[EM] MMPO(IA>MPO) (was IA/MMPO)
Kevin Venzke
stepjak at yahoo.fr
Sun Oct 13 10:02:43 PDT 2013
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 <fsimmons at pcc.edu>
>À : Kevin Venzke <stepjak at yahoo.fr>
>Cc : em <election-methods at 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.
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>
>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.
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>
>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."
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>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),
>
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>In sum, in neither case is the Smith\\IA winner X eliminated by the first step in the method MMPO[IA>=MPO].
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>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.
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>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.
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>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.
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>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.
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>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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