[EM] 3 or more choices - Condorcet

DNOW1 at aol.com DNOW1 at aol.com
Fri Sep 28 13:11:09 PDT 2012

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.
Condorcet applies for legislative bodies and single or multiple 
executive/judicial offices.

Test Winner(s) versus Test Loser - with all others deemed Losers.

For Multiple e/j offices the top M votes count on each ballot -- e.g. 3 
sheriffs or judges.

A Condorcet winner wins in all TW-TL combinations.
A Condorcet loser loses in all TW-TL combinations.
BUT - may be circular ties -- thus the tiebreaker requirement.

For legislative bodies, each final winner would have a voting power equal 
to his/her final votes (i.e. direct and from losers).
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