I don't think that we lose any of the basic properties, and it solves Kevin's 49C, 24B, 27A>B problem without the additional randomness that I was beginning to accept as inevitable.
The method can be described briefly as follows:
List the candidates from top to bottom in order of decreasing total approval.
Eliminate every candidate that is beaten pairwise on more than half of the ballots by some candidate higher up on the approval list.
If more than one candidate remains, then resolve the fundamental ambiguity by democratically giving each candidate a fair chance: the remaining candidate that is ranked highest on a randomly drawn ballot wins.
In Kevin's example the sincere ballots are
49 C>>A=B 24 B>>A>C 27 A>B>>C
The approval order (from top to bottom) is B>C>A .
Candidate C is eliminated because B beats C majority pairwise, as well as in approval.
Random ballot between A and B gives respective probabilities of 27/51 and 24/51.
If the B supporters truncate A, then the approval order is unchanged, and C is still the only candidate beaten from above by a full majority, since the C>A defeat is only on 49 percent of the ballots.
The probabilities remain the same, so the truncation gives no advantage to the B supporters.
Note that if the A supporters didn't like B that much the sincere ballots would be
49 C 24 B>>A>C 27 A>>B>C
In this case the approval order would be C>A>B, and no candidate would be eliminated, since the two full majority defeats go uphill (against approval).
If the B supporters could anticipate this, they might lower their approval cutoffs:
49 C 24 B>A>>C 27 A>>B>C
Then A would be both the approval winner and the ballot CW.
The approvals order would be A>C>B.
Only C would be eliminated by strong defeat, and the probabilities would be the same as in the first example.
It is to the advantage of the B faction to (somewhat insincerely) approve candidate A.
So the method doesn't completely do away with strategy.
Note that A and B are a majority clone set, and that as long as they give a reasonable amount of support to each other they both have a fair chance of winning.
But their combined victory over C is by such a slim margin, that if they do not cooperate, then C also gets a share of the probability.
In other words, a loose clone set doesn't have as much force as a tight clone set.
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