[EM] Puzzle example wasn't zero-info. ZI: vote if U>average
MIKE OSSIPOFF
nkklrp at hotmail.com
Sat Mar 3 20:01:49 PST 2001
>But wait: Is it correct to still use ZI strategy for deciding to vote for
>B?
I think so, though I admit that I'm new to Approval with very few
voters.
Yes, it's the way your votes affect eachother's effects that makes
Pij more difficult to define with very few voters.
But since it's certain that you vote for A & not for C, I don't
think those votes cause a problem.
Isn't it true that we can say that there's a certain probability
that if we vote for B, that will change an AB tie into a B victory
or change an A victory into an AB tie? Given the (unknown) way
the other people voted, and the fact that you voted for A?
And likewisee can't we speak of the probability that by voting for
B, you could change a BC tie into a C victory or change a C victory
into an BC tie? Given the unknown way the other people voted, and
the fact that we didn't vote for C?
So even though we don't know what they are, we have meanings for
Pba & Pbc.
Since we voted for A, voting for B just means not voting A over B,
and so Pba is the probability that A's vote total is equal to B's
or one vote less, before we vote. Since we don't vote for C,
Pbc is the probability that B's vote total is equal to C's vote total
or one vote less, before you vote.
Since we know nothing about the other voters, is there any reason
to expect those 2 P to be different? From our zero info about the
other voters, aren't Pba & Pbc equal?
So Pbc(Ub-Uc) > Pba(Ua-Ub)--the requirement for voting for B,
can be replaced with Ub-Uc > Ua-Ub.
2Ub > Ua+Uc
Ub > (Ua+Uc)/2
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.
With more than 3 candidates I don't know. I don't know how much
of an unsolvable problem for defining Pij is caused by the way
one's votes affect eachother's effects.
Mike Ossipoff
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