[EM] Hey guys, look at this...

[Kevin Venzke]

Hi Kristofer,

Le lundi 20 février 2023 à 05:08:01 UTC−6, Kristofer Munsterhjelm <km_elmet at t-online.de> a écrit :
> > I don't quite follow that since IRV satisfies Condorcet Loser.
> 
> I think the point is this: Robert says that if X beats Y pairwise, then
> Y shouldn't be elected. This requires the Condorcet winner criterion to
> be satisfied, because otherwise X = the CW, Y = anybody else violates it.
> 
> But suppose that a method didn't pass Condorcet loser. Then X = someone
> who's not the Condorcet loser, Y = the Condorcet loser would violate the
> criterion.
> 
> So this leads to a problem when proposing Condorcet,Plurality because if
> a Condorcet loser is elected, then it seems to go against the spirit of
> the criterion.

If I play devil's advocate, the argument says the winner shouldn't lose pairwise, not that
they should win pairwise. Adding a junk candidate for the FPP winner to defeat, so that the
FPP winner is no longer Condorcet Loser, doesn't seem like it would improve the FPP
winner's electability under the argument.

> About the only response I can think of is that if there's
> a cycle, it's impossible to satisfy the criterion (because everyone has
> someone else beating him), thus the criterion is suspended entirely for
> such elections.

It's plausible. "MDD" methods work on this idea, that if every candidate had a full
majority against them then the filter just does nothing instead.

Ideally you'd want an argument that can't disqualify everyone.

> The hard mode interpretation would be that the constraint implies Smith.
> But the problem then is that there are so few ways to make a method
> that's both Smith and simple to express in natural language. There are
> two relatively easy approaches, each with drawbacks: Copeland -- or
> elimination (e.g. Benham); and with elimination you lose monotonicity
> and (usually) summability[1], whereas Copeland seems to in some sense
> always be vulnerable to strategy, no matter what you combine it with.
> That's why it's hard mode :-)

Additionally I don't find this satisfying. The reasonableness of Smith isn't obvious enough
to be a mere footnote on this argument. I think it needs its own arguments.

Barely relevant, but there's a property I use privately called "conservative reform." It
says that the winner must either defeat the FPP winner pairwise, or else *be* the FPP
winner. It brings to my mind the possible sentiment that the FPP winner is special, and can
always be elected if we don't know what else to do.

(This property is probably more conservative than needed, considering that IRV doesn't
satisfy it. However, I do wonder if people would protest if it happened in some IRV
election that the method winner is pairwise defeated by the first-round winner.)

Kevin
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