*“But for the expected gender *ratio*, all *family* outcomes get weighed the same, which means the single-kid families get twice as much “weight per child”, which skews the ratio in favor of boys.”*

To which you responded (22):

*“The part I can’t get through my head is why the expected ratio must weigh all families equally. Why would we take averages of averages?”*

Bennett replies in (25) explaining that he wasn’t saying in multi-family countries you would weight each family equally, he was talking about a single family with various possible outcomes.

If I’m reading your subsequent question about “averages of averages” and “ignoring family size” correctly, this is still stemming from a misunderstanding of Bennett’s original post.

In the original problem there is no equal weighting of families — the problem simply asks for the (expected) ratio of girls in the (entire) population. As math_geek says in (14), this is E(G/(G+B)) where G and B are the total number of girls and boys for the country.

As a brief example, suppose there is a 2-family country and in this particular case the resulting families turn out to be:

1)B

2)GGGGB

Then the ratio we’re interested in, G/(G+B) is 2/3 (in this specific result of one roll of the dice, so to speak). The ratio is not an average of the individual families ratios.

]]>Each country where people follow the “keep having kids till you have a boy” rule is one throw of the dice. Each throw of the dice produces one number (the ratio of girls to (girls+boys) in that country.

We want to know the ratio of girls to girls+boys in the average country, just as we want to know the outcome of an average dice roll. Expected value is a another word for average.

(You might say, well, no, I want to know the ratio in a *particular* country, not the average country. But that’s of course hopeless, just as it’s hopeless to predict the outcome of a particular dice roll. We *can* predict averages; we can’t predict individual outcomes.)

]]>I understand your dice example but I don’t understand why we would do averages of averages ignoring family size. Your dice example doesn’t create different size groups of dice and average the averages of each group. I know what an expected value is but is that the same thing as expected ratio? If so, the ER for dice is 3.5 and the ER for coin with 0 on one side and 1 on the other is 0.5.

I’m not trying to be difficult because I’m sure I’m missing something.

]]>Then I imagine that there is some non-zero probability that the mix of the first batch of children is, say, 80 boys and 20 girls. I then realize that the same probability applies to an alternative state of the world for the first batch being 20 boys and 80 girls. It’s symmetrical for the initial batch. But for the second batch of children, if you have 80 couples that had a boy with the first batch, then that’s 80 couples that have contributed a lot of boys to the ratio that are now lost to contribute anything else. So it is hard for the remaining 20 couples to catch the ratio “back up” so to speak, even IF they have all girls in their second batch, which is highly unlikely. On the other hand, when you look at the mirror image initial batch/scenario of 80 girls and 20 boys, your future states from here are not the mirror image since you are allowing the 80 people who had a girl to keep trying. Since you’re allowed to keep trying based on sex of the child, when you consider all of the possible states of the world, it makes sense that it’s not a 50/50 expectation. It’s this asymmetry that I think is at the heart of this problem.

]]>Post number 100 at the link above really helped me get the gist of this problem mentally, even if I couldn’t really technically do the math on paper. Thank you Harold.

]]>When you throw a die, the expected outcome is always 3.5, though the actual outcome could be 1,2,3,4,5 or 6.

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