[EM] Top Three Condorcet

[Forest Simmons]

On Wed, 9 Jun 2004, Dave Ketchum wrote:

> On Wed, 9 Jun 2004 09:56:24 -0700 (PDT) Forest Simmons wrote:
>
> > On Wed, 9 Jun 2004, Dave Ketchum wrote:
> >
> >
> >>On Tue, 8 Jun 2004 14:06:19 -0700 (PDT) Forest Simmons wrote:
> >>
> >>
> >>>If I understand correctly, Beat Path, Ranked Pairs, MinMax, and all of the
> >>>other serious Condorcet methods are in agreement (except for the
> >>>margins/wv debate) when there are only three candidates: if one of them
> >>>beats each of the others pairwise, then that candidate is the winner.
> >>>Otherwise, the cycle is broken at the weakest link.
> >>>
> >>>So why not take advantage of this agreement by using some simple but
> >>>reasonable method to eliminate all but three candidates and then among
> >>>those three
> >>>
> >>>    If there is a cycle
> >>>       Then break it at the weakest link
> >>>       Else go with the one who beats the other two.
> >>>
> >>>
> >>>Elimination methods that eliminate all the way down to two candidates
> >>>offer too much order reversal incentive, but if there is room for three
> >>>finalists, then that incentive may be negligible.
> >>>
> >>>Here's a more specific proposal along these lines:
> >>>
> >>>Use grade ballots.  The three finalists are A the candidate with the
> >>>greatest number of top grades, B the candidate with the highest grade
> >>>point average, and C the candidate with greatest number of passing grades.
> >>>
> >>>If all three of these turn out to be the same candidate, then this
> >>>candidate wins.
> >>>
> >>>If the set {A,B,C} has only two distinct members, then whichever wins
> >>>pairwise between them is the method winner.
> >>>
> >>>If all three are distinct, and one of them beats the other two pairwise,
> >>>then that one is the winner.
> >>>
> >>>If there is a three way cycle, then the cycle is broken at the weakest
> >>>link.
> >>>
> >>>This method is summable, easy to understand, and hard to criticize, though
> >>>I'm sure the purists will have plenty to say :')
> >>>
> >>>The main disadvantages I see are (1) the controversy over margins versus
> >>>winning votes, and (2) some folks think that it is too hard to grade the
> >>>candidates.
> >>>
> >>>Any other proposals for Top Three Condorcet?
> >>>
> >>>Forest
> >>>
> >>>
> >>What is a grade ballot?
> >>
> >
> > A ballot on which each candidate may be graded on some scale, e.g. A
> > through F or (Steph's idea) A through Z.
> >
> >
> >>Anyway, a method using it should not be called Condorcet.
> >>
> >>
> >
> > Condorcet's method is used to determine the winner from among the three
> > finalists.  As long as there are at least three grades, a ranking can be
> > inferred from the grade ballot.  Actually the name isn't Condorcet, it is
> > "Top Three Condorcet," a name for a class of methods, not aparticular
> > method.
> >
> > If you have a more apt name, I'm willing to consider it.
> >
>
> I am not ready to try to name this creature, which I doubt deserves existence.
>
> Neither do I claim the muscle to be able to demand renaming, unless
> someone else sees what I see and joins in.
>
> What I see as Condorcet is that it is an IMPORTANT method name, and that
> its voting is on a ranked ballot, with NO COMPLICATIONS for the voter such
> as rating or grading.

You can get your three finalists by any method that you like, using any
kind of ballot that you like.  But the simpler the better, because there
are only two justifications for the existence of this class of methods:
(1) The Condorcet step is much easier to describe than Ranked Pairs or
SSD. (2) The winner among the three finalists is the one universally
agreed upon by all Condorcet fans (mod the wv/margins controversy).


>
> I also see it usable in an election for governor, with the arrays of vote
> counts being the ONLY information forwarded by precincts for central
> summing and analysis.
>
> >
> >
> >>Finally:
> >>      If there is no cycle, what is there to brag about here?
> >>
> >
> > Depends on the way that the three finalists are chosen.
> >
>
> Again, if there is no cycle in the voters' ranking, what value would there
> be in grading?

Only if you want to use grades to determine the three finalists.

>
> >
> >>      If there is a three way cycle, why not solve it via Condorcet?
> >>
> >
> > That's exactly what is proposed here.
> >
>
> In context, my saying "in voters' ranking" should clarify.

Are you suggesting that we check for cycles before deciding on the
election method?

In any case, it is much easier to detect and deal with cycles after the
field has been narrowed down to three candidates than it is with twenty or
thirty candidates.  A novice could do it by hand in the case of three
candidates, after a very brief explanation of three candidate Condorcet.

>
> >
> >>      For other cycles, are we not better off dealing with what should be
> >>rarities by agreeing as to how to solve them via ranked ballots?
> >>
> >
> >
> > There cannot be other cycles when there are only three finalists.
> >
>
> Same as above - point is that Condorcet has no trouble disposing of all
> the minor candidates that are not in a cycle.  Here I claim that it should
> be rare, indeed, to have more than three candidates strong enough to be in
> a cycle for Condorcet without your grading.
>

My point is that it is a lot simpler to explain the Condorcet step to the
voters when there can be only three candidates under consideration:

      If there is no cycle, then the candidate that beats both of the
others is the winner.  If there is a cycle, break it at the weakest link.


Try describing Beatpath or Ranked Pairs that simply.


Forest