Hi guys, Last night I was looking for strategy-resistant Condorcet methods not based on IRV. I went to the list of Condorcet methods on the wiki and I stumbled upon Ranked Robin. https://electowiki.org/wiki/Ranked_Robin Proposed by Sass on VotingTheory.org and Reddit just a couple of months ago. It's Copeland but pairwise ties get a score of 0, plus a tiebreaking mechanism. Sass summarizes the method thusly: "Among the candidates who tie for winning the most head-to-head matchups, elect the candidate with the best average rank." His goal is to get IRV supporters to ditch IRV and pick an "RCV" method that is actually good, while being simple enough that they can say "this is also RCV, it's another one, and it's easier". Now, I personally think that the "average rank" is confusing and the tiebreaking mechanism described on the Wiki is a fabulously complicated "4 degree" monstrosity that takes up half the page. And I don't think any of that if needed. If you read this explanation on Reddit you see that what he is trying to do is really simple. I would suggest the following revision: "The candidate that wins the most head-to-head matchups is elected. If there is a tie, grab every finalist and give them a score equal to the sum of all the votes in their favor in every matchup against every other finalist. The finalist with the largest such score is elected. If there's still a tie, conduct a runoff election." Here is an example: 6 votes: D>A>B>C 5 votes: B>C>A>D 4 votes: C>A>B>D This is just a trivial example of a Concorcet cycle to see how to break it. A beats B, 10 vs 5 B beats C, 11 vs 4 C beats A, 9 vs 6 everyone beats D So A,B,C win 1 matchup each, D wins none. ABC are the finalists. Now the tiebreaker round: Score(A) = 10 + 6 = 16 Score(B) = 5 + 11 = 17 Score(C) = 9 + 4 = 13 So B is elected. I'm pretty sure that this should be equivalent to the "1st Degree" tiebreaker described in the wiki, but I think my version is far easier to understand. I would ditch all the other tiebreakers "Degrees" and replace "1st Degree" with my version. That should now produce a system that is as easy to understand as Sass intended. Along those lines, I hope Rob will comment here on whether this system could be used in Burlington. Here's a first stab at legal language: ------------------------------------------------------------------- ... (2) If a candidate receives a majority (over 50 percent of all ballots) of first preferences, that candidate is elected. (3) If no candidate receives a majority of first preferences, then each candidate is compared in turn to every other candidate in a head-to-head match. The candidate that defeats, by a simple majority of voter preferences, the greatest number of other candidates in a head-to-head match, is elected. (3) If there is more than one such candidate, a tiebreaking tabulation is conducted among that group of candidates (from here on, called "finalists"). Each finalist is assigned a vote count equal to the sum of the number of ballots that rank said candidate above the other, for every head-to-head match against another finalist. The finalist with the highest vote count is elected. ... ------------------------------------------------------------------- Cheers, -- Dr. Daniel Carrera Postdoctoral Research Associate Iowa State University

What you are proposing is equuvalent to Borda restricted to the Copeland Set ... at least in the case of complete rankings ... in general the Borda winner is the candidate with the greatest difference between its row average and its column average. With complete rankings the column average is 100 percent minus the row average. Since Borda is well known and can be defined with or without reference to the pairwise score matrix, we can say .. Elect the candidate that pairwise defeats the most other candidates. In case of ties, eliminate those not tied, and elect the candidate with the highest Borda Count relative to the other remaining candidates. If you really want to emphasize the Round Robin Tournament analogy... here's the simplest language that most robustly takes both tied games into account and tied Copeland scores into account: The team with the greatest difference between wins and losses is the tournament winner. In case of ties, the tied team with the greatest difference between total points scored and total points given up, is the tournament winner. Note that the tied team that scores most against opponents could be considered the offensive champ, while the team that gives up the fewest points to its opponents is the defensive champ. The team with the greatest difference should be considered the all around winner. El mié., 19 de ene. de 2022 12:52 p. m., Daniel Carrera <dcarrera@gmail.com> escribió: > Hi guys, > > Last night I was looking for strategy-resistant Condorcet methods not > based on IRV. I went to the list of Condorcet methods on the wiki and I > stumbled upon Ranked Robin. > > https://electowiki.org/wiki/Ranked_Robin > > Proposed by Sass on VotingTheory.org and Reddit just a couple of months > ago. It's Copeland but pairwise ties get a score of 0, plus a tiebreaking > mechanism. Sass summarizes the method thusly: > > "Among the candidates who tie for winning the most head-to-head matchups, > elect the candidate with the best average rank." > > His goal is to get IRV supporters to ditch IRV and pick an "RCV" method > that is actually good, while being simple enough that they can say "this is > also RCV, it's another one, and it's easier". > > Now, I personally think that the "average rank" is confusing and the > tiebreaking mechanism described on the Wiki is a fabulously complicated "4 > degree" monstrosity that takes up half the page. And I don't think any of > that if needed. If you read this explanation on Reddit you see that what he > is trying to do is really simple. I would suggest the following revision: > > "The candidate that wins the most head-to-head matchups is elected. If > there is a tie, grab every finalist and give them a score equal to the sum > of all the votes in their favor in every matchup against every other > finalist. The finalist with the largest such score is elected. If there's > still a tie, conduct a runoff election." > > Here is an example: > > 6 votes: D>A>B>C > 5 votes: B>C>A>D > 4 votes: C>A>B>D > > This is just a trivial example of a Concorcet cycle to see how to break it. > > A beats B, 10 vs 5 > B beats C, 11 vs 4 > C beats A, 9 vs 6 > everyone beats D > > So A,B,C win 1 matchup each, D wins none. ABC are the finalists. Now the > tiebreaker round: > > Score(A) = 10 + 6 = 16 > Score(B) = 5 + 11 = 17 > Score(C) = 9 + 4 = 13 > > So B is elected. I'm pretty sure that this should be equivalent to the > "1st Degree" tiebreaker described in the wiki, but I think my version is > far easier to understand. I would ditch all the other tiebreakers "Degrees" > and replace "1st Degree" with my version. That should now produce a system > that is as easy to understand as Sass intended. > > Along those lines, I hope Rob will comment here on whether this system > could be used in Burlington. Here's a first stab at legal language: > > ------------------------------------------------------------------- > ... > (2) If a candidate receives a majority (over 50 percent of all ballots) > of first preferences, that candidate is elected. > (3) If no candidate receives a majority of first preferences, then each > candidate is compared in turn to every other candidate in a head-to-head > match. The candidate that defeats, by a simple majority of voter > preferences, the greatest number of other candidates in a head-to-head > match, is elected. > (3) If there is more than one such candidate, a tiebreaking tabulation > is conducted among that group of candidates (from here on, called > "finalists"). Each finalist is assigned a vote count equal to the sum of > the number of ballots that rank said candidate above the other, for every > head-to-head match against another finalist. The finalist with the highest > vote count is elected. > ... > ------------------------------------------------------------------- > > Cheers, > -- > Dr. Daniel Carrera > Postdoctoral Research Associate > Iowa State University > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

Better yet use de-cloned Copeland, which has a statistically negligible chance of ties. The offensive score of candidate X is the sum of last place votes of the candidates pairwise defeated by X. The defensive score of candidate X is the sum of first place votes of the candidates that pairwise defeat . In the unlikely case that the offensive and defensive champions are are not the same, elect the pairwise winner of the two. El mié., 19 de ene. de 2022 1:58 p. m., Forest Simmons < forest.simmons21@gmail.com> escribió: > What you are proposing is equuvalent to Borda restricted to the Copeland > Set ... at least in the case of complete rankings ... in general the Borda > winner is the candidate with the greatest difference between its row > average and its column average. With complete rankings the column average > is 100 percent minus the row average. > > Since Borda is well known and can be defined with or without reference to > the pairwise score matrix, we can say .. Elect the candidate that pairwise > defeats the most other candidates. In case of ties, eliminate those not > tied, and elect the candidate with the highest Borda Count relative to the > other remaining candidates. > > If you really want to emphasize the Round Robin Tournament analogy... > here's the simplest language that most robustly takes both tied games into > account and tied Copeland scores into account: > > The team with the greatest difference between wins and losses is the > tournament winner. In case of ties, the tied team with the greatest > difference between total points scored and total points given up, is the > tournament winner. > > Note that the tied team that scores most against opponents could be > considered the offensive champ, while the team that gives up the fewest > points to its opponents is the defensive champ. > > The team with the greatest difference should be considered the all around > winner. > > > > El mié., 19 de ene. de 2022 12:52 p. m., Daniel Carrera < > dcarrera@gmail.com> escribió: > >> Hi guys, >> >> Last night I was looking for strategy-resistant Condorcet methods not >> based on IRV. I went to the list of Condorcet methods on the wiki and I >> stumbled upon Ranked Robin. >> >> https://electowiki.org/wiki/Ranked_Robin >> >> Proposed by Sass on VotingTheory.org and Reddit just a couple of months >> ago. It's Copeland but pairwise ties get a score of 0, plus a tiebreaking >> mechanism. Sass summarizes the method thusly: >> >> "Among the candidates who tie for winning the most head-to-head matchups, >> elect the candidate with the best average rank." >> >> His goal is to get IRV supporters to ditch IRV and pick an "RCV" method >> that is actually good, while being simple enough that they can say "this is >> also RCV, it's another one, and it's easier". >> >> Now, I personally think that the "average rank" is confusing and the >> tiebreaking mechanism described on the Wiki is a fabulously complicated "4 >> degree" monstrosity that takes up half the page. And I don't think any of >> that if needed. If you read this explanation on Reddit you see that what he >> is trying to do is really simple. I would suggest the following revision: >> >> "The candidate that wins the most head-to-head matchups is elected. If >> there is a tie, grab every finalist and give them a score equal to the sum >> of all the votes in their favor in every matchup against every other >> finalist. The finalist with the largest such score is elected. If there's >> still a tie, conduct a runoff election." >> >> Here is an example: >> >> 6 votes: D>A>B>C >> 5 votes: B>C>A>D >> 4 votes: C>A>B>D >> >> This is just a trivial example of a Concorcet cycle to see how to break >> it. >> >> A beats B, 10 vs 5 >> B beats C, 11 vs 4 >> C beats A, 9 vs 6 >> everyone beats D >> >> So A,B,C win 1 matchup each, D wins none. ABC are the finalists. Now the >> tiebreaker round: >> >> Score(A) = 10 + 6 = 16 >> Score(B) = 5 + 11 = 17 >> Score(C) = 9 + 4 = 13 >> >> So B is elected. I'm pretty sure that this should be equivalent to the >> "1st Degree" tiebreaker described in the wiki, but I think my version is >> far easier to understand. I would ditch all the other tiebreakers "Degrees" >> and replace "1st Degree" with my version. That should now produce a system >> that is as easy to understand as Sass intended. >> >> Along those lines, I hope Rob will comment here on whether this system >> could be used in Burlington. Here's a first stab at legal language: >> >> ------------------------------------------------------------------- >> ... >> (2) If a candidate receives a majority (over 50 percent of all ballots) >> of first preferences, that candidate is elected. >> (3) If no candidate receives a majority of first preferences, then each >> candidate is compared in turn to every other candidate in a head-to-head >> match. The candidate that defeats, by a simple majority of voter >> preferences, the greatest number of other candidates in a head-to-head >> match, is elected. >> (3) If there is more than one such candidate, a tiebreaking tabulation >> is conducted among that group of candidates (from here on, called >> "finalists"). Each finalist is assigned a vote count equal to the sum of >> the number of ballots that rank said candidate above the other, for every >> head-to-head match against another finalist. The finalist with the highest >> vote count is elected. >> ... >> ------------------------------------------------------------------- >> >> Cheers, >> -- >> Dr. Daniel Carrera >> Postdoctoral Research Associate >> Iowa State University >> ---- >> Election-Methods mailing list - see https://electorama.com/em for list >> info >> >

On Wed, Jan 19, 2022 at 4:32 PM Forest Simmons <forest.simmons21@gmail.com> wrote: > Better yet use de-cloned Copeland, which has a statistically negligible > chance of ties. > > The offensive score of candidate X is the sum of last place votes of the > candidates pairwise defeated by X. > > The defensive score of candidate X is the sum of first place votes of the > candidates that pairwise defeat . > > In the unlikely case that the offensive and defensive champions are are > not the same, elect the pairwise winner of the two. > I'm not familiar with de-cloned Copeland. I think I'm seriously misunderstanding how it works... So... X's offensive score is the sum of the total number of ballots that favor the OTHER candidate? 6 votes: A>B 5 votes: B>A A beats B, 6 vs 5. So... the last place votes is 5... and B is defeated by A... so A's offensive score is 5? ??? I must have misunderstood. -- Dr. Daniel Carrera Postdoctoral Research Associate Iowa State University

El mié., 19 de ene. de 2022 4:52 p. m., Daniel Carrera <dcarrera@gmail.com> escribió: > On Wed, Jan 19, 2022 at 4:32 PM Forest Simmons <forest.simmons21@gmail.com> > wrote: > >> Better yet use de-cloned Copeland, which has a statistically negligible >> chance of ties. >> >> The offensive score of candidate X is the sum of last place votes of the >> candidates pairwise defeated by X. >> >> The defensive score of candidate X is the sum of first place votes of the >> candidates that pairwise defeat . >> >> In the unlikely case that the offensive and defensive champions are are >> not the same, elect the pairwise winner of the two. >> > I'm not familiar with de-cloned Copeland. I think I'm seriously > misunderstanding how it works... So... X's offensive score is the sum of > the total number of ballots that favor the OTHER candidate? > "Defeated by X..." > > 6 votes: A>B > 5 votes: B>A > > A beats B, 6 vs 5. So... the last place votes is 5... and B is defeated by > A... so A's offensive score is 5? ??? I must have misunderstood. > A's offensive score is the number of last place votes of the candidate defeated by A, namely the six last place votes of B. Here's another way to say it in general: A's offensive score is the number of candidates pairwise defeated by A weighted by their average number of last place votes. So like St Patrick, candidate A gets more credit for driving out the snakes and dragons than for controlling the mole and cricket populations. Similarly, A's defensive score is the number of candidates that defeat A weighted by their average number of first place votes. The smaller this score, the better it speaks for A, .. this score could be small from A suffering few defeats or from the candidates defeating A not being strong enough to get very many first place votes. If there is a Condorcet winner, she will be both best offensive and defensive candidate simultaneously. Otherwise, the head-to-head winner between the two is the natural choice. Those are the heuristic justifications for the two scores, but beyond that it is obvious that if B covers A, then B's offensive score will be higher than A's, and B's defensive score will be lower than A's, so the winner must be in the Landau Set. The weights serve to de-clone standard Copeland. And using weights from opposite ends of the rankings for the offensive and defensive scores allows the right kind of decoupling of mono-raising and lowering to preserve Copeland's monotonicity, which proved to be impossible when we only used first place votes for de-cloning. Thanks for taking an interest! -- > Dr. Daniel Carrera > Postdoctoral Research Associate > Iowa State University >