Alex, I never got around to responding to this message of yours. I just wanted to say that I thought your idea of cancelling all of the symmetries in the Smith set preference orders and then looking again for a CW is a great idea!
It is one of those ideas like Martin Harper's Universal Approval and Richard Moore's Majority Potential that I wish I had thought of. Forest On Thu, 28 Mar 2002, [iso-8859-1] Alex Small wrote: > I'll begin this message concerning Saari's insights by saying that, for the > record, I am AGAINST the Borda Count for real-world elections (although I > understand it has engineering and sports applications). That said, let me > share a thought: > > Saari pointed out that cyclic ambiguities come from a "Condorcet profile" > or "symmetric profile". If the electorate consists of 3 groups > > 35 A>B>C > 33 B>C>A > 32 C>A>B > > we can "decompose" the electorate into > > 32 A>B>C + 3 A>B>C > 32 B>C>A + 1 B>C>A > 32 C>A>B > > The total electorate has no Condorcet winner. The "residual" does. > > It's my understanding that, in general, if there is no Condorcet winner, > subtracting out the symmetric profiles will leave a "residual" profile that > DOES have a Condorcet winner. > > Caveat: If a Condorcet winner does exist, subtracting out the symmetric > profile might yield a different Condorcet winner. > > So, this leads me to propose the following possible election method: > > 1) If a Condorcet winner exists then simply elect him. End of story. > 2) If not, eliminate from all ranking orders any candidate NOT in the > Smith set. So, if A, B, and C are in the Smith set, a person whose > preference order is A>D>C>B is now considered to have the preference order > A>C>B. > 3) Having narrowed down the list, subtract out the symmetric profiles. > 4) There will now be a "Condorcet winner." Elect him. > > A few weeks ago I discussed Saari's insight, and DEMOREP suggested that > discarding ballots like that is of dubious merit. Here's my defense: > > Suppose that there's a conference on the mathematics of voting, and being > the election geniuses that we are we all give invited talks. Ten of us > decide to go out to dinner together. After we've delivered our talks, nine > people are waiting for me to show up (I'm running late, because I had to > call my fiance). > > Joe suggests that those present pick a restaurant. The results are: > > 3 Thai>Pizza>Steak > 3 Pizza>Steak>Thai > 3 Steak>Thai>Pizza > > We have an exactly symmetric profile. Demorep suggests approval voting, > but at that minute I walk through the door. I maintain that _if_ we only > go by preference orders, then my vote should decide all. Not because I'm > special, but because if you're only one vote away from an exact tie, that > single vote must be the deciding factor for any reasonable method. > > Since my preference order is Thai>Pizza>Steak I maintain that Thai should > be the winner. With the method I proposed above Thai would win. With SSD > the weakest defeat is Thai vs. Steak (margin of 2). All other defeats have > a margin of 4. SSD would hence select Thai. Not knowing much about other > methods I can't say which others would pick Thai. > > So, it isn't such a stretch to conclude that cyclic ambiguities should be > resolved by whatever ballots are in excess of a tie. > > Of course, Demorep's idea of using approval to pick the winner (presumably > from among the members of the inner-most unbeaten set in the general case) > is far easier. I merely throw this out for examination. I haven't thought > out what flaws it might have. > > Alex > > ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
