[EM] 3 or more choices - Condorcet
rbj at audioimagination.com
Fri Sep 28 13:31:29 PDT 2012
On 9/28/12 4:11 PM, DNOW1 at aol.com wrote:
> A > B
> Choice C comes along.
> C may - head to head ---
> 1. Beat both
> C > A
> C > B
> 2. Lose to both
> A > C
> B > C
> 3. Beat A ---- BUT lose to B
> C > A > B > C
> Thus, obviously, a tiebreaker is needed in case 3.
> Obviously perhaps Approval.
> i.e. BOTH number votes and YES/NO Approval votes.
> Obviously much more complex with 4 or more choices.
> ANY election reform method in the U.S.A. has to get past the math
> challeged appointed folks in SCOTUS.
> i.e. ANY reform must be REALLY SIMPLE.
well, there is Ranked-Pairs and there is MinMax. for a cycle involving
only 3 candidates, both will elect the same candidate and the same
candidate as Schulze (which appears to be the "best" method according to
a few different criteria, but is *not* simple).
the methods may diverge when there are 4 or more candidates in the Smith
set, but i would bet that in a public, governmental election using a
ranked ballot decided by a Condorcet compliant method, that this would
never ever ever happen. a cycle will rarely happen and a cycle greater
than a simple ring of 3 is far more unlikely.
nonetheless, if it were to happen, any of these methods will resolve the
election unless it truly is a tie, like 3 groups of voters of exactly
the same size all voting A>B>C (33%), or B>C>A (33%), or C>A>B (33%).
this is comparable to a simple two-person FPTP election with a dead tie
(and we actually had that in Burlington for the Democratic mayoral
caucus last December: there were 4 candidates and 3 rounds and the last
round came out to be a 540-540 tie). then the law needs to deal with
that with either another election or with drawing lots or perhaps a
legislative tie-breaking decision. but that would be no different
regarding a perfectly symmetrical Condorcet cycle.
r b-j rbj at audioimagination.com
"Imagination is more important than knowledge."
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