[herodotus]

Because we might score -1, -2 or worse if 2 or 3 busses went in the other direction before ours came, but if ours came first, we score 1. We get on the bus and thus don’t know if another one or more arrives first on our side.

[repsilat]

This reminds me of a mathematical paradox that makes me doubt your conclusion: "In this country, every couple wants to have one daughter. They keep having children until they have a daughter, and then they stop. What gender balance should we expect?"

Couples can have any number of sons, and every couple has exactly one daughter. Still, the accepted mathematical solution is an equal gender ratio for the couples' children.

[Terr_]

I think the "paradox" comes from how people implicitly assume "any number of sons" is somehow distributed or weighted in a way that favors towards numbers of 1 or above.

In contrast, "0 sons" is going to describe a full half of all marriages.

[jon_richards]

Same situation with the bus.

[pwagland]

Not really. In the son/daughter case, the calculations are: expected daughters: 1 expected sons: 1/20 + 1/41 + 1/82 + 1/163 + 1/324 + …

So number of expected daughters = 1, number of expected sons = 1. In practice since women can't have an infinite number of children, then this wouldn't be an infinite series, so the real number of expected boys would be lower than one, but there you go…

Now, for the bus case, you get +1 if your bus turns up first, and -1 for every other bus that turns up first. Assume that it is completely random, then: expected + score is: 1/2 1 expected - score is: 1/2 * -1 + 1/4 * -2 + …

Expected + is 0.5, expected - is -1.

[repsilat]

I guess it would be balanced if the rule was,

> 1 when your bus turns up, -1 for every bus going the other way.

[selestify]

Isn't that the same as what the OP's rules were?

[pwagland]

Correct.

[DalekBaldwin]

The expected number of sons is 1, and the expected number of daughters is 1 (by the framing of the problem, in every possible scenario, there is exactly one daughter), but the expected value of the ratio is not 1:1. E[X]/E[Y] = E[X/Y] is not a valid identity.

http://www.thebigquestions.com/2010/12/22/a-big-answer-2/

[Pristina]

I read that and it seems wrong. The question asked "what fraction of the pop is female" but his argument is that 3 families of 4 girls and 1 family of 12 boys make the fraction of girls in the average family 75% (the average of 100% x3 and 0% x1) which is non-sensical to me.

>E[X]/E[Y] = E[X/Y] is not a valid identity.

is completely irrelevant here because it is being used to point out that a non-answer is wrong.

[DalekBaldwin]

He starts with a simpler problem to elucidate the principle that is key to the original problem. Do continue reading.