[Condorcet] RE:[EM] MinMax(ao)

Rob Lanphier robla at robla.net
Sat Sep 10 01:25:12 PDT 2005

On Fri, 2005-09-09 at 18:58 -0700, Simmons, Forest wrote: 
> If MinMax(AO) turns out to satisfy clone winner, then I think it would
> be the best AWP proposal, because it would be simpler to explain than
> Beatpath or Ranked Pairs, and no harder to explain than Copeland,
> especially when you take into account that MinMax has just been
> proposed by Rob Lanphier as a Copeland tiebreaker.

Whoa there.  The whole layer of a separate Approval election on top of a
Condorcet election is VERY complex for people new to Condorcet methods.
Yes, Copeland//Minmax(wv) sounds complicated when you use all of the
jargon to describe it (e.g. "Copeland", "Minmax" and "wv").  However,
when you use simple language, it's no more complicated than what many
people are already used to in sports.

The problem with Minmax(ao) (or any of the proposed Condorcet/Approval
hybrids) is that there are two scores, used at different times.  Let's
take the following example similar to one Kevin recently used (49:A,
24:B, 27:C>B).  However, let's add a couple of minor candidates to the
results, and for purposes of this example, we'll use explicit Approval
(forgetting the ballot complexity issues for a moment).


I'm going to explain the Copeland//Minmax(wv) result.  I'd like you to
explain, in layman's terms, the Minmax(ao) result.
A:3-1-0  (beats everyone, except loses to B 49-51)
B:3-1-0  (beats everyone, except loses to C 24-27)
C:3-1-0  (beats everyone, except loses to A 27-49)
D:1-3-0  (loses to everyone but E, worst loss to B 5-51)
E:0-4-0  (loses to everyone but E, worst loss to B 8-47)

Needing a tiebreaker between A, B, and C, we scan the right column for
the fewest votes against one of the three candidates.  "B" only has 27
votes against it, so that's the least consequential loss.  Throw that
result out, and B wins.

Okay, your turn.


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