## Averaging Averages

Jorge Tierno sent me a link to the following puzzle:

There is a certain country where everybody wants to have a son. Therefore, each couple keeps having children until they have a boy, then they stop. What fraction of the children are female?

If we assume that a boy is born with probability 1/2 and children do not die, then every birth will produce a boy with the same probability as a girl, so girls will comprise half of all children.

Now, I wonder why everyone would want a boy? Y-chromosomes are much shorter than X-chromosomes. If a man wants to pass his genes to the next generation, a daughter should be preferable as she keeps more genes from the father. I am a mother of two boys, so my granddaughters will have my X-chromosome while my grandsons will have my ex-husband’s Y-chromosome, so to keep my genes in the pool I should be more interested in granddaughters.

But I digress. I started writing this essay because in the original puzzle link the answer was different from mine. Here is how the other argument goes:

Half of all families have zero girls, a quarter have 1/2 girls, 1/8 have 2/3 girls, and so on. If we sum this up the expected ratio of girls to boys is (1/2)0 + (1/4)(1/2) + (1/8)(2/3) + (1/16)(3/4) + … which adds to 1 − ln 2, which is about 30%.

What’s wrong with this solution?

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## LSK:

The incorrect solution uses rations where it should use the actual value. No families have 1/2 girls – the 1/4 of families referred to has 1 girl. Similarly, 1/8 of families have 2 girls, 1/16 have 3 girls, and so forth. Adding the infinite series:

n

——

2^(n+1)

from n=1 to infinity, we get a total of 1, for an average of 1 girl per family. And, by definition, there is 1 boy per family. Hence, the ratio is 50-50.

26 May 2011, 7:54 pm## Anonymous Rex:

See also this MO question:

26 May 2011, 7:55 pmhttp://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys

## anonymous:

The solution giving 1-ln(2) does not weight by the number of children when finding the expected number of girls. The result is the expected proportion of girls in a randomly chosen family – that is, if you first uniformly choose a family and then uniformly choose a child in that family, that is the probability you find a girl. If you uniformly choose a child in the country, then the probability of finding a girl is 50%.

In the first method, a member of a large family has less chance of being chosen. Since large family also has many girls, girls are less likely to be chosen, accounting for the probability of less than 50%.

26 May 2011, 9:30 pm## Eugene Krokhalev:

I’t very simple. The second solution does not consider next fact: families with 50% girls has two times less children then families with 75% girls and we need account families only with corresponding multiplier in the common sum.

27 May 2011, 1:32 am## Christ Schlacta:

There is a country where Pa and Ma are the king and queen. their country extends as far as the eye can see. they’re both nearly blind and can see about 30 yards, which is a nice sized home and yard. One year they decide to have a child. it’s a girl. The next year they try again and have another girl. This procedure continues for five more years, until on the eighth year they have a boy. The ratio in this country is 7:1 girls to boys, and my solution is correct.

27 May 2011, 2:13 am## Pratik Poddar:

LSK solution is correct

27 May 2011, 5:12 am## Sergei Bernstein:

The way I see it, the basic idea of the wrong solution is using the total probability theorem. The probability of picking a girl is equal to the sum over n of the probability of picking a family of size n times the probability of picking a girl given that we picked a family of size n. That idea is correct. The error is made when calculating the probability of picking a family of size n. They assume that the probability of picking any specific family is equal to the probability of picking any other family, but this does not take into account the sizes of the families. For example, we see that there are twice as many families with one child than there are with two children, but the two children families are twice as large, so the probability of picking a one child family and a two child family should be the same. Taking this into account, we can calculate that the probability of picking a family of size n is actually n*(1/2)^(n+1). Then, the probability of picking a girl is equal to the sum over n of (n*(1/2)^(n+1))*((n-1)/n), which equals 1/2.

27 May 2011, 8:36 am## Jenny:

I did it this way:

27 May 2011, 3:56 pmThere’s a 1/2 chance of having no girls (B), a 1/4 chance of having 1 girl (GB), a 1/8 chance of having two girls (GGB), etc. In general, there is a 1/(2^n) chance of getting n – 1 girls.

To find the expected value, you can add up the sum of the series (n – 1)/2^n. This is a sum of an infinite number of geometric series:

1/2 + 1/4 + 1/8 +…

+ 1/4 + 1/8 + 1/16 +…

+ 1/8 + 1/16 + 1/32 +…

and so on. The sum each series is 1, 1/2, 1/4, 1/8, etc. The sum of the sums is therefore [2].

## Jonathan:

What’s wrong with all of this?

When do we sample? This is frustratingly obvious. If we sample when everyone is done reproducing, but ah, no, we have a problem with that.

So we have to pretend that we have a finished, stable society. Hmmm, but ok.

Let’s take a better simplification. We have 2 families. Probably 1 will have one boy. And the other is likeliest to have a girl and a boy, or 2 girls and a boy. There are 3 or 4 children in this society, and 2/3 or 2/4 are boys.

Let’s go bigger. Four families. 2 have one boy. 1 has a girl and a boy. The last has 2 or 3 girls and a boy. So we have 4 boys out of 7 or 8 children.

Bigger: Eight families. 4 with one boy. 2 with a girl and a boy. 1 with 2 girls and a boy. and 1 with 3 or 4 girls and a boy. 8 boys out of 15 or 16 children.

Clearly we are moving toward LSK’s series.

30 May 2011, 12:42 pm