[EM] [Fwd: Re: Participation examples]

Wed Sep 17 08:29:03 PDT 2003

-------- Original Message --------
Subject: Re: Participation examples
Date: 2003年9月17日 15:16:04 +0930
From: Chris Benham <chrisbenham at bigpond.com>
To: Kevin Venzke <stepjak at yahoo.fr>
References: <20030916221203.95947.qmail at web13807.mail.yahoo.com>
Kevin,
Thanks, very interesting. See comments below.
Kevin Venzke wrote:
>Chris,
>> --- Chris Benham <chrisbenham at bigpond.com> a écrit : 
>>>>Kevin,
>>Regarding the 3 methods I have recently posted: "Approval Runoff" 
>>(AERWE), "Improved Generalised Bucklin" (featuring "forwards-backwards 
>>runoff"), and Condorcet completed by reduced-rank Condorcet.
>>I would be very interested in seeing any examples of Participation failure.
>>>>>>I've come up with one each. You may want to check my IGB work.
>>Condorcet completed by utility-based compression:
>40 b 100, c 80, a 0
>35 c 100, a 25, b 0
>>B is the CW and wins.
>>Add in:
>25 a 100, b 25, c 0
>>Now there's a cycle. When the ballots are compressed to two-rank, C is the winner.
>CB: A pity, but is there a better Condorcet completion method? It is 
certainly a better way to count high-resolution
CR ballots than just adding them up (and yet, with sincere voting, 
likely to get the same result.)
>"AERWE" (Approval Runoff):
>>6 A>D>C | B (The pipes are the approval cutoffs.)
>5 B>D>C | A
>A has a majority and wins.
>>Add in:
>4 C>A | B>D
>First B and then A are shunted, at which point D has a majority.
>CB: Agreed. My/our version of Condorcet-Approval hybrid gives the 
same result.
>Improved Generalised Bucklin (IGB):
>>6 A>E>B>C>D
>5 B>E>A>C>D
>The final pairing is A vs. E, and A wins, clearly.
>>CB: I think it is safe to say that this method meets the Majority 
criterion.
>Add in:
>4 C>D>A>B>E
>Now the final pairing is E vs. himself, so E wins.
>>No, I get A as the second finalist. Then A>E 10-5, and so A (the 
CW) wins.
IGB second finalist.
Elimination mini-election 1.
Rnd.1: A:0 B:0 C:0 D:11 E:4 Eliminate D.
Elim. m-e 2.
Rnd.1: A:0 B:0 C:11 E:4 Eliminate C . Now the ballots 
equate to:
6:A>E>B
5:B>E>A
4:A>B>E
Elm. m-e 3.
Rnd.1: A:5 B:6 E:4
Rnd.2: A:5 B:10 E:15 All ballots contribute to either or both of 
E and B, so eliminate E.
Then A>B 10-5 and so A is the second finalist.
Chris Benham.
>>Please tell me if your results differ.
>>>Kevin Venzke
>stepjak at yahoo.fr
>>>___________________________________________________________
>Do You Yahoo!? -- Une adresse @yahoo.fr gratuite et en français !
>Yahoo! Mail : http://fr.mail.yahoo.com
>>>>
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