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

[Forest Simmons]

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 solves the first problem because in any moderate to large sized
election, it would be rare for two candidates to have the same minimum sum.

It also solves the second problem because if a candidate is cloned, the
first place votes of the cloned candidate are divided up among the clones.

[End of Introduction to Improved Copeland for the lay voter.]

Now, as mentioned in my last post, it is more general to replace the phrase
"first place votes" with "random candidate probabilities," i.e. benchmark
lottery probabilities.   Even other suitable lottery probabilities could be
used.

This method (at least under the top rank counts or  benchmark lottery)
always elects from Landau, since if X covers Y, then only a subset of the
candidates that beat Y will beat X, yielding X a smaller probability sum
than Y.

Also since if candidate X is raised on a ballot it can only decrease the
benchmark probability of any other candidate, and the set of candidates
that now beat X will be a (possibly proper) subset of the candidates that
did before raising X on the ballot; i.e. this method is monotonous (if not
monogamous).

And it seems tp satisfy the Chicken Defense criterion:

49 C
26 A>B
25 B  (sincere is B>A)

C>A is the only pairwise defeat of A, so the A sum is 49.
A>B is the only pairwise defeat of B, so the B sum is 26.
B>C is the only pairwise defeat of C, so the C sum is 25.

Candidate C (with the smallest sum) is elected, thus thwarting the
threatened chicken attack

What's not to like?

Now think in terms of "Yee BoLson Diagrams":

A candidate's score is the sum of the Dirichlet Cell probabilities (i.e.
Voronoi Polygon probabilities).  These are the Dirichlet/Voronoicells of
the candidates that are closer to the center of the distribution than the
given candidate. [the respective cells represent the voters that top rank
the respective candidates.]

So the winning candidate is the candidate for which the mass of cells
closer to the center than the candidate has the smallest total probability.

In the case of the standard centrally symmetric distribution used in Yee
Bolson diagrams, the candidate closest to the center will be the winner
with no defeats, so the "mass of defeating cells" will be empty.

Not bad!

Is it good enough and simple enough to propose?
.
Forest




> ----------------------------------------------------------------------
>
> Message: 1
> Date: Wed, 5 Jun 2019 19:28:48 -0700
> From: Forest Simmons <fsimmons at pcc.edu>
> To: Kristofer Munsterhjelm <km_elmet at t-online.de>, EM
>         <election-methods at lists.electorama.com>,  "C.Benham"
>         <cbenham at adam.com.au>, Kevin Venzke <stepjak at yahoo.fr>
> Subject: [EM] A New Spinoff of Our Recent Discussions
> Message-ID:
>         <
> CAP29onc__B876PaCWDT-d_T7BwQi5eR3yDR6dV8PDGOYOto4ew at mail.gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> I was mulling over Kristofer's ideas abbout using first place counts in new
> ways.
>
> It reminded me about what we called "Borda Done Right" where we decloned
> Borda by use of the first place counts.
>
> Then a light bulb turned on: Why not de-clone Copeland in rthe same way?
> After all, Copeland always chooses from the Landau set, the set of
> uncovered candidates.  [I first realized this a few years ago when Marcus
> expressed doubt that there was a monotonic method that would always choose
> from Landau. After a little thought Copeland was the most obvious example
> to clear up the question.]
>
> So here is Improved Copeland:
>
> For each candidate X, let S(X) be the sum of first place votes of the
> candidates that do not pairwise beat X.
>
> [Note that this sum includes the number of first place votes received by X,
> since X does not pairwise beat X.]
>
> Elect from argmax(S(X)).
>
> Note that in large public elections argmax(S(X)) will almost surely consist
> of only one candidate.
>
> That version works best when the ballots have have easily discerned
> favorites.
>
> Here's a version that works better for a greater variety of ballots,
> especially where equal top votes are allowed:
>
> For each candidate Y let P(Y) be the probability that Y would be chosen by
> a random ballot lottery. [Actually, any other decent lottery would work
> just as well.]
>
> For each candidate X, let S(X) be the sum (over all candidates Y that do
> not pairwise beat X) of P(Y).
>
> Elect from argmax(S(X)).
>
> Note that if Z covers X, then S(Z) is greater than or equal to S(X),
> because every P(Y) in the sum for S(X) will also be a term in the sum
> defining S(Z).
>
> Therefore the max(S(X)) candidate is uncovered.
>
> Like Copeland the method is also monotone, and unlike Copeland the method
> is clone proof.
>
> Since Copeland is one of the most familiar Condorcet methods, and has an
> obvious appeal until the clone dependence probblem is pointed out, ithis
> new method can be presented as a simple , easily understandable solution to
> that problem.
>
> How does it hold up on our favorite examples?
>
> Try
>
> 49 C
> 26 A>B
> 25 B
>
> S(C)=49+26=75
> S(A)=26+25=51
> S(B)=25+49=74
>
> Arrgmax(S(X))={C}
>
> The method passes the CD criterion.
>
> Is this too good to be true?
> ***************************
>
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