Before I start--A few days ago I posted an example in which the margins methods' only equilibria were ones in which order-reversal was being used. They were equilibria in which the C voters were using defensive order-reversal to protect B, the sincere CW, against offensive order-reversal by the A voters. That example showed that the margins methods are falsifying. Winning-votes: By the premise of these definitions, there's a sincere CW, and there's no indifference between the CW & the other candidates. That means that, for each candidate other than the CW, there's a majority who prefer the CW to that other candidate. Suppose that everyone sincerely ranks everyone down to the CW, but doesn't rank anyone whom they like less than the CW. Of course the CW is the winner. Suppose some people who like X better than the CW order-reverse by ranking Y over the CW. A majority of the voters prefer the CW to X, and so they rank the CW, but not X. That means that X has a majority against him. That majority of the voters don't rank X, and so it's impossible for X to have a majority against anyone, since a majority aren't ranking him over anyone. If the order-reversal caused some candidate, Y, to beat the CW, and X has a beatpath to Y, then conceivably X could win with some circular tie solutions. But with BeatpathWinner, the CW has a majority-strength beatpath (a 1-defeat beatpath) to X, and X can't have that strong a beatpath against anyone, since he can't beat anyone by majority. So X can't win in BeatpathWinner, since he has at least one beatpath win against him, from the CW. RP never drops or declines to keep a defeat unless it's the weakest defeat in a cycle, according the procedure that defines RP. The defeat of X, by the CW, is stronger than any defeat by X, and is therefore never the weakest defeat in any cycle containing X. So the CW>X defeat never gets dropped or passed-up for keepting. X doesn't win in RP. I showed in early 2000 that any defeat among the members of the Schwartz set is in a cycle. If the defeat from A to B isn't in a cycle, then there's no beatpath from B to A. By the beatpath definition of the Schwartz set, then, B isn't in the Schwartz set. So any defeat that's among members of the Schwartz set must be in a cycle. I showed some time ago the equivalence of the beatpath definition and the unbeaten set definition of the Schwartz set. So SSD never drops a defeat unless it's the weakest defeat in a cycle. If it wasn't initially the weakest defeat in a cycle, then dropping other defeats won't make it the weakest defeat in a cycle, since dropping defeats can't create a stronger beatpath that wasn't already there. So X can't win in SSD for the same reason that X can't win in RP. In PC, X has a majority defeat, and the CW never has a majority defeat, and so X can't win in PC, since it always has a stronger defeat than the CW does. X can't win by order-reversal. If truncation could make some Y beat the CW, it would be a weaker defeat, and wouldn't improve X's chance of winning over order-reversal. Obviously extending one's ranking past the winner, to someone less-liked can't improve a voter's outcome. No change in strategy--order-reversal, truncation, extension of a ranking--can improve anyone's outcome. I have to finish up right now, but if I've left anything out, I'll add it in a subsequent posting. Mike Ossipoff _________________________________________________________________ Chat with friends online, try MSN Messenger: http://messenger.msn.com
