[EM] Random Ballot Condorcet
Forest Simmons
fsimmons at pcc.edu
Tue May 20 16:53:17 PDT 2014
Whenever Smith has fewer than four members the result is the same as random
ballot Smith, but your method does not require a computation of the Smith
set, the method is seamless, like Majority Enhancement.
Majority Enhancement of a random ballot order gives the same result in the
case of fewer than four Smith members, but it requires computing the
entire pairwise win matrix.
Date: Tue, 20 May 2014 15:55:03 -0700 (PDT)
> From: Ross Hyman <rahyman at sbcglobal.net>
> To: "election-methods at electorama.com"
> <election-methods at electorama.com>
> Subject: Re: [EM] Random Ballot Condorcet
> Message-ID:
> <1400626503.16920.YahooMailNeo at web181601.mail.ne1.yahoo.com>
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> A better random ballot Condorcet method is: Chose a random ballot (and if
> it is not complete, draw others to break ties until there is a complete
> ranking). ?Eliminate the pair-wise loser of the two lowest ranked
> candidates. ?Repeat until one candidate remains. ?Elect that candidate.
>
> I believe it has the following desired properties: monotonic, clone
> independent, only Smith candidates get a non-zero probability of being
> elected, independence of zero probability alternatives, and it requires the
> fewest number of pair comparisons and chooses the candidate that tends to
> be higher ranked than the previous version.?In the three candidate case, if
> there is a cycle, it will always choose the top ranked candidate from the
> random ballot. ?
>
> One can form a complete social ranking by starting from the lowest ranked
> candidate and moving candidates down if they lose to the one below it. ?The
> social ranking from the previous method is equivalent to starting from the
> highest ranked candidate and moving candidates up if they beat the one
> above it.
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> On Wednesday, May 7, 2014 6:51 PM, Ross Hyman <rahyman at sbcglobal.net>
> wrote:
>
>
>
> Random Ballot Condorcet: ?Choose a random ballot. ?Elect the lowest ranked
> candidate that pairwise beats all higher ranked candidates.
>
> Has this method been discussed before? ?I believe that the following are
> true: ?It will always elect a Condorcet candidate if there is one.
> ?Otherwise it will elect a member of the Smith set with some nonzero
> probability for each member of the Smith set. ?Non-Smith set candidates
> will have zero probability of being elected. ?It is monotonic in that
> raising a candidate on some ballots cannot decrease its probability of
> being elected. ?It is clone proof in that the probability of electing from
> the clone set is independent of the number of clones in the set. It is
> independent of irrelevant alternatives in that deleting a candidate with
> zero probability of winning cannot effect the probabilities for electing
> other candidates. ?
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