# [EM] Puzzle example X 1000

MIKE OSSIPOFF nkklrp at hotmail.com
Thu Mar 8 20:33:51 PST 2001

```> > What the
> > strategy is for the new example has no bearing on 0-info strategy.
>
>Well, you seem to be missing the point. I'm arguing that your own
>small numbers of voters.

My argument that it didn't depended on the assumption that all ties
are 2-way. Yesterday I admitted that they aren't, when there are
very few voters, and that therefore I don't have a demonstration
that expectation maximizing strategy with very few voters & 0-info
is the above-mean strategy.

>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.

Still, as far as vote totals are concerned, all 3 candidates are
looking just as promising before you vote, since each has 1/2

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 != j) are identical, and I'm saying
>that your first choice has little effect on those Pij, so you can ignore
>that effect in deciding your second vote. (You said the same in your
>last message: "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.") Since the Pij are identical, the strategy in this case
>is identical to ZI.

That's right. That's true with or without 0-info when there are
lots of voters, for the reason that you quote there.

Before I realized that there can be n-way ties with very few voters
(your puzzle brought that out), I thought that Pij didn't change
when we voted for j or didn't vote for j. Now I realize that isn't
so.

>
>I'm then extending the example down to small populations. When the
>population reaches four voters, you have my original example. But
>we've already seen that ZI strategy fails for that example.

But you don't
need that example to show that I haven't demonstrated that above-mean]
is valid with very few candidates, because I declared yesterday that

To show that above-mean isn't valid with few voters, a more direct
demonstration would be better.

>So we
>have to conclude that the Pij calculated from the external data only
>will diverge from the Pij calculated from external data plus what you

That's true with very few voters. I believed it wasn't, because
I hadn't considered n-member ties. That's why I withdrew my claim
yesterday.

>
>Perhaps you think it is too much of a leap to say that this sort of
>divergence would take place between ZI calculations of Pij and
>choice and no external data.

No, I realize that that's true, due to the possiblity of larger ties.
Pij, with very few voters & 0-info can be differnt depending on whether or
not you vote for j.

>But all I'm saying is that if you're
>deciding whether to vote for B, then any information about any
>other votes cast (whether by you or someone else) is relevant to
>the statistics of the situation; only in a large population the
>information about a single vote is so diluted as to be insignificant.

True. But it's still an open question whether or not above-mean
is valid with very few voters. You've told why my demonstration wasn't
valid, which I told you yesterday. I agree that my demonstration
wasn't valid, because it ignored ties with more than 2 candidates,
as I said yesterday.

>The latter was intended. The voters have similar tendencies but act
>independently of each other.

>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.

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.

>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.

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.

I make no claim one way or the other about whether above-mean is
valid for very few voters, though it seems more likely than not.

Mike Ossipoff

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