[EM] Full Majority as Part of Strong Defeat
Forest Simmons
simmonfo at up.edu
Thu Mar 17 11:11:43 PST 2005
Jobst, the more I think about it, the more I like your idea (influenced by
Kevin) of requiring full majorities for strong defeat.
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.
Forest
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