rob brown wrote: > On 12/5/05, *Dan Bishop* <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > If Margins is used as the measure of defeat strength, then an A=B ballot > is equivalent to the combination ½ A>B + ½ B>A, as they both have zero > net contribution to the defeat strength. This is what Kevin meant. > > So a method that used margins would be unaffected (as I mentioned in my > post, "clearly some methods will be unaffected"). Others *would* be > affected, to some degree or another. > > Isn't it reasonable to discuss *how* the methods that would be affected, > would be affected?
They're affected in exactly the same way as if the measure of defeat strength were changed to Margins. > And if you do have a > reason for half-votes, you can always create a half-vote matrix from a > traditional matrix, whereas the converse is not true. > > > I'm not sure I buy this. How would you do that? For example, consider an election with 12 voters and the Condorcet matrix A B C A 0 4 8 B 4 0 8 C 4 4 0 This matrix tells you the number of people who expressed a preference in each pairwise contest: A-B: 4+4 = 8 votes A-C: 8+4 = 12 votes B-C: 8+4 = 12 votes And thus, you also know the number of voters *without* a pairwise preference: A=B: 12-8 = 4 abstensions A=C: 12-12 = 0 abstensions B=C: 12-12 = 0 abstensions You can treat the 4 A=B nonpreferences as 2 A>B + 2 B>A, thus producing the half-vote matrix: A B C A 0 6 8 = 14 Borda points B 6 0 8 = 14 Borda points C 4 4 0 = 8 Borda points ---- election-methods mailing list - see http://electorama.com/em for list info
