2011/6/16 Kristofer Munsterhjelm <[email protected]> > 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? >
Let me be clear that in a generic center-squeeze scenario such as Gore/Bush/Nader, there is an honest CW, and so both FPC and C//IRV get the "right" answer with honest votes (unlike plurality or IRV). Thus, if the "Nader" (weaker extreme) voters decided to use favorite betrayal strategy in FPC or C//IRV, it would be as a defense against burial by the "Bush" (stronger extreme) voters. This is a valid defensive strategy in both cases, but I do think that FPC's not-explicitly-Condorcet rules do make this strategy more obvious than with C//IRV, especially to voters accustomed to plurality.
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