Rob,
--- rob brown <[EMAIL PROTECTED]> a écrit :
> I keep revisiting this problem, and each time, I seem to get closer and
> closer to something that I feel would work well. My general approach has
> *not* been to find a way to take existing methods (beatpath or ranked pairs
> or what-have-you) and then work backwards to produce scores, but instead to
> come up with a brand new method that produces scores first, with the top
> scoring candidate being considered the winner. Meanwhile the system must
> still meet the condorcet criterion...so if there is a condorcet winner, that
> candidate must have the highest score. Of course it must do a reasonable job
> of selecting a winner when there is a condorcet tie.
> Also it is important
> that the scores do a good job of showing how the other candidates did
> comparatively. For instance, if the #2 candidate's score is very close to
> the #1 candidate, that would indicate that a relatively small number of
> additional ballots could cause #2 to surpass #1 and win instead. Of course,
> the more stable the scores, the better.
I'd just use the beatpath ordering (that is, candidate A has a
stronger beatpath to every other candidate than vice versa and so
wins; candidate B has a stronger beatpath to everyone but A, etc.).
But you seem to not want this because it doesn't consider any
specific contests.
Additional ballots aren't a bad idea. Say that the beatpath(wv)
winner has a score of 0. Everyone else's score is the number of
bullet votes they'd neat to become the Condorcet winner. (You
could also say "to become the beatpath winner," but that would be
harder to calculate.) That gives a good idea of how close everyone
was, and still picks a good winner.
Some examples:
7 A>B
5 B
8 C
Scores are B 0, A 2, C 5.
If A gets those two votes:
2 A
7 A>B
5 B
8 C
Scores are A 0, B 5, C 5. (Tie for second.)
49 A
24 B
27 C>B
Scores are B 0, A 2, C 23.
40 A>B>C
35 B>C>A
25 C>A>B
Scores are A 0, B 31, C 50.
I think that's a decent method.
Kevin Venzke
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