[EM] Puzzle example X 1000

[Richard Moore]

MIKE OSSIPOFF wrote:

> > > 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?
>
> In an actual voting situation we can't rule out the possibility that
> 1 or more voters could abstain completely, and so, now that you mention
> it, that situation  should be considered.

We can't rule it out, but without some basis for assigning a probability to this

event, how can we factor it in? Maybe in the real world you could look at
historical patterns. For an isolated example, there's nothing to go on. Might
as well solve the problem for the case where everybody is certain to vote
for at least one candidate.

> >Do we assume nobody will cast an ABC ballot?
>
> That seems a reasonable assumption, since with very few voters it's
> easy to tell everyone that they gain nothing from such a ballot.

Agreed.

> >Do we assume they
> >are using the above-mean strategy
>
> That question hadn't occurred to me. If we assume they're using it,
> and, when they do, it benefits everyone who uses it, including me,
> then that shows that above-mean works in that example. We could assume
> that everyone perceives the same position-scale and is positioned with
> their favorite. I'd thought that we had to assume that we know nothing
> at all about the other voters, but it does seem reasonable to assume
> that they're using above-mean strategy, and then if that works for
> everyone then above-mean is right, and if it doesn't, then above-mean
> is wrong with that number of voters & candidates.

If we assume above-mean strategy for everyone, and assume a uniform
probability distribution for the utility of each person's middle candidate,
then we would get a 50-50 split between each person's probability of
voting one or two preferences. That assumption makes the problem
simpler. In the real world, some people will vote rationally, and some
may vote irrationally (like the Gore-Buchanan voters, though that
irrationality was unintentional). The rational voters may apply
above-the-mean strategy, or if they believe they are in a situation
where that strategy is not optimal (or they are unaware of it), they may
apply some other strategy (but since they are rational and it is approval,
at least they will vote sincerely). Given zero information about the
likelihood of these scenarios, I think it would be impossible to improve
on the 50-50 assumption in any foolproof way.

> I've meant to add up your benefit by voting for B in addition to A,
> with all 729

I'm not sure about that number. If you add the no-vote possibility, but
not the ABC vote, each person can vote one of 7 ways, so it becomes
7x7x7 or 343. Did I miss something?

> of the ways that those 3 people could vote, to find out
> if that summed benefit is positive or negative. But it's also worthwhile
> checking whether above-mean strategy works for everyone if everyone uses it,
> and no one votes ABC, but some people might abstain completely.
>
> But it seems to me that I've shown that above-mean strategy maximizes
> utility with 3 candidates, even with very few voters, if we know nothing
> about how those other people will vote.

I reserve any further judgement until I see the actual values worked out.
I probably won't take this any further for at least the next several days,
but I did think of some notation that might help:

P(xyz) = probability that x, y, and z are in a 3-way tie
P(xy,z) = probability that x and y are in a 2-way tie, with z trailing by 1 vote

P(xy,,z) =  probability that x and y are in a 2-way tie, with z trailing by more
than 1 vote
P(x,yz) = probability that x is in the lead, with y and z tied and trailing by 1
vote
P(x,,yz) = probability that x is in the lead, with y and z tied and trailing by
more than 1 vote
P(x,y,z) = probability that x is in the lead, with y trailing by 1 vote and z
trailing by more than 1 vote

Obviously, for ZI, before you add in your known first-place vote,

P(ab,c) = P(ac,b) = P(bc,a)
P(ab,,c) = P(ac,,b) = P(bc,,a)
P(a,bc) = P(b,ac) = P(c,ab)
P(a,,bc) = P(b,,ac) = P(c,,ab)
P(a,b,c) = P(a,c,b) = P(b,a,c) = P(b,c,a) = P(c,a,b) = P(c,b,a)

At least it's a start.

 -- Richard