[EM] An ABE solution

[C.Benham]

Forest Simmons, responding to questions from Mike Ossipff, wrote (19 Nov 
2011):

> > 4. How does it do by FBC? And by the criteria that bother some
> > people here about MMPO (Kevin's MMPO bad-example) and MDDTR 
> (Mono-Add-Plump)?
>
> I think it satisfies the FBC.

Forest's definition of the method being asked about:

> Here’s my current favorite deterministic proposal: Ballots are Range 
> Style, say three slot for simplicity.
>
> When the ballots are collected, the pairwise win/loss/tie relations are
> determined among the candidates.
>
> The covering relations are also determined. Candidate X covers 
> candidate Y if X
> beats Y as well as every candidate that Y beats. In other words row X 
> of the
> win/loss/tie matrix dominates row Y.
>
> Then starting with the candidates with the lowest Range scores, they are
> disqualified one by one until one of the remaining candidates X covers 
> any other
> candidates that might remain. Elect X.


Forest,

Doesn't this method meet the Condorcet criterion? Compliance with 
Condorcet is incompatible with FBC, so
why do you think it satisfies FBC?


http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html

> Hello,
>
> This is an attempt to demonstrate that Condorcet and FBC are incompatible.
> I modified Woodall's proof that Condorcet and LNHarm are incompatible.
> (Douglas R. Woodall, "Monotonicity of single-seat preferential 
> election rules",
> Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
>
> I've suggested before that in order to satisfy FBC, it must be the case
> that increasing the votes for A over B in the pairwise matrix can never
> increase the probability that the winner comes from {a,b}; that is, it 
> must
> not move the win from some other candidate C to A. This is necessary 
> because
> if sometimes it were possible to move the win from C to A by increasing
> v[a,b], the voter with the preference order B>A>C would have incentive to
> reverse B and A in his ranking (and equal ranking would be inadequate).
>
> I won't presently try to argue that this requirement can't be avoided 
> somehow.
> I'm sure it can't be avoided when the method's result is determined solely
> from the pairwise matrix.
>
> Suppose a method satisfies this property, and also Condorcet. Consider 
> this
> scenario:
>
> a=b 3
> a=c 3
> b=c 3
> a>c 2
> b>a 2
> c>b 2
>
> There is an A>C>B>A cycle, and the scenario is "symmetrical," as based on
> the submitted rankings, the candidates can't be differentiated. This means
> that an anonymous and neutral method has to elect each candidate with 
> 33.33%
> probability.
>
> Now suppose the a=b voters change their vote to a>b (thereby 
> increasing v[a,b]).
> This would turn A into the Condorcet winner, who would have to win 
> with 100%
> probability due to Condorcet.
>
> But the probability that the winner comes from {a,b} has increased 
> from 66.67%
> to 100%, so the first property is violated.
>
> Thus the first property and Condorcet are incompatible, and I contend 
> that FBC
> requires the first property.
>
> Thoughts?
>
> Kevin Venzke
>
>
Chris Benham