[EM] Puzzle example wasn't zero-info. ZI: vote if U>average

[MIKE OSSIPOFF]

>So the best strategy is to vote A only.
>
>This is different from the usual rule of using the mean utility as a 
>threshold,

But this isn't a zero-info election. We know everyone's sincere rankings,
and have probability information about how many candidates each person
will vote for.


>because the electorate is so small that one voter has a considerable chance 
>of becoming a tie-breaker, and therefore has a lot of power. The mean 
>utility rule is a valid approximation for large populations.

At least with only 3 candidates, it seems to me that the above-mean
rule is valid no matter how few voters there are.

A candidate's strategic value is the sum, over all j, of:

Pij(Ui-Uj)

What Pij really means is the probability that by voting for i, you'll
make or break a tie between i & j, the probability that either you'll
change a tie between i & j into an outright win for i, or that you'll
change an outright win for j into a tie between i & j.

What it takes, from the other voters, for that condition to exist
depends on whether you're voting for j. If you aren't voting for j,
then it's the probability that, without your ballot, i's vote total is
equal to j's, or one vote less. If you are voting for j, it's the
probability that i's vote total is equal to j's or one vote more.
(because then it amounts to a matter of whether or not you're voting
j over i).

With lots of voters, we can feel safe that those 2 probabilities are
about the same for all practical purposes. With only a few voters
simplifying assumptions like that might not always be valid. I don't
know if that causes a problem for defining Pij when there are very
few voters, or, if it does, how serious a problem it is. That's because
I don't have much experience with Approval with very few voters.

But when there are only 3 voters, it's assured that you're voting for
Favorite and not for Worst, and so the only question is whether to vote
for Middle. So that problem isn't there when there are only 3 voters.

So, how do we arrive at the above-mean strategy? As I said, strategic
value is the sum, over all j, of Pij(Ui-Uj). We vote for everyone whose
Si > 0.

With zero info, all the Pij are the same, and so it's just the sum,
over all j, of (Ui-Uj). That's the sum, over all j, of Ui minus the
sum over all j of Uj. The sum, over all j, of Ui is just Ui multiplied
by the number of candidates other than i: Ui(N-1). So it's
Ui(N-1) - (Sum, over all j, of Uj). That must be greater than zero.
So, Ui must be greater than (Sum, over all j, of Ui)/(N-1).

In other words, Ui must be greater than the average of the other
candidates' utilities. That's the same as saying that Ui must be
greater than the average of all the candidates' utilities.

There's nothing about this that excludes cases where there are very
few voters, except that, when there are more than 3 candidates, and
very few voters, I don't really have a clear definition of Pij, and
so it could be objected that I'm using something that I don't have
a definition of.

But the puzzle example just has 3 candidates, and so, unless I've
missed something, it seems certain to me that the above-the-mean
strategy would be valid if we had no information about the other voters.
In that case we'd maximize our expectation by voting for B as well as A.

Mike Ossipoff








>
> > At what B utility (keeping A and C constant) do your A and > AB votes' 
>strategic values become equal?
>
>(19/24)*100 + (5/48)*X = (23/48)*(100+X)
>
>(1500/48) - (18/48)*X = 0
>
>X = 1500/18 = 83.33
>
>This could explain why some IRV fanatics complain that approval voting will 
>lead to bullet voting. This opinion is only validd for elections involving 
>very small populations, where each voter has a lot of power. Anyone who 
>claims it will happen with a large electorate has failed to understand what 
>happens in approval voting with a large electorate: the break-even 
>utilities move towards the mean, which motivates most voters to cast their 
>ballots for more than one candidate.
>
>-- Richard
>
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