Post by Steve Eppley

Post by Forest Simmons

Has anybody ever proposed minimizing the maximum opposition

rather than minimizing the maximum defeat?

Has anybody ever proposed minimizing the maximum oppositionrather than minimizing the maximum defeat?

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

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.

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.

Post by Forest Simmons

I know that theoretically this could elect the Condorcet Loser, but it

seems very unlikely that it would do so.

I know that theoretically this could elect the Condorcet Loser, but itseems very unlikely that it would do so.

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.

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

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It's a reasonably good method, although it doesn't satisfy somecriteria I consider important such as Minimal Defense (which issimilar to Mike Ossipoff's Strong Defensive Strategy Criterion) andClone Independence. (That's why I think the best method is avariation of Ranked Pairs which I call Maximize Affirmed Majorities,or MAM. See the web pages at www.alumni.caltech.edu/~seppley formore info and rigorous proofs. The website is still underconstruction, not yet friendly for people who aren't socialscientists, and most of the web pages requires a web browser thatsupports HTML 4.0 and Microsoft's "symbol" font to be viewedproperly.)Minimax(pairwise opposition) even satisfies a criterion promoted byLet w denote the winning alternative given some setof ballots. If one or more ballots that had only whigher than bottom are changed so some other"compromise" alternative x is raised to secondplace (still below w but raised over all the otheralternatives) then w must still win.The proof that Minimax(pairwise opposition) satisfies Uncompromisingis simple: Raising x increases the pairwise opposition for allcandidates except w and x, and does not decrease the pairwiseopposition for any candidate, so w must still have the smallestmaximum pairwise opposition.Changing pairwise indifferences to strict preferences in ballots thatranked w top cannot increase w's pairwise opposition or decrease anyother alternatives' pairwise opposition.-snip-Minimax(pairwise defeat) can also elect a Condorcet Loser. Supposethere are 4 alternatives, 3 of them in the top cycle but involved ina "vicious" cycle. If the pairwise defeats of the 4th are slimmajorities, 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'reright. But what if supporters of the Condorcet Loser votestrategically to create a vicious cycle?Minimax(pairwise opposition) can also defeat a "weak" CondorcetWinner, one that wins all its pairings but does not have more thanhalf the votes in each of its pairings. The CW may defeat candidatex pairwise by a plurality such as 48% (less than half the votes) andthat might be x's largest opposition, whereas the CW may have alarger opposition, such as 49%, in some pairing.If there is a "strong" Condorcet Winner (an alternativethat wins each of its pairings by more than half of thevotes) then it must be elected.Thus anyone who claims no Condorcet-consistent method satisfiesUncompromising is incorrect.-- Steve Eppley_______________________________________________Election-methods mailing listhttp://lists.electorama.com/listinfo.cgi/election-methods-electorama.com

How about this modification of this variant?Start with the pairwise matrix P in which the number in the rowcorresponding to candidate X and the column corresponding to candidate Yis the number of ballots ranking (or rating or grading) X strictly aboveY.Order the candidates according to the maximum entry in their columns, i.e.their maximum pairwise oppositions. [Of the two possible orders, choosethe one that puts the max max opposition candidate at the bottom of thelist and the min max opposition candidate at the top.]Bubble sort this order to get its local Kemenization.This is done by starting at the top and working your way down through theorder, letting each candidate percolate up as far as possible bytranspositions of adjacent candidates whenever dictated by the pairwisewin matrix. [This pairwise win matrix is obtained from the pairwise matrixP by subtracting its transpose from it, replacing all resulting positivenumbers by ones, and zeroing out the rest of the matrix.]I like this method because it comes close to satisfying the FBC.Why would you rank Compromise over Favorite? Only if you thought thatFavorite would get in the way of Compromise during the percolationprocess, i.e. only if ...(1) Favorite can beat Compromise head-to-head, and (2) Favorite's maxopposition is smaller than Compromise's max opposition, AND (3) Compromisecan beat all the candidates who at worst give up fewer points thanFavorite, including at least one that Favorite cannot beat.You would have to be pretty fickle to abandon Favorite on accountof some poll pretending to be accurate enough to predict all of that.Forest