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    Condorcet Score

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    • J
      Jack Waugh last edited by Jack Waugh

      This is a proposal for determining a single winner. Collect Score-style ballots. If there is a Condorcet winner, elect that candidate. Otherwise, there is a Condorcet cycle. Expand the ballots linearly so that they give full range to the Condorcet cycle and no further mention of the candidates who weren't in it. Elect the Score winner based on those results.

      Approval-ordered Llull (letter grades) [10], Score // Llull [9], Score, STAR, Approval, other rated Condorcet [8]; equal-ranked Condorcet [4]; strictly-ranked Condorcet [3]; everything else [0].

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        cfrank @Jack Waugh last edited by cfrank

        @jack-waugh I think one issue with this is that, for any more than 4 candidates, there can be more than one distinct Condorcet cycle. (Think of a pentagram, with the outer edges giving one cycle and the inner star giving another).

        This is a theorem about Hamiltonian cycles in tournaments. Although, for up to 17 candidates, there can be no more than 2 Hamiltonian cycles. For 18 candidates, the maximum number is unknown other than being at least 2 and at most 5. For 19 it’s known to be 5.

        Therefore this method generally depends on which Condorcet cycle you choose. But if we can think of a good way to deal with these 2 Condorcet cycles, we may be able to come up with an interesting (though complicated) Condorcet method that works for up to 17 (and maybe 18) candidates. And admittedly, having more than one Condorcet cycle for up to 17 candidates seems pretty unlikely, just in terms of the possible tournament configurations. I don’t know the combinatorics of that though.

        score-stratified-condorcet [10] cardinal-condorcet [9] ranked-condorcet [8] score [7] approval [6] ranked-bucklin [5] star [4] ranked-irv [3] ranked-borda [2] for-against [1] distribute [0] choose-one [0]

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          Jack Waugh @cfrank last edited by

          @cfrank, Ouch!

          OK, then, how about this one?

          Collect Score-style ballots. If there is a Condorcet winner, elect her. Otherwise, elect the straight Score winner.

          Approval-ordered Llull (letter grades) [10], Score // Llull [9], Score, STAR, Approval, other rated Condorcet [8]; equal-ranked Condorcet [4]; strictly-ranked Condorcet [3]; everything else [0].

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            cfrank @Jack Waugh last edited by cfrank

            @jack-waugh I think it’s definitely reasonable. I think that’s the usual cardinal Condorcet concept. It will still fail participation annoyingly. Steadily I might be gravitating toward approval voting, somehow. Simply because there is no other method I know of that satisfies independence of clones and participation while having game-theoretical convergence to the Condorcet criterion.

            score-stratified-condorcet [10] cardinal-condorcet [9] ranked-condorcet [8] score [7] approval [6] ranked-bucklin [5] star [4] ranked-irv [3] ranked-borda [2] for-against [1] distribute [0] choose-one [0]

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              Toby Pereira @Jack Waugh last edited by

              @jack-waugh said in Condorcet Score:

              @cfrank, Ouch!

              OK, then, how about this one?

              Collect Score-style ballots. If there is a Condorcet winner, elect her. Otherwise, elect the straight Score winner.

              I think Smith//Score is better than Condorcet//Score. Condorcet//Score could elect the Condorcet loser.

              Or what I was discussing here, which I think is just Landau//Score. But maybe that's getting a bit complex.

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