> Date: Fri, 17 Dec 2004 04:49:45 +0100 (CET) > From: Kevin Venzke <[EMAIL PROTECTED]> > Subject: Re: [EM] Defection, nomination disincentive, MMPO
> Also, every example I've seen of MMPO's Majority failure involves > the use of four slots. It's always this scenario: > > 20 A>B>C>D > 20 B>C>A>D > 20 C>A>B>D > 13 D>A>B>C > 13 D>B>C>A > 13 D>C>B>A > > Majority (as well as Smith) requires that A, B, or C win, but D > has the lowest MMPO score. As Smith//MinMax exists, this could be extended to Smith//MMPO. Though, I think this will make the method fail Later-No-Harm. > > The most easily presentable tie-breaker I thought of using was to find > > the next highest opposing votes for each of the tied candidates. The > > candidate with the lowest number of these opposing votes is the > > winner. > > > > If there is still a tie here, then you go on to the next highest > > opposing votes for the tied candidates. And so on, if required. > > This occurred to me, but I'm worried about a Clone-Loser problem. It > seems to me that a party could benefit from running clones, so that the > opposition votes from the party's candidates have to be plowed through > one-by-one during the tiebreaker. Do you really mean Clone-Loser here? If so, why not just use MMPO on the tied candidates? If you meant Clone-Winner here then may be something like which of the tied candidates had the better pairwise result against the bottom most MMPO candidate can be used? Unfortunately, this one is a bit artificial. Or may be for each tied candidate, find their Pairwise Proposition? The number of ballots that ranked Y>X is the Pairwise Opposition of candidate X used for MMPO. Therefore, the number of ballots that ranked X>Y is the Pairwise Proposition of candidate X. The candidate with the highest proposition from among the tied candidates is the winner. However, I think this is bit naff as I think it would encourage Approval style voting. Thanks, Gervase. ---- Election-methods mailing list - see http://electorama.com/em for list info