[EM] Improved Copeland (was "A New Spinoff of Our Recent Discussions")

[Kristofer Munsterhjelm]

On 06/06/2019 23.11, Forest Simmons wrote:
> Here's another slightly simpler approach aimed at the lay voter:
> 
> Tell the audience that the Condorcet ideal is a candidate that is not
> pairwise beaten by any other candidate.
> 
> When that is not possible, it is natural to consider a candidate that is
> beaten pairwise by the fewest other candidates.  This idea is the basis
> of the Copeland Method.
> 
> There are two problems with the Copeland method: (1) It has a strong
> tendency to produce ties,  and (2) More subtle problems created by
> cloning certain candidates to increase the number of defeats suffered by
> certain other candidates without increasing the number of defeats of the
> cloned candidates.
> 
> Because of these two problems, Copeland is not considered a serious
> contender for use in public elections.
> 
> But what if there were a simple modification of Copeland that would
> totally resolve these two problems in one fell swoop?
> 
> There is; instead of counting the number of candidates that defeat
> candidate X, (and electing the candidate with the smallest count), we
> add up all of the first place votes of all of the candidates that defeat
> X, and elect the candidate with the smallest sum.

This sounds like what I called first preference Copeland and you called
Clone-proofed Copeland back in 2006:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2006-December/117180.html

In the examples below, I'll call the candidates' sums "penalties".

It's not cloneproof. Example outline by Warren Smith
(https://groups.yahoo.com/neo/groups/RangeVoting/conversations/topics/2934
):

6: A>B>C
3: C>A>B
4: B>C>A

ABCA cycle, so A's penalty is 3, B's penalty is 6, and C's penalty is 4.
A wins.

Now clone A in a Condorcet cycle (into D,E,F below):

2: D>E>F>B>C
2: E>F>D>B>C
2: F>D>E>B>C
3: C>E>F>D>B
4: B>C>F>D>E

B is beaten by D, E, and F, so his penalty is 6.
C is beaten by B, so his penalty is 4.
D is beaten by C and F, so his penalty is 5.
E is beaten by C and D, so his penalty is 5.
F is beaten by C and E, so his penalty is 5.

Cloning A made A lose and C win.

Also, compared to ordinary Copeland, it loses monotonicity:

3: A>B>C
3: B>C>A
3: C>A>B

Everybody has a penalty of 3, so we have a tie.

Now raise A to the top on one of the B>C>A ballots:

4: A>B>C
2: B>C>A
3: C>A>B

A is beaten by C: penalty 3
B is beaten by A: penalty 4
C is beaten by B: penalty 2

so C wins.

The problem here is that putting A top may harm A by concealing a
candidate (here, B) who would otherwise penalize someone else (here, C)
enough to keep A from losing.

The fpA - fpC method avoids nonmonotonicity by crediting A with A-top
ballots. If raising A to the top conceals some penalizer's first
preferences, that doesn't matter because whatever A loses to the third
candidate, he regains by the fpA term.

So the fpA-fpC method seems better than first preference Copeland for
three candidates. If only we could cloneproof it! (And hopefully retain
its anti-strategy properties, like CD compliance and unmanipulable majority)