[EM] three-slot methods

[Kevin Venzke]

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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