Forest Simmons wrote:
And the obvious PAV solution would be 60 A's and 40 C's. Membership in the cycle didn't guarantee B any positive probability.
It is interesting to me that in my other example
34 ABC
> 33 BCA > 33 CAB
some respondents thought it was obvious that A, B, and C should have nearly equal chances, while others thought it was obvious that the deterministic winner A should be chosen 100 percent of the time.
Some thought my hypothetical coin toss was way out of line in one direction, while others thought it was way out of line in the other.
What I meant by my earlier question is, suppose the above ballot example is the result of strategic voting in which the CAB voters' sincere preference order was really CBA, where C is the sincere Condorcet loser. Any deterministic or non-deterministic method that gives C a nonzero probability of winning also gives this faction incentive to create a cycle. It's possible that the nonzero probability of an A win would be enough to counter this incentive, but probably not if this faction preferred B only slightly to A.
OTOH, the sincere CW might have been A, and the BCA ballots the result of an insincere BAC faction attempting to create a cycle. In this case any method that gives B a nonzero probability of winning also provides incentive for this behavior.
The same for the ABC ballots, if C was the sincere CW.
Of course this is not solely a problem for non-deterministic methods, but I don't see how non-determinism provides the solution.
Sorry if this rehashes questions that have already been covered. I haven't been following the "sprucing up" thread closely; I'm mainly thinking about general three-candidate examples like the one above.
Regards, Bart
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