On April 9 last year I suggested the "Double Defeat, Hare" method (nominating it as an alternative in a poll). > *Voters strictly rank from the top however many candidates they wish and also may specify an approval cutoff. > > Default approval is only goes to top-ranked candidates. > > All candidates that are pairwise beaten by a more approved candidate are disqualified. > > If that leaves more than one qualified (i.e. not disqualified) candidate, commence eliminations according to Hare rules until only one qualified candidate remains.* I now withdraw that not-so-clever idea because of this scenario: 43 A| 03 A<B| 44 B>C|  (sincere is B| or B|>A or B>A|) 10 C|>A A is both the normal Hare winner and the sincere Condorcet winner. Approvals:   C 54     B 47    A 46. C>A 54-47,     A>B 56-44,     B>C 47-10. Only A is disqualified by Double Defeat  and  C is eliminated by the Hare rule leaving B the winner, rewarding the outrageous Burial strategists. My favourite Condorcet method, Margins-Sorted Approval(explicit) would punish them by electing C   (as would the not-too-bad Smith//Approval). It would first look at the BA pair because they are adjacent to each other in the approval order and have a smaller margin of difference in their approval scores (47-46=1) than the CB adjacent pair (54-47=7) and notice that they are pairwise (by the rankings) out of order and so flip that order, making it  C>A>B.  Neither of the two adjacent pairs are now pairwise out of order so that order is final  and C is on top of it so C wins. Smith//Approval(explicit)  sees that all three candidates are in the Smith set and so elects the one with the most approval. Hare, being completely immune to Burial strategy (because it meets Later-no-Help) elects (in this example)  the sincere Condorcet Winner. Chris  Benham