Accordingly, they continue reproducing until they have a boy. Or until the mother can't give birth anymore. To make things easier, suppose each mother is able to produce ten children.

In such a society, will there be more boys or more girls? (than in a standard society)

Now, a new behaviour has arisen. If parents have a girl as first-born, they kill her. Let us suppose that they can't do it for next-borns, so that they continue the above tactic of reproducing until a boy is born.

How does this change affect your answer?

Then every mother has 1 boy.

1/2 of all mothers have 0 girl, 1/4 of all mothers have 1 girls, 1/8 of all mothers have 2 girls....

Number of girls per mother is then sum(n=1 to infinity) (n-1)/2^n = 1.

In this scenario, there will be same number of girls than boys in this scenario.

If only 10 children are allowed (sum n to 10 instead of infinity), then there will be 1013/1024 girls per mother, and 1023/1024 boys per mother.

The ratio of boys to girls is then indeed 1023/1013 =~ 1.01, or about 1% more boys.

If parents kill their first-born girl (but not their second born girl), then half of all mothers kill a girl, and hence there will be 501/1024 girls per mother.

The ratio of boys to girls is then 1023/501 =~ 2.04.

Why do you think this problem is not that simple ?

Actually, I am trying to make a connection between this problem and the one I ask about in Blackjack/Effect of position.

Normally, whatever the bound (10 here), the ratio should be still 1 on 1. Or p if you don't assume equiprobability to begin with.

^{n}. But the conditional distribution is a truncated geometric: pq

^{n}/q

^{10}, n=0 to 10.

Yet there are some families with ten girls and no boy.

Quote:kubikulannThis is a country where culture makes parents prefer to have a boy than a girl.

Accordingly, they continue reproducing until they have a boy. Or until the mother can't give birth anymore. To make things easier, suppose each mother is able to produce ten children.

In such a society, will there be more boys or more girls?

I am going for a new record for the number of times in a day that I can point out that E(X) + E(Y) = E(X+Y).

It will be 50-50. And, no betting system can make money in a 0EV game.

Quote:Now, a new behaviour has arisen. If parents have a girl as first-born, they kill her. Let us suppose that they can't do it for next-borns, so that they continue the above tactic of reproducing until a boy is born.

How does this change affect your answer?

Ooooh, a loss rebate on your first bet, if it loses. Now we are +EV!

If we assume that mothers can have infinitely many children, the question requires very little arithmetic, so that's the one that I'm going to answer.

Each mother has 2 children on average. Each mother kills 1/2 of 1 girl on average. Half the girls are killed (!!!) We are left with 2 boys for every 1 girl.

Nature often gives a young mother a female because they are easier to take care of and more experienced mothers get males who roam around more and are adventuresome.

After major population shakeups, females tend to be valued more. Post Black Death women were not put on a pedestal, they were needed.

Interesting. Do you have a reference?Quote:FleaStiffNature often gives a young mother a female because they are easier to take care of and more experienced mothers get males who roam around more and are adventuresome.

What does it have to do with the problem here?Quote:AxiomOfChoiceI am going for a new record for the number of times in a day that I can point out that E(X) + E(Y) = E(X+Y).

Ditto?Quote:AxiomOfChoiceAnd, no betting system can make money in a 0EV game.

Proof?Quote:AxiomOfChoiceIt will be 50-50.

This kind of sentence costs points to my students. Shows a misunderstanding of the concept of average. "Each" and "average" are contradictory. The correct form is "Half the mothers, on average, kill one girl."Quote:AxiomOfChoiceEach mother kills 1/2 of 1 girl on average.

Quote:kubikulannWhat does it have to do with the problem here?

The expected chance of boy or girl doesn't change if you have 1 child or 100. Stopping early doesn't matter. Every birth is a fresh roll of the dice. Kinda neat way to look at the problem (that I've seen before).

Quote:Proof?

E(Boy) = 0.5 and E(Girl) = 0.5

Hence 50% boys, 50% girls. The only wrinkle is the capping out at 10 boys.

Half the families have no girls, a quarter one girl, an eighth two, etc., one in 1,024 nine girls and one in 1,024 ten girls. That makes (256+256+192+128+80+48+28+16+9+10)/1024 = 1023/1024. There are the same number.

It's the same sort of thinking behind betting systems, as becomes obvious when you put the mothers in a line: the mothers change, but the babies stay the same.

The second obviously changes things.

Putting the mothers in a line again, a girl born after a boy is killed, and a girl born after a boy and 10n girls is killed. The first takes out half the girls; the second, .5 * Σ (1/1024)^k (k starting at 1), which is equal to one girl in 2,046. Add it up and we get 1024/2046 girls killed, or 512/1023 killed, leaving 511 for every 1023 who would have been born, which since the birthrates were equal, means 511 for every 1023 boys.