[EM] Decloned Copeland Borda Hybrid

[James Faran]

I may not understand the method, but consider the following with candidates X, Y, Z:

51 X<Y<Z
49 Y>Z>X

Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on those ballots B in the group of 51. Thus N(X)=51.

Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.

If I've understood the method  correctly, it is not Condorcet efficient.

Jim Faran
________________________________
From: Election-Methods <election-methods-bounces at lists.electorama.com> on behalf of Forest Simmons <forest.simmons21 at gmail.com>
Sent: Wednesday, April 6, 2022 2:57 AM
To: EM <Election-methods at lists.electorama.com>; Kristofer Munsterhjelm <km_elmet at t-online.de>
Subject: [EM] Decloned Copeland Borda Hybrid

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


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