[EM] [Fwd: Re: Participation examples]

[Chris Benham]

-------- Original Message --------
Subject: Re: Participation examples
Date: Wed, 17 Sep 2003 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
>
>
>___________________________________________________________
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>
>  
>

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