----- Original Message ----- From: "C.Benham" > Forest, > > Your suggested method fails both the Minimal Defense and > Plurality criteria. > > 49: A > 24: B > 27: C>B > > "Forest scores" > A: 51-49 = 2, C: 49-27 = 22, B: 49-24 = 25. > > A has the lowest score and is uncovered and so wins, violating > Minimal > Defense (which says that A can't win because on more than > half the ballots A is ranked below B and not above equal bottom). > > 7: A>B > 5: B > 8: C > > "Forest scores" > A: 8-7 = 1, B: 8-5 = 3, C: 12-8 = 4. > > A has the lowest score and is uncovered and so wins, violating > the > Plurality criterion (which says that A can't win because C has more > top-preference votes than A has above-bottom votes). > > > Chris Benham
Chris, Both of these problems go away if we count tied ranks as 1/2 point (the same way that ties are counted in Copeland) and consider a candidate tied with itself so that the diagonal entry for alternative X in the pairwise matrix has half the number of ballots ono which X was ranked. In your first example the points become A: 51-24.5=26.5 B: 49-24=25 C:49-13.5=35.5 so B wins. In the second example we get A: 8-3.5=4.5 B: 8-5=3 C: 12-4=8 So B wins again. The reason for including the self ties is that if we included a twin for each of the candidates, the old winner and its twin should be tied as the new winners. Forest. ---- Election-Methods mailing list - see http://electorama.com/em for list info
