> From: "C.Benham" > To: election-methods-electorama....@electorama.com > Subject: [EM] Enhanced DMC
> Forest, > The "D" in DMC used to stand for *Definite*. Yeah, that's what we finally settled on. > > I like (and I think I'm happy to endorse) this Condorcet method > idea, > and consider it to be clearly better than regular DMC > > Could this method give a different winner from the ("Approval > Chain > Building" ?) method you mentioned in the "C//A" thread (on 11 > June 2011)? Yes, I'll give an example when I get more time. But for all practical purposes they both pick the highest approval Smith candidate. > > >Initialize a variable X to be the candidate with the most approval. > > > >While X is covered, let the new value of X be the highest > approval candidate that covers the old X. > > > >Elect the final value of X. > > > >For all practical purposes this is just a seamless version of > C//A, i.e. it avoids the apparent > >abandonment of Condorcet in favor of Approval after testing for > a CW. > > > > > >Assuming cardinal ballots, candidate A covers candidate B, iff > whenever B is rated above C on more > >ballots than not, the same is true for A, and (additionally) A > beats (in this same pairwise sense) some > >candidate that B does not. > > > > > > Your newer suggestion ("enhanced DMC") seems to have an > easier-to-explain and justify motivation. > > Chris Benham > > > Forest Simmons wrote (12 July 2011): > > >One of the main approaches to Democratic Majority Choice was > through the idea that if X beats Y and > >also has greater approval than Y, then Y should not win. > > > >If we disqualify all that are beaten pairwise by someone with > greater approval, then the remaining set P > >is totally ordered by approval in one direction, and by > pairwise defeats in the other direction. DMC > >solves this quandry by giving pairwise defeat precedence over > approval score; the member of P that > >beats all of the others pairwise is the DMC winner. > > > >The trouble with this solution is that the DMC winner is always > the member off P with the least approval > >score. Is there some reasonable way of choosing from P that > could potentially elect any of its members? > > > >My idea is based on the following observation: > > > >There is always at least one member of P, namely the DMC > winner, i.e. the lowest approval member of > >P, that is not covered by any member of P. > > > >So why not elect the highest approval member of P that is not > covered by any member of P? > > > >By this rule, if the approval winner is uncovered it will win. > If there are five members of P and the upper > >two are covered by members of the lower three, but the third > one is covered only by candidates outside > >of P (if any), then this middle member of P is elected. > > > >What if this middle member X is covered by some candidate Y > outside of P? How would X respond to > >the complaint of Y, when Y says, "I beat you pairwise, as well > as everybody that you beat pairwise, so > >how come you win instead of me?" > > > >Candidate X can answer, "That's all well and good, but I had > greater approval than you, and one of my > >buddies Z from P beat you both pairwise and in approval. If Z > beat me in approval, then I beat Z pairwise, > >and somebody in P covers Z. If you were elected, both Z and > the member of P that covers Z would have > >a much greater case against you than you have against me."
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