# [EM] error

MIKE OSSIPOFF nkklrp at hotmail.com
Fri Mar 9 12:45:26 PST 2001

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I've just realized that the probability that voting i over j will
change the winner from j to i isn't independent of whether I'm voting
for k.

Say, with my vote for A, the 3 candidates are in a 3-way tie. Then
when I vote for B, I make B the unique winner. The probability that
I've changed the winner from C to B is 1/3, because it's 1/3 that
B would have won the tie.

But if I hadn't voted for A, then the tie would have just been between
B & C, and so it's only 1/2 that I've changed the winner from C to
B when I vote for B and make B the unique winner.

In the matter of B vs C, I'm voting for A, but in the matter of A vs
B, I'm not voting for C. And so the probability that my vote for A over
B changes the winner from B to A isn't the same as the probability that
voting B over C changes the winner from C to B.

If that's correct, then Pbc(Ub-Uc) > Pab(Ua-Ub), and Pab > Pbc,
together mean that Ub - Uc must still be greater than Ua - Ub,
and that Ub must be greater than (Ub + Uc)/2.

If there's a situation where Pij could be less because I didn't vote for
k, then my conclusion in the previous paragraph isn't warranted. But
I don't know of such a situation as yet. So far, then, it looks as
if I've shown that, with 3 candidates, even with very few voters,
the above-mean strategy is valid.

Mike Ossipoff

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