Gervase, On Tues.Dec.21 you wrote:
Monotonicity to me seems to be a very fundamental requirement for ranked election methods. If I had to choose between Clone Independence and Monotonicity, but not both, then I think I would go for Monotonicity.Why? I live in Australia, where IRV's failure of Mono-raise (lack of Monotonicity) goes virtually completely
unnoticed, much less worried about. Failure of Clone Independence on the other hand, tends to have obvious
pernicious effects which begin with the nominations. (There is either a split-vote problem, or a Rich Party problem).
Mono-raise has fallen off my list of "fundamental requirements", and I now rate it as merely highly desirable.
It is somewhat hard to meet, and the best methods that meet it seem to be much more vulnerable to Burying than the
best methods that don't (as evidenced by a lot of good examples from James Green-Armytage).
Your remark was prompted by my opinion that Raynaud is better than MinMax Pairwise Oppostion (MMPO).
MMPO meets Mono-raise and Later-no-harm. Raynaud loses those but gains (Mutual) Majority, Condorcet, and
Clone Independence. Given that it seems to give similar results in the 3-candidate, lots of truncating, scenarios that
Kevin Venzke was concerned about, surely that is a great trade!
It seems to me that three versions of Raynaud are possible ( one that meets Symmetric Completion and two that don't).
The obvious one that does is Raynaud (Margins). The other two could be called Raynaud (Pairwise Opposition),
which eliminates the candidate which loses the pairwise comparison in which the winner has the highest gross score
(explicit winning votes); and Raynaud (Gross Loser), which eliminates the candidate with the lowest gross score in any
pairwise comparison.
In this example of Kevin Venzke's:
49: A 24: B 27: C>B
A>C 49-27 (m 22) C>B 27-24 (m 3) B>A 51-49 (m 2)
The three versions each give a different winner. PO eliminates A and elects C, failing Plurality.
GL eliminates B and elects C, failing Minimal Defense. Margins eliminates C and elects B.
Douglas Woodall gives this demonstration that Raynaud fails Mono-raise:
7 abc 7 bca 6 cab
"There are no trucated ballots, so the three methods are identical. I think c has the most decisive defeat here, by b, and so c is eliminated and a is elected. But if you replace two of the bca ballots by abc, then b has the most decisive defeat, by a, and so b is eliminated and c is elected."
Chris Benham
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