[EM] mproved Copeland
Forest Simmons
fsimmons at pcc.edu
Thu Jun 6 14:32:06 PDT 2019
I forgot to mention that the method is efficiently summable over precincts:
Make a pairwise matrix where the off diagonal elements give the information
for pairwise defeats, and the diagonal elements keep track of the first
place totals.
In particular, if a certain ballot ranks the first three candidates equal
top, then that ballot adds the fraction 1/3 to each of the first three
diagonal elements of the summary matrix.
>
>
>
> Message: 2
> Date: Thu, 6 Jun 2019 14:11:44 -0700
> From: Forest Simmons <fsimmons at pcc.edu>
> To: EM <election-methods at lists.electorama.com>
> Subject: [EM] Improved Copeland (was "A New Spinoff of Our Recent
> Discussions")
> Message-ID:
> <CAP29onczeXL4533KDUN_ttziV2-0=
> AjCD26Ly60cHHR42pNZww at mail.gmail.com>
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>
> 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
>
>
>
>
> ************
>
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