[EM] Puzzle example X 1000

[Richard Moore]

MIKE OSSIPOFF wrote:

> >I'm giving a non-ZI case with perfect symmetry
> >in the probabilities
>
> I don't call that perfect symmetry, when there's .5 probability that
> the A voter will vote for B, and there's 0 probability that the B
> voter will vote for A.

Different types of symmetery. It's a circular symmetry, not a bilateral one.
A voters have a 50% chance of helping B, B voters a 50% chance of
helping C, and C voters a 50% chance of helping A.

> Still, as far as vote totals are concerned, all 3 candidates are
> looking just as promising before you vote, since each has 1/2
> probability of getting a 2nd vote, not counting your votes.
>
> Maybe, then, that's "like" a 0-info election, even though it isn't
> one. Maybe you can say that if above-mean doesn't maximize your
> expectation in that example, it wouldn't in a 0-info election
> eitlher, with the same number of voters & candidates. I'd prefer a more
> direct demonstration, though.
>
> , so all Pij
>
> We were using Pij with a meaning that was about 2-way ties, a meaning
> that isn't so useful when there are very few voters. I suggest
> that now Pij should mean the probability that my vote for i will
> change the winner from j to i. And let's say that my vote arrives
> late, when the count is concluded, and any tie solved, and a winner
> declared. Then when my vote arrives the count is redone. That way,
> we can say for sure whether or not my ballot changed the winner from
> j to i.

I think I tried to argue this when I spoke of an array of delta-P values
a number of days ago.

> >If your B vote has a probability k of helping B
>
> ...of making B win when he wouldn't otherwise have won?
>
> , it has a probability of
> >k/2 of hurting A and a probability k/2 of hurting C (B is just as likely
> >to be tied with A as with C).
>
> But you said yourself it isn't meaningful to talk about how likely
> your vote is to change the winner from someone else to B without
> saying whether you're voting for that other candidate too.

Maybe in the small-population context. In the large-population context
your A vote doesn't make a significant change in the delta-P created by
your B vote. You will recall that I wrote the paragraph quoted above in
the context of the 3001-voter example.

> A is more favored than C is. Two people are voting for A for sure,
> now that you're joining the election. It doesn't seem reasonable
> to say that voting for B has to equally reduce A's & C's chances of
> winning.

You're arguing the small-population case here, aren't you?

> >So your B vote's strategic value
> >(assuming B's utility is 70 as in the original example) would be:
> >
> >-100k/2 + 70k = -50k + 70k = 20k
> >
> >which is positive since k is positive. So vote AB.
>
> That seems a questionable conclusion for the reason that I
> stated above.

Again: Large or small population?

> Anyway, as I said before, even if your above argument were valid,
> it would be better to give us a more direct demonstration that
> above-mean isn't valid with very few voters.

Actually I started working out the case for three other voters, three
candidates, and zero info, on paper. I didn't finish it. First, a few basic
assumptions are needed: Are all three voters guaranteed to cast votes?
Do we assume nobody will cast an ABC ballot? Do we assume they
are using the above-mean strategy (even though it may not be the best
strategy for this small population)? Assuming yes for all three of these
questions, there are 6x6x6 equally probable permutations to consider,
so it's a non-trivial problem. Interestingly, these assumptions aren't
particularly important when dealing with large populations, because the
random actions of a large population (such as not making it to the
polling place on election day) tend to cancel each other out.

If I get some time to finish the solution I'll happily post it here. But don't

let that stop you from trying the same. Maybe you'll beat me to it.

 -- Richard