls)= P(population will have equal number of boys and girls)= 1/2

Consider there are 16 couples

• 8 give birth to girls and hence stop. 8 give birth to boys, so they give another chance.
• 4 give birth to girls and hence stop. 4 give birth to boys, so they give another chance.
• 2 give birth to girls and hence stop. 2 give birth to boys, so they give another chance.
• 1 give birth to a girl and hence stop. 1 give birth to a boy, so they give another chance.
• Note till now there Number of boys = Number of girls
• Now probability that a single remaining couple give birth to a girl is 1/2. In that case they will stop and there will be one more girl than boys in the population. Probability that a single remaining couple give birth to a boy is 1/2, in which case they will give another chance in which they will again have a 1/2 probability of giving birth to boys, thus again balancing girl-boy ratio.

So am I correct with two facts:

Fact 1:

P(population will have more girls than girls)= P(population will have equal number of boys and girls)= 1/2

Fact 2:

P(population will have more boys than girls) = 0

• it must be possible to have more boys, maybe one couple has 17 boys in a row . With 16 couples you must always end up with 16 girls
– WW1
Sep 30 at 22:09
• Check out the Geometric Distribution for example here en.wikipedia.org/wiki/Geometric_distribution Sep 30 at 22:25
• No, both facts are clearly wrong. What the quoted answer shows is that in average there will be as many boys as girls. It says northing about those probabilties you consider. Sep 30 at 22:25
• In particular, if we call $A,B,C$ the probabilities that there are more/equal/less boys than girls, the above does not imply that $A=C$ nor that $A=B$ . In particular $B$ should tend to zero. Sep 30 at 22:31
• Side comment: I think, if you're in doubt, it's best not to call these propositions "facts." It's not entirely a trivial comment: Readers are likely to react negatively to statements presented boldly as "facts" if they are in fact incorrect (as they are here, as it turns out). Oct 1 at 0:21

I'll give an alternative solution to the original problem.

For each $$i\in \{1,\ldots,N\},\$$ let $$X_i\$$ be the random variable, "number of babies a couple has, up to and including their first girl." Then, for each $$i,\ X_i\sim \text{Geo}\left(\frac{1}{2}\right),\$$ where $$\text{Geo}\$$ denotes the Geometric distribution. We know that $$\text{Exp}(X_i) = \frac{1}{\frac{1}{2}} = 2\$$ for every $$i.$$

By the algebra of linear combination of random variables, the expected total number of births (in the population) such that every couple in the population has a girl, is given by $$\text{Exp}(X_1 + X_2 + \ldots + X_N) = \text{Exp}(X_1) + \text{Exp}(X_2) + \ldots + \text{Exp}(X_N) = 2 + 2 + \ldots + 2 = 2N.$$ But each couple has exactly one girl, so that's $$N$$ girls in total, and therefore $$N$$ boys in total also. So the expected boy to girl ratio is $$1:1.$$

As for your "Facts", they are both clearly wrong, and to me show a lack of understanding of discrete probability distributions (Like Geometric or Binomial), but I'm tired and I'll let someone else have the burden of explaining why your facts don't make sense.

• (+1) You're totally right :). I was writing a very similar solution. Sep 30 at 22:30

If there are N couples, the probability that couple 1 has $$k$$ boys and all the other couples have no boys is given by ... $$P_{k, 0, 0,0...}= \frac{1}{2^{N+k} }$$ A moment's reflection should convince you that this same probability applies to any possible way of distributing the $$k$$ boys among the $$N$$ couples.

By stars and bars, the number of possible $$N$$-tuples of whole numbers summing to $$k$$ is given by... $$N_k = \binom{N+k-1 }{N-1 }$$ So $$P(k \text { boys }) = \frac{\binom{N+k-1 }{N-1}}{ 2^{N+k}}$$

In particular, if there are $$N=16$$ couples and there are equal numbers of girls and boys

$$P(16 \text{ boys and} 16 \text{ girls })=\binom{31}{15}2^{-32}\approx 0.07$$