[EM] Relative vs. Majority Condorcet

[Closed Limelike Curves]

*Does "Majority-Condorcet" mean the CW needs to have a majority over every
othercandidate?*

Yes: a CW needs more than 50% of the vote, including tied ranks, to defeat
every other candidate. This version of Condorcet is compatible with FBC.

On Wed, Apr 24, 2024 at 3:43 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> On 2024-04-24 12:32, Kevin Venzke wrote:
>
> > The second option doesn't offer Smith, but if it did, I would note that
> Smith is a
> > poor guarantee of quality. Here's a 1025-voter election where a 2-vote
> candidate is
> > in the Smith set (along with all other candidates):
> >
> > 2: A>B>C>D>E>F>G>H>I>J>K
> > 1: B>C>D>E>F>G>H>I>J>K
> > 2: C>D>E>F>G>H>I>J>K
> > 4: D>E>F>G>H>I>J>K
> > 8: E>F>G>H>I>J>K
> > 16: F>G>H>I>J>K
> > 32: G>H>I>J>K
> > 64: H>I>J>K
> > 128: I>J>K
> > 256: J>K
> > 512: K
> >
> > While this is not realistic, I do think it is realistic that a candidate
> of limited
> > interest to most voters would sometimes manage to pairwise defeat a more
> viable
> > candidate. And we should be ready to interpret this as noise.
>
> That was phrased a bit oddly in the context of the rest of your post,
> but I understand you to be saying "the worst method that passes Smith
> may still be pretty bad", not necessarily that proposed methods passing
> Smith are actually bad. Is that right?
>
> -km
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