OK, then could we call it the "First-level-strategic Approval Winner set" or the 1SAW set for short? I suspect better names are possible, but I can't think of one.
As an aside: I think exploring good ranked methods like this is worthwhile from a theoretical point of view. But from a practical perspective, I suspect that this kind of thing will always be too complicated to explain to voters. That's why I prefer things like MAV (using a Bucklin-like motivation) or SODA (whose rules, though they aren't simple, at least "make sense" and give each candidate a clear approval score at each step, with the highest final approval winning). Second aside: a while back I gave an example which purported to show that SODA was not monotonic, but I missed a (rationally dominant) way to get the right result on that example within the SODA framework; so at the present, I strongly believe that SODA is (rationally) monotonic after all, and in fact I'm working on a proof. Jameson. 2013/10/14 Forest Simmons <fsimm...@pcc.edu> > Kevin and Jameson, > > thanks for the insights and suggestions. It's kind of you to suggest my > name, Jameson, but I would rather something more descriptive similar to > "the potential approval winner set" of Chris and Kevin or more public > relations friendly like the Democratically Acceptable Set. My original > motivation (that eventually led to IA/MPO as an approximate solution) was > to find a candidate most likely to win two approval elections in a row > (going into the second election as front runner) without a change in > sincere voter preferences, but with an opportunity to adjust their ballot > approval cutoffs. > > Remember when we were looking at DMC from various points of view? One was > to think of the DMC winner as the beats all candidate relative to the set P > of candidates that were not doubly defeated, i.e. not defeated both > pairwise and in approval by some other candidate. In other words each > member of P defeats pairwise every candidate with greater approval. The > approval winner, the DMC winner, the Smith\\Approval winner, and the lowest > approval candidate that covers all higher approval candidates, etc. are > some of the members of P. > > Also, no member of P has greater MPO than IA (assuming we are talking > about Implicit Approval in the definition of P). So every member of P has > a fair chance at winning MMPO[IA>=MPO], although the winner is not > guaranteed to come from P. The main advantage of MMPO[IA>=MPO] over > MMPO(P) is that the former satisfies the FBC while the latter does not. > > For the record, here's why the entire set P survives step one: > > Let X be a member of P, and let Y be a candidate whose pairwise opposition > against X is maximal, i.e. is MPO(X). If IA(Y) is greater than IA(X), then > (by definition of P) X beats Y pairwise, and so X is ranked above Y more > than MPO(X), the number of ballots on which Y is ranked over X. In other > words, in this case X is ranked on more ballots than the number MPO(X), i.e > IA(X)>MPO(X). If IA(Y) is no greater than IA(X), then MPO(X) is no greater > than IA(X), since MPO(X) is no greater than IA(Y). Since the cases are > exhaustive and in neither case is MPO(X) greater than IA(X) we are done. > > Personally, I still prefer IA-MPO over MMPO[IA>=MPO] because of the > superior participation properties, but I recognize the importance of the > Majority Criterion in public proposals. Ironically, in reality Approval > satisfies the ballot version of the Majority Criterion, while IA-MPO does > not, yet in the face of disinformation or other common sources of > uncertainty IA-MPO is at least as likely to elect the actual majority > favorite as Appoval is. > > We need Chris to search for the chinks in the armor of these methods. > Where are you Chris? > > Forest > > > > > > > > > On Sun, Oct 13, 2013 at 10:02 AM, Kevin Venzke <step...@yahoo.fr> wrote: > >> Hi Forest, >> >> I read your first message: At first glance I think the new method (elect >> the MMPO winner among those candidates whose IA>=MPO) is good. It doesn't >> seem to gain SFC, which is actually reassuring, that this might be a >> substantially different method from others. It seems like it is mainly an >> MMPO tweak (since the MMPO winner usually will not be disqualified) with >> corrections for Plurality and SDSC/MD. >> >> Off the top of my head I can't see that anything is happening that would >> break FBC. >> >> >> > De : Forest Simmons <fsimm...@pcc.edu> >> >À : Kevin Venzke <step...@yahoo.fr> >> >Cc : em <election-meth...@electorama.com> >> >Envoyé le : Samedi 12 octobre 2013 13h58 >> >Objet : Re: MMPO(IA>MPO) (was IA/MMPO) >> > >> > >> >Kevin, >> > >> >In the first step of the variant method MMPO[IA >= MPO] (which, as the >> name suggests, elects the MMPO candidate from among those having at least >> as much Implicit Approval as Max Pairwise Opposition) all candidates with >> greater MPO than IA are eliminated. >> > >> >I have already shown that this step does not eliminate the IA winner. >> Now I show that this step does not eliminate the Smith\\IA winner either: >> > >> >Let X be the Smith candidate with max Implicit Approval, IA(X), and let >> Y be a candidate that is ranked above X on MPO(X) ballots. There are two >> cases to consider (i) Y is also a member of Smith, and (ii) Y is not a >> member of Smith. >> > >> > >> >In both cases we have MPO(X) is no greater than IA(Y), because Y is >> ranked on every ballot expressing opposition of Y over X. >> > >> > >> >Additionally in the first case IA(Y) is no greater than IA(X) because X >> is the Smith\\IA winner. So in this case MPO(X) is no greater than IA(X) >> by the transitive property of "no greater than." >> > >> > >> >In the second case, X beats Y pairwise since X is in Smith but Y is >> not. This entails that X is ranked above Y on more ballots than Y is >> ranked above X. In other words, X is ranked on more ballots than MPO(X). >> Therefore IA(X) > MPO(X), >> > >> > >> >In sum, in neither case is the Smith\\IA winner X eliminated by the >> first step in the method MMPO[IA>=MPO]. >> > >> > >> >We see as a corollary that step one never eliminates a (ballot) >> Condorcet Winner. In particular, it does not eliminate a (ballot) majority >> winner. And since MMPO always elects a ballot majority unshared first >> place winner when there is one, and MMPO is the second and final step of >> the method under consideration, this method satisfies the Majority >> Criterion. >> > >> > >> >Also worth pointing out is this: since step one eliminates neither the >> IA winner nor the Smith\\IA winner, if there is only one candidate that >> survives the first step, then the IA winner is a member of Smith, and the >> method elects this candidate. >> > >> >> I think this is right, though the method as a whole doesn't satisfy >> Smith, which is probably damning for one who finds it crucial. >> >> >> > >> >Also in view of this result, I suggest a strengthening of the Plurality >> Criterion as a standard required of any method worthy of public proposal. >> > >> > >> >A method (involving rankings or ratings) satisfies the Minimum Ranking >> Requirement MRR if it never elects a candidate whose max pairwise >> opposition is greater than the number of ballots on which it is rated above >> MinRange or (in the case of ordinal ballots) ranked above at least one >> other candidate. >> > >> > >> >What do you think? >> > >> >> You could. Chris and I discussed a "pairwise Plurality" criterion by >> which a winner can't have MPO exceeding their maximum "votes for" in some >> pairwise contest. In contexts where one uses pairwise considerations to >> make proofs regarding Plurality, SDSC, or SFC etc., you're basically using >> a stronger, pairwise-based criterion anyway. "Pairwise Plurality" implies >> both Plurality and SDSC. >> >> >> The motivation suggests to me a "Potential Approval Winner" criterion. >> Basically, the information on the cast ballots don't admit any >> interpretations by which the disqualified candidate(s) might have been the >> winner under Approval. >> >> >> >> > >> >Also we need a nice name for the set of candidates that is not >> eliminated by step one. >> > >> >Any suggestions? >> >> If not the set then at least the combined method. I'm not sure how many >> uses the set has. I'll give the method some more thought. >> >> Thanks. >> >> Kevin Venzke >> >> >
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