Our first attempts at improved Copeland ended up losing monotonicity without fully achieving clone independence. I will repeat that version here for comparison: Elect the candidate with the fewest top rank ballot totals for the candidates that beat her pairwise. It turns out that we have to replace the top rank totals with something that counts a few additional votes beyond the top tallies: For each candidate X let T(X) be the number of ballots on which candidate X is ranked above all of the candidates that she beats pairwise. This total includes all of the unique top votes of candidate X, but also includes some others. So here's the method: elect the candidate Y that minimizes S(Y) defined as the sum of T(X) (over all X that beat Y pairiwise). Here's an example: 4 A>B 2 B>C 3 C>A T totals in the form of top votes plus extras from second ranks: T(A) = 4 + 3 = 7 T(B) = 2 + 4 = 6 T(C) = 3 +2 = 5 S(A) = T(C) = 5 < 6 = T(B)=S(C) < 7 = T(A)=S(B), so S(A) < S(C) < S(B) A is the winner, B is the loser, and C is in the middle of the social order according to this method. I would appreciate it being tested on your favorite examples. If it needs clarification I will use your examples to clarify it, as long as it holds up to scrutiny. Thanks, Forest

Hi Forest, Try this example: 98: Abby > Cora > Erin > Dave > Brad 64: Brad > Abby > Erin > Cora > Dave 12: Brad > Abby > Erin > Dave > Cora 98: Brad > Erin > Abby > Cora > Dave 13: Brad > Erin > Abby > Dave > Cora 125: Brad > Erin > Dave > Abby > Cora 124: Cora > Abby > Erin > Dave > Brad 76: Cora > Erin > Abby > Dave > Brad 21: Dave > Abby > Brad > Erin > Cora 30: Dave > Brad > Abby > Erin > Cora 98: Dave > Brad > Erin > Cora > Abby 139: Dave > Cora > Abby > Brad > Erin 23: Dave > Cora > Brad > Abby > Erin Abby defeats all candidates except Brad, and Brad defeats all candidates except Dave. So S(Abby) is the total number of ballots on which Brad is ranked above all other candidates except possibly Dave. So T(Brad) = Brad > Abby votes , 463 minus the 23 ballot where Cora > Brad. So S(Abby) = 440. Similarly, S(Brad) = T(Dave). Dave defeats all candidates except Abby and Erin. So T(Dave) = total number of ballots on which Dave is ranked higher than all candidates except possibly Abby or Erin. So T(Dave) = 21 + 30 + 98 + 139 + 23 = 311. I haven't worked out the rest, but I believe they are all higher. So Brad would beat Abby. I think in this case I would prefer Abby to Brad, so I'm not entirely happy. I am also not seeing an obvious way to make this summable. What's wrong with using Equal-Rated-Zero for pairwise votes, and then minimizing the sum of defeating scores against a candidate? In other words, add up all the defeating scores in a candidate's column in the pairwise array. I'm sure that has a name already. In this example, Abby would win with a total of 463 votes Brad>Abby, which is less than the total defeating scores for any other candidate. It also has the advantage of being summable with no other information than the pairwise array required. On Tue, Jun 11, 2019 at 5:02 PM Forest Simmons <fsimmons@pcc.edu> wrote: > Our first attempts at improved Copeland ended up losing monotonicity > without fully achieving clone independence. > > I will repeat that version here for comparison: > > Elect the candidate with the fewest top rank ballot totals for the > candidates that beat her pairwise. > > It turns out that we have to replace the top rank totals with something > that counts a few additional votes beyond the top tallies: > > For each candidate X let T(X) be the number of ballots on which candidate > X is ranked above all of the candidates that she beats pairwise. > > This total includes all of the unique top votes of candidate X, but also > includes some others. > > So here's the method: elect the candidate Y that minimizes S(Y) defined as > the sum of T(X) (over all X that beat Y pairiwise). > > Here's an example: > > 4 A>B > 2 B>C > 3 C>A > > T totals in the form of top votes plus extras from second ranks: > T(A) = 4 + 3 = 7 > T(B) = 2 + 4 = 6 > T(C) = 3 +2 = 5 > > S(A) = T(C) = 5 < 6 = T(B)=S(C) < 7 = T(A)=S(B), > > so S(A) < S(C) < S(B) > > A is the winner, B is the loser, and C is in the middle of the social > order according to this method. > > I would appreciate it being tested on your favorite examples. If it needs > clarification I will use your examples to clarify it, as long as it holds > up to scrutiny. > > Thanks, > > Forest > > > > ---- > Election-Methods mailing list - see https://electorama.com/em for list > info >

As I just wrote, What's wrong with using Equal-Rated-Zero for pairwise votes, and then > minimizing the sum of defeating scores against a candidate? In other > words, add up all the defeating scores in a candidate's column in the > pairwise array. I'm sure that has a name already. In this example, Abby > would win with a total of 463 votes Brad>Abby, which is less than the total > defeating scores for any other candidate. It also has the advantage of > being summable with no other information than the pairwise array required. Thinking about this, it doesn't make sense to count the same ballots twice. That opens up too many strategic options. So consider the S score S(Y) = total number of ballots on which Y is ranked strictly below any candidate who pairwise defeats Y. The candidate with the minimum S score is the winner. When a candidate has only a single defeat, their S score is the pairwise winning votes of the candidate who defeated them. This is not easily summable (at first glance), but it has a certain kind of intuitive sense to it, similar to MMPO. On Wed, Jun 12, 2019 at 11:14 AM Ted Stern <dodecatheon@gmail.com> wrote: > Hi Forest, > > Try this example: > > 98: Abby > Cora > Erin > Dave > Brad > 64: Brad > Abby > Erin > Cora > Dave > 12: Brad > Abby > Erin > Dave > Cora > 98: Brad > Erin > Abby > Cora > Dave > 13: Brad > Erin > Abby > Dave > Cora > 125: Brad > Erin > Dave > Abby > Cora > 124: Cora > Abby > Erin > Dave > Brad > 76: Cora > Erin > Abby > Dave > Brad > 21: Dave > Abby > Brad > Erin > Cora > 30: Dave > Brad > Abby > Erin > Cora > 98: Dave > Brad > Erin > Cora > Abby > 139: Dave > Cora > Abby > Brad > Erin > 23: Dave > Cora > Brad > Abby > Erin > > > Abby defeats all candidates except Brad, and Brad defeats all candidates > except Dave. So S(Abby) is the total number of ballots on which Brad is > ranked above all other candidates except possibly Dave. So T(Brad) = Brad > > Abby votes , 463 minus the 23 ballot where Cora > Brad. So S(Abby) = 440. > > Similarly, S(Brad) = T(Dave). Dave defeats all candidates except Abby and > Erin. So T(Dave) = total number of ballots on which Dave is ranked higher > than all candidates except possibly Abby or Erin. So T(Dave) = 21 + 30 + > 98 + 139 + 23 = 311. > > I haven't worked out the rest, but I believe they are all higher. So Brad > would beat Abby. I think in this case I would prefer Abby to Brad, so I'm > not entirely happy. > > I am also not seeing an obvious way to make this summable. > > What's wrong with using Equal-Rated-Zero for pairwise votes, and then > minimizing the sum of defeating scores against a candidate? In other > words, add up all the defeating scores in a candidate's column in the > pairwise array. I'm sure that has a name already. In this example, Abby > would win with a total of 463 votes Brad>Abby, which is less than the total > defeating scores for any other candidate. It also has the advantage of > being summable with no other information than the pairwise array required. > > On Tue, Jun 11, 2019 at 5:02 PM Forest Simmons <fsimmons@pcc.edu> wrote: > >> Our first attempts at improved Copeland ended up losing monotonicity >> without fully achieving clone independence. >> >> I will repeat that version here for comparison: >> >> Elect the candidate with the fewest top rank ballot totals for the >> candidates that beat her pairwise. >> >> It turns out that we have to replace the top rank totals with something >> that counts a few additional votes beyond the top tallies: >> >> For each candidate X let T(X) be the number of ballots on which candidate >> X is ranked above all of the candidates that she beats pairwise. >> >> This total includes all of the unique top votes of candidate X, but also >> includes some others. >> >> So here's the method: elect the candidate Y that minimizes S(Y) defined >> as the sum of T(X) (over all X that beat Y pairiwise). >> >> Here's an example: >> >> 4 A>B >> 2 B>C >> 3 C>A >> >> T totals in the form of top votes plus extras from second ranks: >> T(A) = 4 + 3 = 7 >> T(B) = 2 + 4 = 6 >> T(C) = 3 +2 = 5 >> >> S(A) = T(C) = 5 < 6 = T(B)=S(C) < 7 = T(A)=S(B), >> >> so S(A) < S(C) < S(B) >> >> A is the winner, B is the loser, and C is in the middle of the social >> order according to this method. >> >> I would appreciate it being tested on your favorite examples. If it needs >> clarification I will use your examples to clarify it, as long as it holds >> up to scrutiny. >> >> Thanks, >> >> Forest >> >> >> >> ---- >> Election-Methods mailing list - see https://electorama.com/em for list >> info >> >