[EM] Decloned Copeland Borda Hybrid

[Kevin Venzke]

Hi Forest,

I'm not sure if you're still pursuing this method but as I understand it I don't
think it's monotone.

0.394: B>C>A
0.366: A
0.151: C
0.087: A>B>C

A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.

Switch last faction to
0.087: A>C>B

Cycle remains. Effective last preferences are now A, B, B, B. This causes C to net
lose a few points, and A to gain a lot due to the increased relevance of the A>B
contest, so that A wins.

The mechanism also doesn't seem to ensure majority favorite, but maybe you know that.

I do think the "anti-favorite" concept is interesting.

Kevin


Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons <forest.simmons21 at gmail.com> a écrit :
> I intended to say each candidate ranked above or pairwise beating a' gets a point, and two points for both.
> 
> The monotonicity proof goes through unchanged.... as z goes down, y gains a point, but so does X as it goes up.

> > On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
> >
> > [A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]
> >
> > Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).
> >
> > In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots
> >
> > {B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
> >
> > The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.
> >
> > The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it. 
> >
> > This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
> >
> > In any case, N(X)-N(Y) does not decrease.
> >
> >  Does this method satisfy the Condorcet Criterion?
> >
> > Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet loser.
> >
> > Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW  X?
> >
> > Perhaps ... if so, then this method is not a stand alone Condorcet method.
> >
> > It would be a valuable task for someone to test its Condorcet efficiency experimentally.
> >
> > Thanks!
> >
> > -Forest