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@pcc.edu> > To: EM <election-methods@lists.electorama.com> > Subject: [EM] Improved Copeland (was "A New Spinoff of Our Recent > Discussions") > Message-ID: > <CAP29onczeXL4533KDUN_ttziV2-0= > AjCD26Ly60cHHR42pNZww@mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > 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 > > > > > ************ >