1. The set P cannot be empty.
2. Any candidate that has more approval than some member of P must also be a member of P.
3. Any candidate that beats some member of P pairwise must also be a member of P.
From these conditions it would follow that the Smith set is contained inP, and every candidate that is more approved than some member of the Smith set would also be a member of P.
In particular, the Approval winner is a member of P, as is the CW if there is one.
Hereafter assume that P is the smallest set that satisfies the three conditions above.
If P is a singleton, then its only member is both the CW and the Approval winner, and according to the unique lottery based on P, this Approval/Condorcet candidate wins the lottery with 100 percent certainty.
In general (i.e. even if P is not a singleton), if W wins and the supporters of some other candidate C complain, how can the supporters of W respond?
Well, they can say that there is a sequence of candidates W=C1, C2, ... Ck=C, such that each candidate in the sequence beats the next either pairwise or approval-wise.
Furthermore, if there is a similar sequence from C to W, then C also had a positive chance of winning but just didn't have the luck of the draw.
It seems to me that this is a reasonable way to answer loser complaints.
It remains to be seen how probabilities should be assigned to the members of P in such a way that monotonicity and clone independence are satisfied, as well as the all important discouragement of insincere order reversals.
Uniform probability on P would violate clone independence because large clone sets would have a disproportionate share of the probability compared to small ones.
Perhaps random ballot restricted to the members of P would be the way to go.
Alternately, one could construct the beat clone hierarchy for P, and assign the probabilities accordingly, but this is potentially messy.
One simple example:
55 A>B>>C 45 C>B>>A
We get P={A,B} since A beats B pairwise, but B has more approval than A.[Candidate C has no beat path of any kind to either member of P.]
If we assign probabilities according to random ballot restricted to P, then the probabilities for A and B are .55 and .45, respectively.
Next time if the first faction thinks that it still has a solid majority (and still prefers A to B), they will probably vote A>>B>C. If so, then A will become the unique member of P, thereby winning the lottery with certainty.
A majority favorite always has a positive probability of winning, and as we see in the above example, a persistent majority favorite will eventually win with certainty.
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