[EM] 3 or more choices - Condorcet

[robert bristow-johnson]

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."