[EM] Condorcet cyclic drop rule
Tom Ruen
tomruen at itascacg.com
Tue Mar 27 17:01:28 PST 2001
I see I'm interpreting weakest defeat differently. I assumed it was smallest
difference between winner and loser among all pair elections.
About interpreting ties, my opinion is:
Tied rankings give 0.5 vote among 2.
Two unranked candidates get no votes.
This can be important depending on how weakest defeat is determined.
----- Original Message -----
From: "LAYTON Craig" <Craig.LAYTON at add.nsw.gov.au>
To: <election-methods-list at eskimo.com>
Sent: Tuesday, March 27, 2001 6:38 PM
Subject: RE: [EM] Condorcet cyclic drop rule
> Tom wrote:
>
> >However if some voters bullet vote:
> >Example ballots: AC=3, A=2, BA=4, CB=3
> >
> >Pair elections:
> >A:B=5:7 - diff=2, ratio=1.4
> >B:C=4:6 - diff=2, ratio=1.5
> >A:C=9:3 - diff=6, ratio=3
> >
> >Which is the weakest defeat? (With some work I could come up with a case
> >that the difference and ratio pick different defeats, but a tied
difference
> >is still interesting.)
> >
> >I would judge the weakest defeat as the one in the election with more
> voters
> >participating. Therefore the 5:7 defeat is "weaker" than the 4:6 defeat.
> >
> >Do these examples look correct? Might weakest ratio be a better criterion
> >than weakest difference?
>
> Some people interpret the weakest defeat to be the one in which the
smallest
> number of voters voted for the winning candidate. By this definition, the
> 4:6 defeat is weaker. The rationale is that you are overruling less
voters
> when you ignore this result.
>
> However, I'm not sure I agree with this. How should we interpret tied
> preferences (eg A=B>C>D=E)? There is one view (with which I'm inclined to
> agree) that gives each candidate in a pairwise tie 0.5 votes. Truncating
> your vote is the same as tieing all of the unranked candidates - the
> previous example should be equivalent to A=B>C. If you count the ballots
in
> this fashion, all of the ways to interpret pairwise defeats are
equivalent -
> ratios, winning votes, margins.
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