A while ago, a community I often visit (http://museumofzzt.com) ran a game creation contest where the objective was to create the best game in 144 minutes. The judges, who were also the game creators, chose the winner by the Borda count (https://museumofzzt.com/file/view/summer22bkzzt/?file=Results%2FOverall%20Results.png ). One interesting bit about this, from a voting perspective, is that they also calculated a divisiveness score per candidate, which provides information about how consistent that candidate's rank is. They used the mean absolute deviation (the most obvious choice would be the MSE, but perhaps the MAD is better since it's more robust). It's an interesting concept. For Borda and Range it's pretty easy: use some variance or robust variance measure. But figuring it out for Condorcet methods seems much harder. A possible quick and dirty version could go like this: For any method where the candidate that maximizes or minimizes some score is elected, use bootstrapping to create a distribution of that score, per candidate. Let the divisiveness measure be the standard deviation (or some other dispersion measure) of a random variable of that distribution. It's not particularly elegant, however! Any better ideas? -km

On Mon, Jul 25, 2022 at 5:51 AM Kristofer Munsterhjelm <km_elmet@t-online.de> wrote: > It's an interesting concept. For Borda and Range it's pretty easy: use > some variance or robust variance measure. But figuring it out for > Condorcet methods seems much harder. > > A possible quick and dirty version could go like this: For any method > where the candidate that maximizes or minimizes some score is elected, > use bootstrapping to create a distribution of that score, per candidate. > Let the divisiveness measure be the standard deviation (or some other > dispersion measure) of a random variable of that distribution. > > It's not particularly elegant, however! Any better ideas? > For any ranked method, a natural score for each candidate is their rank normalized by the number of candidates. I don't like using standard deviations for anything that is not known to be Gaussian. Even other bell curves can really mess up the stdev. MAD is better, and it's reasonably intuitive ("half the points are at most this far from the median"), but I'm boring and I like quantiles. Divisive Score = (95th percentile approval score) - (5th percentile approval score) or Divisive Score = (85th percentile approval score) - (15th percentile approval score) Cheers, -- Dr. Daniel Carrera Postdoctoral Research Associate Iowa State University

On 7/25/22 11:33 PM, Daniel Carrera wrote: > For any ranked method, a natural score for each candidate is their rank > normalized by the number of candidates. > > I don't like using standard deviations for anything that is not known to > be Gaussian. Even other bell curves can really mess up the stdev. MAD is > better, and it's reasonably intuitive ("half the points are at most this > far from the median"), but I'm boring and I like quantiles. > > Divisive Score = (95th percentile approval score) - (5th percentile > approval score) > > or > > Divisive Score = (85th percentile approval score) - (15th percentile > approval score) You're right, that makes a natural choice, and for range and approval, it's also rather easy (like Borda). I was just playing with the idea that if the method can be cast as "elect max over c in Candidates: f(c)", then a natural choice is also some kind of dispersion based on f, so that the method is consistent in that sense. -km