This is what is intuitively distatefull to us non-specialists about these methods.
Why should the 33 percent of the voting population who most dislike "A" be the cause of A to win, whereas 56 pecent rank C>A ? > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] > ] On Behalf Of Steve Eppley > Sent: Wednesday, October 06, 2004 12:05 PM > To: [EMAIL PROTECTED] > Subject: Re: [EM] Does MAM use the Copeland method? > > Hi, > Gervase L asked: > > Just a quick question that should clear up my understanding > of MAM. > > Is it the same as Copeland (i.e. count each candidate's number of > > wins) except that any pairwise wins that are inconsistent with the > > Rank Pairs ranking are dropped before the Copeland score is tallied > > up? > > No, that's not MAM. > > The question itself isn't very clear, since some people use > the name Ranked Pairs as a synonym for MAM, while others, > including me, use the name Ranked Pairs to refer only to > Tideman's 1987 method and Zavist-Tideman's 1989 method. > Those two methods differ from MAM in several ways, the most > important difference being that MAM measures the size of each > majority primarily by the number of votes that rank the > majority's more-preferred candidate over their less-preferred > candidate, whereas Tideman and Zavist-Tideman subtract from > that the size of the opposing minority (that is, the number > of votes that rank the majority's less-preferred candidate > over their more-preferred candidate). Because of this, MAM > satisfies some criteria that Ranked Pairs does not. > > MAM constructs the order of finish of the candidates a piece > at a time, by considering the majorities one at a time from > largest to smallest, adopting into the finish order each > majority's preference that does not conflict > (cycle) with the partial order of finish already constructed. > The winner is the candidate that tops the order of finish so > constructed. (In a multiwinner context, the N winners are the > N candidates that top the order of finish.) > > Here's a simple example to illustrate: > > 9 voters, 3 candidates > > The votes: > 4: A>B>C > 3: B>C>A > 2: C>A>B > > There are three majorities: > 7: B>C > 6: A>B > 5: C>A > > MAM first adopts the preference of the largest majority, > for B over C. Thus B finishes over C. Then MAM adopts > the preference of the second largest majority, for > A over B. Thus A finishes over B. Since MAM is > constructing an ordering and A finishes over B and > B finishes over C, this implies A finishes over C. > So when MAM considers the preference of the third > largest majority, for C over A, it is not adopted > since it has already been established that A finishes > over C. Thus the complete order of finish is A>B>C. > Since A tops the finish order, A is elected. > > I've oversimplified a bit. When the number of voters is > small, it's not implausible that two or more majorities will > be the same size and it's not implausible that one or more > pairings won't have a majority (in other words, a pairing may > be a "tie.") For more details about MAM, see my web pages at > <www.alumni.caltech.edu/~seppley>. > > --Steve > > ---- > Election-methods mailing list - see http://electorama.com/em > for list info > ---- Election-methods mailing list - see http://electorama.com/em for list info
