[EM] Condorcet cyclic drop rule

[Tom Ruen]

Mike,

Let me verify: Is this what you want?

I will compute for each pair election ballot:
A>B: 1 point for A
B>A: 1 point for B
A=B: no points for either
A ranked, B unranked: 1 point for A
B ranked, A unranked: 1 point for B
A unranked, B unranked: no points for either

(This simplifies to 3 cases if you give all unranked choices rank "N" for N
candidates.)

Tom

Original Message -----
From: "MIKE OSSIPOFF" <nkklrp at hotmail.com>
To: <election-methods-list at eskimo.com>
Sent: Tuesday, March 27, 2001 9:33 PM
Subject: Re: [EM] Condorcet cyclic drop rule


>
>
>
> >>About interpreting ties, my opinion is:
> >Tied rankings give 0.5 vote among 2.
> >Two unranked candidates get no votes.
>
> That isn't how we do it here. Again, partly because it eliminates
> the desirable properties of Condorcet. But also partly because it's
> undemocratic to record a pairwise preference vote that the voter
> never expressed. If I vote A & B equal, voting them both over C,
> then I'm casting A>C and B>C, but I'm not voting any preferences
> between A & B. Not half preferences. Not equal & opposite preferences.
> No preferences.
>
> Mike Ossipoff
>
>
>
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