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Ballots with cycles (reposts of excerpts + new thoughts)

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  • warren_d_smith31
    Andrew Myers: Suppose that in a Condorcet system, we allow people to submit a ballot that has an arbitrary preference relation, so any two alternative A and B
    Message 1 of 2 , Mar 5, 2008
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      Andrew Myers:
      Suppose that in a Condorcet system, we allow people to submit a
      ballot that has an arbitrary preference relation, so any two
      alternative A and B can have either A<B, A=B, or A>B. There can
      therefore be cycles in the graph of preferences, like A<B<C<A.

      One reason why we might want to set up the system this way is that we
      can protect voter privacy better by separating different preferences
      during the tallying process.

      The question is whether this creates new strategic voting
      opportunities. I have not been able to construct a scenario where it
      makes strategic voting more powerful.

      Juho Laatu:
      This makes it a bit easier to intentionally generate a loop among say
      three candidates (A,B,C) of the competing party. My vote could be
      X>A, X>B, X>C, A>B, B>C, C>A, where X is my own party candidate. If
      many X supporters vote systematically this way there is a chance that
      the candidates of the competing party will all lose to each others,
      and that might make X the winner in some Condorcet methods like
      [Simpson-Kramer] minmax if the race is otherwise very tight between
      the two parties.

      WDS illustrates to make it more concrete:
      compare
      #voters their vote
      3: A>B, B>C, C>A
      with each candidate defeated by a 3:0 majority,
      versus

      1: A>B>C
      1: B>C>A
      1: C>A>B
      with each candidate defeated by a 2:1 majority (not as severe).

      The former is more powerful in the Simpson-Kramer minmax system
      (in which the only thing that matters is the strength of your
      most-severe defeat) as a way to hurt the ABC party.
      Indeed if there were 49% X-voters and 51% ABC voters then
      X would normally lose to each of ABC, indeed is the Condorcet loser.
      However with the X-voters strategically playing the cycle game, X
      suffers only tiny-margin defeats while A,B,C each suffer severe
      pairwise defeats -- therefore X now wins the election under
      Simpson-Kramer.


      WDS:
      permitting cycles in ballots has the big advantage that it makes
      Condorcet possible even on a "dumb" totalizing machine (it is beyond
      the capability of such a machine to detect cycles and thus forbid
      illegal votes, but if we make them be LEGAL votes, then no problem).
      If there are few enough candidates that the labor of providing an
      all-pairs vote is not ridiculously large, then Condorcet would
      gain adoptability advantages.

      CONCLUSION:
      Hence it is worth investigating this further - can you prove a theorem
      - for various natural kinds of Condorcet methods that are based on
      pairwise matrices (Schulze? Tideman?) - that cyclic votes are never
      strategically useful?
    • warren_d_smith31
      I m not totally sure, but think ballots with cycles might also be strategically useful with Schulze-beatpath style of Condorcet voting. Here s a sketch of
      Message 2 of 2 , Mar 5, 2008
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        I'm not totally sure,
        but think ballots with cycles might also be strategically useful
        with Schulze-beatpath style of Condorcet voting.

        Here's a sketch of artificial scenario to which may be an example.

        Again, suppose it is again X on one side, versus A,B,C
        (and perhaps D,E,F... too) in the opposed party, and
        normally there would be a moderate-strength cycle X>A>B>X.

        Hence, normally, it would not be clear a priori who is going to win:
        A,B, or X.

        And aside from that:
        Due to random fluctuations and a small fraction of
        independent voters, normally X would
        slightly defeat some of the ABCDEF party candidates pairwise,
        while being defeated by a slight margin by the others.

        However, suppose a big chunk of the ABCDEF-voters systematically vote
        ABCDEF in a <-directed cycle.

        In that case, Schulze says "aha, I see there are strong beatpaths
        leading from each of {A,B,C...} to each of the others, and hence there
        is a beatpath from each of A,B,C... to X whose strength in each case
        is the same as the strongest pairwise defeat of X."

        Schulze also says "X has a beatpath to each of A,B,C,... whose
        strength is the same as the strongest pairwise victory by X, EXCEPT
        since the B-A margin is weak, this only works for some not
        others.

        Hence Schulze concludes (with better probability) that X does not win.

        hence the cylcing strategy paid off for the ABC.. party.


        Warning: I'm not sure this really works.
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