# [EM] MMPO(IA>MPO) (was IA/MMPO)

Jameson Quinn jameson.quinn at gmail.com
Sun Oct 13 09:27:07 PDT 2013

```The Simmons set?

2013/10/12 Forest Simmons <fsimmons at pcc.edu>

> 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.
>
> 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?
>
> Also we need a nice name for the set of candidates that is not eliminated
> by step one.
>
> Any suggestions?
>
> Forest
>
>
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>
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