An observation on the ordinal version:
1 A>C>B
1 B>C>A
If the method is clone free and neutral, then the clone sets B'={A,C}
and A'={B,C} must have equal probability with B and A respectively.
This implies that C must have zero probability.
In the ratings version that Jobst specified in his simple poll, it
would be a stretch to say that either {A,C} or {B,C} was a clone set,
since their ratings are so distinct.
So I would think that a ratings method that gave some positive
probability to C might still be considered "clone free."
Forest
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