Date: Sat, 12 Mar 2005 16:43:21 -0800 From: Russ Paielli <[EMAIL PROTECTED]> Subject: Re: [EM] Re: Chain Climbing --> Chain Filling To: [EMAIL PROTECTED] Message-ID: <[EMAIL PROTECTED]> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Ted Stern tedstern-at-mailinator.com |EMlist| wrote:
Is it possible for the least-approved candidate to be the Condorcet winner?
I reply, Yes:
Here's the example that has inspired so much ingenuity:
49 C>>A=B 24 B>>A>C 27 A>B>>C
Candidate A is the CW, but has the least approval, only 27, compared to 49 for C and 51 for B.
Perhaps it is this weakness that makes A vulnerable to B's offensive truncation:
49 C 24 B 27 A>B
Jobst and I are convinced that B cannot confidently do this insincere truncation in the face of uncertainty introduced by a certain amount of randomization.
In my opinion, one key is to give both the CW and the Approval Winner (AW) some of the probability. Part of Jobst's approach is to perturb the approval order through randomization so that B cannot count on all of the benefits of being the approval winner. How best to combine these ideas in general is the subject of our quest.
My current tentative solution is to give every candidate that is not defeated pairwise by any other candidate with greater (randomly perturbed) approval a piece of the probability pie by random ballot.
In cases where candidate A is both the solid CW and the solid AW, no other candidate qualifies for a piece of the pie.
By "solid" I mean, "not easily perturbed."
The easier they are perturbed, the more the probability should be spread around.
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