[EM] Defensive strategy for Condorcet methods KM

[Kristofer Munsterhjelm]

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?