cbenham at adam.com.au
Tue Sep 27 07:21:34 PDT 2016
Your "MAM-like strategy" example:
30: A (sincere may be A>B)
And you are happy that WV and Smith//MMPO elects B (the supposed/maybe
But WV (including MAM) and Smith//MMPO both have a random-fill
incentive. From a recent email:
> C: If the method has a random-fill incentive then it is more logical
> to assume that truncators are more sincere than those
> who rank more candidates.
> M: True.
Therefore any assumption that B is the "sincere CW" has a very flimsy
basis. A, being positionally dominant (if these are 3-slot
ratings ballots then A is the most Top-Rated and the most Approved) and
uncovered is arguably the prettiest winner. Most likely
there is no "sincere CW" and A is the highest Social Utility candidate.
In any case under WV and Smith//MMPO the A supporters can make A the
winner by voting A>C. I don't think that "thwarting"
a bad "strategy" that is probably sincere or not very insincere at the
cost of allowing a more insincere strategy to work fine is
B>A 45-30, A>C 50-25, C>B 55-20. Max PO scores: A45 < C50 <B55.
A method that I like that does elect B in your example is Approval
It initially orders the candidates from most approved to least approved.
If any adjacent pair is out of order pairwise
it switches the order of the pair with the smallest difference in their
approval scores. If there is a tie for that it switches
the tied pair lowest in the order. At the end of the process it elects
the highest-ordered candidate.
The method meets Plurality and Minimal Defense and fails Weak CD. In the
second example (with 30 A>C) it narrowly elects C:
Approval scores: C55 > A50 > B45. Both C>A and A>B are pairwise out
of order and in both cases the difference in approval scores
is 5, so we switch the lower-ordered of the two pairs to give C>B>A. Now
no adjacent pair is pairwise out of order so that order is final
and C wins.
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