rob brown wrote: > On 12/5/05, *Kevin Venzke* <[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>> > wrote: > > > Has anyone ever considered also adding one-half a "point" to each > of the > > equals? > > Yes. It's equivalent to using margins. > > Not sure I understand what you mean. "Using margins" in what sense?
Several Condorcet-completion methods have the idea of "defeat strength". When defeats form a cycle, defeat strength is used to decide which defeats to keep and which to discard. Kevin was referring to the use of the pairwise margin of victory as the measure of defeat strength. Other measures are winning votes (i.e., the number of votes for the pairwise winner) and relative margins (margin of victory as a proportion of the number of preferences in the pairwise contest). For example, given that A pairwise beats B 54 to 12, then the defeat strength of A over B is 54-12=42 if measured by margins, 54 if measured by winning votes, and (54-12)/(54+12)=63.64% if measured by relative margins. 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. This list has had lots of discussion on defeat strength; check the archives. > I guess my question was, for all of the above, doesn't it make more > sense to convert the ballots into a matrix as I described? Nope. Even if you're using Margins, in which case it doesn't make a difference, your proposal, well, doesn't make a difference, so why spend the effort of adding half-votes to your matrix? 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. ---- election-methods mailing list - see http://electorama.com/em for list info
