[EM] Copeland's criteria

Rob Lanphier robla at robla.net
Sun Sep 11 22:52:42 PDT 2005

Hi Kevin,

You're right.  Copeland has some pretty big issues.

It appears to suffer the same deficiencies as margins-based methods,
since Copeland ends up treating a victory of 3%-1% with 96% abstaining
as being just as good as some other candidate's 51%-49% victory with 0%

Below is a proposal I'm throwing out there not as a serious proposal,
but as food for thought.  It may evolve into a serious proposal, but I'm
not optimistic.

As Abd alluded to in at least one email, it's possible to have a revised
version of Copeland that works differently.  For example, it could be
possible to not credit a candidate with a victory if they don't receive
majority support (called "Copeland Majority" for purposes of this mail).
We'll say one point for a win, no points for a loss, tie, or
"draw" (where neither candidate gets a majority).

Here's what the results of some of the examples we've discussed:
Example 1
49: A
24: B>E
27: C>D>B>E

Ordinary Copeland result 
C: 3-1-0
A: 2-2-0
B: 2-2-0
D: 2-2-0
E: 1-3-0

Winner is C

Copeland Majority (Win-Loss-Draw)
B: 2-0-2 (beats A and E) 
E: 1-1-2 (beats A)
C: 0-0-4
D: 0-0-4
A: 0-2-2

Winner is B

Example 2
49: A>F
24: B
27: C>G>B

Ordinary Copeland result 
A: 3-1-0
B: 2-2-0
C: 2-2-0
F: 2-2-0
G: 1-3-0

Winner is A

Copeland Majority:
A: 0-1-3
B: 2-0-2 (beats A and F)
C: 0-0-4
F: 0-1-3
G: 0-0-4

Winner is B

It's something to think about.  Myself, I'm not inclined to advocate
Copeland Majority in absence of serious analysis, and I'm not inclined
to work on that analysis at this time.  I suspect it actually has some
nice properties when combined with a good tiebreaker, but don't have
anything provable to back up my hunch.


On Mon, 2005-09-12 at 00:44 +0200, Kevin Venzke wrote:
> Hi,
> I thought about this a bit. Consider this election:
> 49 A
> 24 B>E
> 27 C>D>B>E
> C has 3 wins, and is the only Copeland winner.
> Woodall's plurality criterion is violated, since there's no way to raise C 
> in rankings including C so that C has even the first preference count that 
> A starts with.
> Consider this:
> 49 A>F
> 24 B
> 27 C>G>B
> A has 3 wins, and is the only Copeland winner.
> Eppley's minimal defense criterion is violated, because there is no way
> for the C>G>B voters (with the B voters) to at least elect B, without
> insincerely ranking B above G (for a tie with A) or both C and G (to win).
> So when I told Rob I couldn't advocate Copeland unless the tie-breaker
> satisfied minimal defense, I was talking about something impossible.
> Kevin Venzke
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