On 7 Mar 2003 at 10:42, Forest Simmons wrote: > Has anybody ever proposed minimizing the maximum opposition > rather than minimizing the maximum defeat?
It's a reasonably good method, although it doesn't satisfy some criteria I consider important such as Minimal Defense (which is similar to Mike Ossipoff's Strong Defensive Strategy Criterion) and Clone Independence. (That's why I think the best method is a variation of Ranked Pairs which I call Maximize Affirmed Majorities, or MAM. See the web pages at www.alumni.caltech.edu/~seppley for more info and rigorous proofs. The website is still under construction, not yet friendly for people who aren't social scientists, and most of the web pages requires a web browser that supports HTML 4.0 and Microsoft's "symbol" font to be viewed properly.) Minimax(pairwise opposition) even satisfies a criterion promoted by some advocates of Instant Runoff, which I call "Uncompromising": Let w denote the winning alternative given some set of ballots. If one or more ballots that had only w higher than bottom are changed so some other "compromise" alternative x is raised to second place (still below w but raised over all the other alternatives) then w must still win. The proof that Minimax(pairwise opposition) satisfies Uncompromising is simple: Raising x increases the pairwise opposition for all candidates except w and x, and does not decrease the pairwise opposition for any candidate, so w must still have the smallest maximum pairwise opposition. That criterion can be strengthened somewhat and still be satisfied: Changing pairwise indifferences to strict preferences in ballots that ranked w top cannot increase w's pairwise opposition or decrease any other alternatives' pairwise opposition. > I know that theoretically this could elect the Condorcet Loser, but it > seems very unlikely that it would do so. -snip- Minimax(pairwise defeat) can also elect a Condorcet Loser. Suppose there are 4 alternatives, 3 of them in the top cycle but involved in a "vicious" cycle. If the pairwise defeats of the 4th are slim majorities, the 4th wins. As for the likelihood, this may depend on how you model voters' preferences. In a spatial model with sincere voting, I think you're right. But what if supporters of the Condorcet Loser vote strategically to create a vicious cycle? Minimax(pairwise opposition) can also defeat a "weak" Condorcet Winner, one that wins all its pairings but does not have more than half the votes in each of its pairings. The CW may defeat candidate x pairwise by a plurality such as 48% (less than half the votes) and that might be x's largest opposition, whereas the CW may have a larger opposition, such as 49%, in some pairing. But it satisfies a weaker Condorcet-consistency criterion: If there is a "strong" Condorcet Winner (an alternative that wins each of its pairings by more than half of the votes) then it must be elected. Thus anyone who claims no Condorcet-consistent method satisfies Uncompromising is incorrect. -- Steve Eppley _______________________________________________ Election-methods mailing list [EMAIL PROTECTED] http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com