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

[Jameson Quinn]

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 <fsimmons at 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 <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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