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

Richard Moore rmoore4 at home.com
Sat Mar 3 17:44:20 PST 2001

MIKE OSSIPOFF wrote:

> >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.

You're right that it isn't a zero-info situation. But the probabilities are so
evenly distributed that it resembles one.

I do think that the reason this example fails the mean utility strategy is
related to the small population, rather than the presence of information
about other voters. Consider the situation if we really did have zero
information in this example. Then if we check the probability changes
caused by our voting for A only, we get +1 (for A) and -1/2 (for B
and C). The probabilities are scaled for convenience.

Now we want to know what adding a B vote to our A vote will do.
If you apply ZI strategy at this point, the probability changes would still
be +1 (for B) and -1/2 (for A and C). Then since B has a higher-than-
mean utility, you would still include B.

But wait: Is it correct to still use ZI strategy for deciding to vote for B?
You've already decided to vote for A, and that is information you
can use. Given the small population, your vote for A significantly
affects the outcome probabilities. So it isn't a ZI situation when you
are trying to decide whether to make a second (or later) choice.

With a large population, your A vote is so diluted you can ignore its
effect, and use the mean utility strategy. With a small population,
even with zero information about the other voters, you should
at least consider the information you have about yourself.

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.

-- Richard