I've discovered that this method fails both Clone-Winner and Clone-loser. Say we have this example of the scenario specified in the "Chicken Dilemma" criterion definition: 34 C 33 A>B 32 B MMLV(erw) scores: C34, A33, B32. Pairwise C>A>B, so Margins-Sort is happy with that order and elects C. Now say we add two clones of C (X and Y) to give: 13 C>Y>X 11 X>C>Y 10 Y>X>C 33 A>B 32 B C>Y 24-10, Y>X 23-11, X>C 21-13. C/X/Y >A 34-33, A>B 33-32, B>C/X/Y 65-34 MMLV(erw) scores: A33, B32, C13, X11, Y10. Pairwise X>C and Y>X. The MMLV(erw) score margin between the latter pair is the lowest, so Margins-Sort gives the order A>B>C>Y>X and so elects A. To not fail Clone-Winner the winner has to be one of C, X, Y. Say instead of cloning the winner we have X and Y be clones of A: 34 C 12 A>Y>X>B 11 X>A>Y>B 10 Y>X>A>B 32 B A>Y 23-10, Y>X 22-11, X>A 21-12. C>A/X/Y 34-33, A/X/Y>B 33-32, B>C 65-34. MMLV(erw) scores: C34, B32, A12, X11, Y10 This is pairwise out of order, and Margins-Sort breaks the score-margins tie between X>A and Y>X by demoting the lower-ranked candidate, so it goes C>B>A>Y>X, then B>C>A>Y>X and then is content and elects B. To meet Clone-Loser the winner has to remain C. For these failures to occur there had to be a sub-cycle inside the Smith-set, which I think would be extremely unlikely in any real election. I think the problem can be fixed. I suggest replacing MMLV(erw) Margins-Sort with *MMLV(erw) Margins-Sort Elimination*. We order the candidates as before (with MMLV(erw) Margins-Sort) but instead of simply electing the candidate highest in that order, we eliminate (drop from the ballots) the lowest-ordered candidate, recalculate the MMLV(erw) scores, and repeat until one candidate remains (or giving the same result and saving time, until 3 candidates remain and then elect the highest-ordered candidate). In all the example elections I gave in earlier posts on MMLV(erw) Margins , this method gives the same winners. To take the clone-winner example I gave, MMLV(erw) Margins-Sort gives the order A>B>C>Y>X. After eliminating X, C's smallest number of votes in a pairwise defeat (C's MMLV(erw) score) changes from 13 back to 34. Then Margins-Sort gives the order C>A>B>Y, and then we eliminate Y and C wins. In the clone-loser example I gave MMLV(erw) Margins-Sort gives the order B>C>A>Y>X. After eliminating X, A's MMLV(erw) score changes from 12 back to 33. Then Margins-Sort gives the order C>A>Y>B and then we eliminate B and C wins. Chris Benham On 4/26/2014 4:10 AM, C.Benham wrote: > This is my new idea for a Condorcet method that meets Mono-raise and > Chicken Dilemma and is relatively resistant > to Burial strategy. > > *Voters rank from the top however many candidates they wish Truncation > and equal-ranking is allowed. > > A pairwise matrix is created, giving normal gross scores except that > ballots that explicitly equal rank (not truncate) any two > candidates X and Y give a whole vote to each in that pairwise contest. > > Using this information, give each alternative a score that equals the > smallest number of votes it received in a pairwise loss. > > Henceforth we are only concerned with the direction of the pairwise > defeats and these individual candidate scores. > > Use the Schulze algorithm, weighing each pairwise "defeat" by the > absolute margin of difference between the two candidates' > scores. (Or use Ranked Pairs or River in the same way if you prefer). > > Or use the candidate scores for the Margins Sort algorithm.* > > This method uses the same type of pairwise matrix as a Schulze > (Losing Votes) variant I suggested earlier. I think this is much > better. > > 46 A > 44 B>C (sincere is B or B>A) > 05 C>A > 05 C>B > > A>B 51-49, B>C 44-10, C>A 54-46. MinMax (Losing Votes) scores: > B49, A46, C10. > > The method I suggested earlier elects the buriers' candidate B, but my > new method elects A (the "sincere CW"). > > The Margins Sort version begins with the MM(LV) order B>A>C, then > notices that the two adjacent candidates with the two most > similar scores are B and A and that A pairwise beats B, so flips that > order and then considers A>B>C and then sees that there for each > pair of adjacent candidates, the one higher in the order pairwise > beats the one lower in the order and so is content and elects the now > highest-ordered candidate. > > The other versions look at the pairwise results weighted thus: A > B > (46-49 = -3) B>C (49-10 = 39) C>A (10-46 = -36). > > A's pairwise defeat score (negative 36) is by far the lowest so A wins. > > Chris Benham > ---- > Election-Methods mailing list - see http://electorama.com/em for list > info >