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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

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  • $\begingroup$ 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 $\endgroup$
    – WW1
    Sep 30 at 22:09
  • $\begingroup$ Check out the Geometric Distribution for example here en.wikipedia.org/wiki/Geometric_distribution $\endgroup$ Sep 30 at 22:25
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    $\begingroup$ 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. $\endgroup$
    – leonbloy
    Sep 30 at 22:25
  • $\begingroup$ 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. $\endgroup$
    – leonbloy
    Sep 30 at 22:31
  • $\begingroup$ 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). $\endgroup$
    – Brian Tung
    Oct 1 at 0:21

2 Answers 2

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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.

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  • $\begingroup$ (+1) You're totally right :). I was writing a very similar solution. $\endgroup$
    – bluemaster
    Sep 30 at 22:30
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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

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

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