[EM] 3 candidates, few voters, 0-info

[MIKE OSSIPOFF]

Richard wrote:

>MIKE OSSIPOFF wrote:
>
> > It seems to me that the above-the-mean strategy is still valid
> > no matter how few voters there are, as long as there are only
> > 3 candidates.
>
>Then can you explain the following? I posted this at the end of my
>previous message:
>
>"If we replace voters 1, 2, and 3 with three groups of 1000 voters
>each, with the preferences and voting probabilities indicated in the
>example, then I think the mean utility strategy would work (because
>of the equal probabilities) even though it is not a true ZI case."

I can't explain it because I don't believe that it's so. Write the
example. Show that your expectation is better if you vote for A & B
than it would be if you voted only for A.

But you needn't, because I've shown that, even with very few voters,
0-info strategy is to vote for the above-mean candidates.

If you disagree, then tell specifically what you believe is incorrect
in that demonstration.

>
>I think since all the Pij are equal in this case, the mean utility
>strategy
>will work. So why doesn't it work with the same probabilities in the
>small-population case?
>
>As the population shrinks, won't the Pij start to diverge (because of
>the effect of your certain A vote),

No, because Pij has nothing to do with your vote. Pij is about how
_other people_ vote. Pij is about a certain tie or near-tie existing
before you vote.

Now yes, when there are more than 3 candidates, Pij has a different
meaning if you vote for j. As I said, with 4 candidates, A, B, C, & D,
which you like in that order, if you don't vote for C then Pbc is
the probability that B's vote total is equal to C's, or is one vote less.

That's because Pij is defined as the probability that voting for i
will change an ij tie into an i victory, or change a j victory into an
ij tie.

If you're voting for C, then Pbc is the probability that C's vote total
is equal to B's, or is one vote less.

So Pbc means something different when you're voting for C.

But, since we know nothing about the other voters or how they'll vote,
those 2 Pbc that I described are the same. And for the same reason,
they're the same as Pba & Pbd.

So even though the meaning of Pij is different if you vote for j,
the value of Pij doesn't change if you vote for j. All the Pij remain
equal. Because we know nothing about how the other voters will vote.

Pij is only about the vote totals configuration that exists before
you vote.

Now, with very few voters, and nonzero-info, and more than 3 candidates, 
then yes, it then makes
a difference if you vote for j. In that case, it's no longer possible
to determine each Pij independent of whether you voted for j. It's
then necessary to calculate the benefit to you of each combination
of candidates you could vote for, because Pij can have two distinct values, 
depending on whether you vote for j. But with 0-info, there's
no reason to believe that any of those near ties that I described are
any more or less probable than eachother. They're equally probable.

But if there are lots of voters, then those 2 Pij can be assumed to be
the same, because a difference of 2 votes isn't enough to change the
probability density function by any appreciable percentage.

And, even with nonzero-info, with very few voters, if there are just 3 
candidates, there
is only 1 Pba and there's only 1 Pbc, and the problem described in
the previous paragraph doesn't exist.

p.s. In my recent demonstration about 0-info strategy, I mistakenly
said that Sum over all j of Pij(Ui-Uj) = Sum over all j of (Ui-Uj).

They aren't equal, but, for the matter of whether the sum is negative
or positive, that isn't affected by dividing each of its terms by
the same number. Whether it's positive or negative depends on differences
among its terms, and the relative magnitudes of those differences are
unaffected by dividing all the terms by the same number.

Of course, if by voting for i, you'll change an ij tie to an i victory,
or change a j victory to an ij tie, then the value of that for you
is (Ui-Uj)/2.

I've shown why 0-info strategy is always the above-mean strategy.


Mike Ossipoff



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