[EM] strategy-free Condorcet method after all!

[Warren Smith]

Jobst Heitzig's "Reverse Llull" method was intended to
(1) elect Condorcet winners (under the assumption one exists)
(2) cause strategic and honest voting to be the same thing (at least,
under perfect info assumptions)

It would be better to replace his goal #1 by this better goal
(1') elect the max-summed-utility winner
but unfortunately I cannot see any way to accomplish that.

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To make it crystal clear that (as I said before, but with less
clarity) the Reverse Llull winner can depend on the candidate
pre-ordering, consider an A>B>C>A Condorcet cycle.

Assume the Heitzig ordering is A,B,C.  Clearly with no C, the winner would be A.
So voters will "approve" C (since C>A says a majority) and it will win.
Arguing symmetrically, any one of the three will win, depending on
the Heitzig pre-ordering.

So it appears that with random pre-ordering, every member of a
top-cycle will be equally likely to be the winner with Reverse Llull.

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I believe it is shown in one of the puzzles on the CRV puzzle page,
that with random votes in the limit of a large number of candidates
and voters,  the probability-->1
that EVERY candidate is a member of a top cycle ("pan-cyclicity")
so that with probability-->1 for a "random election" Reverse Llull just becomes
"random winner."  This, of course, is very poor in terms of Bayesian Regret
compared with (say) approval voting.

So all in all, despite Reverse Llull's beauty, a case could be made that
plain approval voting is better...

-- 
Warren D. Smith
http://RangeVoting.org  <-- add your endorsement (by clicking
"endorse" as 1st step)
and
math.temple.edu/~wds/homepage/works.html