3 ms·

Very cool- thank you for getting me to think about this. I get your results if you answer the question for any given sibling group (say, M M F, for example)

by dhimes 10y ago

Very cool- thank you for getting me to think about this. I get your results if you answer the question for any given sibling group (say, M M F, for example) Number of male brothers of males = 1 each * 2 Males = 2 Number of male brothers of females = 2 each * 2 Females = 2 and proceed to add all of those. In other words, if you weight the count by the number of siblings that have that count. If you just add up, for any group, how many brothers a female in that group has vs how many brothers a male in that group has, then the females win. In the case above, for example, males have 1 brother, females have 2. So, said another way, odds are that a given female has more brothers than a given male. But your argument led me to a result that was not at all intuitive (for me)- which is very cool.

tempestn 10y ago
In any given family of siblings that has both boys and girls, by definition the females will have more brothers than the males, yes. But it doesn't follow that "odds are that a given female has more brothers than a given male". The reason is that not all families have both. Families with multiple boys and no girls will bring the average up for boys. Families with multiple girls and no boys will bring it down for girls. The end result is that any given person has the same expected number of brothers as any other. That also makes sense if you think about it intuitively. As the best (IMO) answer to the SO post put it: "I'll write p for the gender ratio, i.e. the proportion of all people who are men. In real life p is close to 0.5, but this doesn't make any difference. In any random set of n persons, the expected (average) number of men is n⋅p. 1. Take an arbitrary child x, and let n be the number of children in x's family. 2. Let S(x) be the set of x's siblings. Note that there are no gender-related restrictions on S(x): It's just the set of children other than x. 3. Obviously, the expected number of x's brothers is the expected number of men in S(x). 4. So what is the expected number of men in this set? Since x has n−1 siblings, it's just (n−1)⋅p, or approximately (n−1)÷2, regardless of x's gender. That's all there is to it."
- dhimes 10y ago
  Nice. Thanks.

by dhimes 10y ago

tempestn 10y ago

In any given family of siblings that has both boys and girls, by definition the females will have more brothers than the males, yes. But it doesn't follow that "odds are that a given female has more brothers than a given male". The reason is that not all families have both. Families with multiple boys and no girls will bring the average up for boys. Families with multiple girls and no boys will bring it down for girls. The end result is that any given person has the same expected number of brothers as any other. That also makes sense if you think about it intuitively. As the best (IMO) answer to the SO post put it: "I'll write p for the gender ratio, i.e. the proportion of all people who are men. In real life p is close to 0.5, but this doesn't make any difference. In any random set of n persons, the expected (average) number of men is n⋅p. 1. Take an arbitrary child x, and let n be the number of children in x's family. 2. Let S(x) be the set of x's siblings. Note that there are no gender-related restrictions on S(x): It's just the set of children other than x. 3. Obviously, the expected number of x's brothers is the expected number of men in S(x). 4. So what is the expected number of men in this set? Since x has n−1 siblings, it's just (n−1)⋅p, or approximately (n−1)÷2, regardless of x's gender. That's all there is to it."

dhimes 10y ago
Nice. Thanks.