[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 <[EMAIL PROTECTED]>
To:	Kevin Venzke <[EMAIL PROTECTED]>
References:	<[EMAIL PROTECTED]>

Kevin,
Thanks, very interesting. See comments below.

Kevin Venzke wrote:

Chris,

 --- Chris Benham <[EMAIL PROTECTED]> 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
[EMAIL PROTECTED]


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