[EM] three-slot methods

[Forest Simmons]

I haven't had time to give these new 3-slot ideas too much thought yet,
but my first question is, "Which of them satisfy the (weak) FBC?"

MCA and 3-slot CR both satisfy the (weak) FBC, but do any of these newer
methods?

Forest

On Sat, 11 Oct 2003, [iso-8859-1] Kevin Venzke wrote:

> Gervase, and Forest,
>
>  --- Gervase Lam <gervase at group.force9.co.uk> a écrit :
> > > Date: Thu, 9 Oct 2003 00:50:47 +0200 (CEST)
> > > From: =?iso-8859-1?q?Kevin=20Venzke?= <stepjak at yahoo.fr>
> > > Subject: Re: [EM] three-slot methods
> >
> > > The voter places each candidate in one of three slots.
> > > The ballots are counted such that each voter gives a vote to every
> > > candidate placed in either the first or second slot.
> > > If no more than one candidate has votes from a majority, the candidate
> > > with the most votes wins.
> >
> > I assume here that you mean a majority of the votes, which can be in
> > either the first or second slot.
>
> Maybe; I mean "votes from a majority of the voters."  And yes, first or second
> slot rankings are the definition of "votes."
>
> > > Otherwise, eliminate the candidates who don't have votes from a
> > > majority,
> >
> > This is where I get lost.  If no "...candidate has votes from a
> > majority..." that means that all of the candidates have been eliminated.
>
> The "otherwise" means if the "if no more than..." condition is not true.
> So if more than two candidates have majority approval, the others are eliminated.
>
> However, I've found a scenario with a pretty unsatisfactory result.  MAR can
> fail to elect a Majority Favorite.  Consider:
>
> 25 A>B>C
> 30 A>C>B
> 45 B>C>A
>
> All three candidates have a majority.  No one is eliminated, and no voter is
> able to move their cutoff.  Thus C wins.  We can't even fix this problem by
> eliminating the lowest scorer one at a time, because A would be eliminated first!
>
> "MAFP" would actually do a better job.  The second round looks only at the first
> preferences, so A would win easily and C would receive no votes.  (Whenever MCA
> would elect a candidate for being Majority Favorite, MAFP also picks that
> candidate.)
>
>
> I wonder if this is a graceful compromise between solving the later-no-harm
> problem of Approval, avoiding cycles and CWs, and avoiding an arbitrary cutoff:
>
> "Elect the winner of the pairwise contest between the candidate with the most 1st
> rankings, and the candidate with the most 1st+2nd rankings."
>
> Not sure what such a method should be called.  It could be summable with a pairwise
> matrix, or the votes could simply be counted twice.
>
> In the example above, A and C would be the finalists, and the 30 voters would
> not regret that they put C in the middle slot.
>
> 40 A>B>C
> 35 B>C>A
> 25 C>A>B
>
> Finalists A and B, with A winning.  MCA and MAR pick B; MAFP and Schulze pick A.
>
> 12 A>B>C
> 7 B>>AC
> 11 C>>AB
>
> A vs. B, A wins.  MCA, MAR, and MAFP pick B; Condorcet methods pick A.
>
> change the 7 voters to:
> 7 B>C>A
>
> Same result: A wins.  MCA, MAR, and Schulze(wv) pick B (later-no-help issue with the
> latter); MAFP actually picks C now (both B and C have a majority, and C is the favorite
> of more voters).
>
>
> I'm having difficulty coming up with a scenario where this new method doesn't elect
> a CW, but it must be possible.  The low resolution of the ballot makes the task
> tough.
>
> Also, I'm not sure yet what kinds of candidates one would tend to place in the middle
> slot.
>
>
> Kevin Venzke
> stepjak at yahoo.fr
>
>
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