[EM] 3 or more choices - Condorcet

[DNOW1 at aol.com]

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.
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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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