From: Markus Schulze <[EMAIL PROTECTED]> Subject: Re: [EM] How to break this tie?
Dear Chris,
here is an example to illustrate my reservations about the uncovered set.
Suppose the defeats are (sorted according to their strengths in a decreasing order):
D > A A > B B > C C > A C > D B > D
The uncovered set is {A,B,C}. When the MinMax method (or any other method that is identical to the MinMax method in the 3-candidate case) is applied to the uncovered set, then candidate A is the winner.
Now suppose some voters rank candidate A higher (without changing the order in which they rank the other candidates relatively to each other), so that the defeats are (sorted according to their strengths in a decreasing order):
A > C D > A A > B B > C C > D B > D
Now the uncovered set is {A,B,D}. When the MinMax method is applied to the uncovered set, then candidate D is the winner. Thus by ranking candidate A higher, candidate A is changed from a winner to a loser.
It is interesting that "Condorcet Lottery" (in keeping with its monotonicity) assigns probability 1/3 to A both before and after the change.
However, it seems to me that your example could be elaborated without much trouble into an example that show that Random Ballot Uncovered fails monotonicity, since when D becomes part of the uncovered set it reduces the number of ballots on which A is the highest ranked uncovered candidate. [so increased support for A would actually reduce A's winning probability]
It might be fairly easy to cook up an example in which A's probability goes from positive to zero under Random Ballot Uncovered. Suppose for example, that A were part of a subcycle such that on every ballot where A is ranked higher than the other members of the subcycle, candidate D is ranked even higher. Then when D becomes part of the uncovered set, A loses all of its probability. In this kind of example at least one of A's clones would retain some positive probability. Is there a way of doing it without the clones?
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