Kevin Venzke wrote:
er... Is this right? I thought your penalty in the 3c case would have
to be just a single candidate's first preferences.
I think I am probably right here. If you draw a triangle with ABC,
you have two cycle possibilities. In both cycles IRV elects the winner
between A and B. And in both cycles, that same candidate is beaten
by the "FPL" C.
I'll just have to say oops again. I'm not very well acquainted with
cycles, and it shows :-) If you have A>B>C>A, then C isn't beaten by
both A and B, because if that was the case, who would C be beating?
Instead, C is beaten by B and beats A. So the penalties are
for A: f(C)
for B: f(A)
for C: f(B)
Doesn't that mean A must win, since C is the candidate with the least
FPP votes?
And if we have A>C>B>A, then
for A: f(B)
for B: f(C)
for C: f(A)
and B wins?
Which I suppose is what you're saying.
At least I know that this equivalence won't be the case for more than
three candidates, because IRV isn't summable and FPC is.
In an earlier post, Jameson said that the compromise strategy is too
obvious for FPC in something like the Gore-Nader-Bush scenario. If FPC
is C//IRV in the three-candidate case, does that similarly indict
C//IRV, or does C//IRV seem complex enough that the voters don't see that?
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