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

[Forest Simmons]

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 <stepjak at 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 <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.
> >
> >
> >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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