James G-A, You recently wrote:
My answer to your first question is that it seems to me that it does, and it also seems to meet Minimal Defense.The topic of this posting is a version of IRV-completed Condorcet that seems to pass Steve's truncation resistance criteria, aka Mike's strategy free criteria.
I had proposed something else back in July, involving something I called a "UMID" set, but I think Chris has shown me that it's not good. So how about this:
1. Eliminate non-members of the minimal dominant set. 2. Eliminate all candidates who are pairwise-beaten by a full majority UNLESS this doesn't leave anyone at all. 3. Hold an IRV tally between remaining candidates.
Am I correct in thinking that this meets the criteria mentioned above?
Does this seem like a sensible way to do IRV-completed Condorcet in
general?
But it shares Condorcet (Winning Votes)'s zero-information random-fill incentive, which in my view is silly and unfair
and therefore not really acceptable.
My answer to your second question is "No". I assume we all agree that two completely essential criteria that a method
must meet are Woodall's "Mono-add-plump" and "Mono-append". The former says that if x wins, and then some ballots
are added that do nothing except vote x alone in first place, then x must still win. The latter says that if x wins, and then no
change is made except that some of the ballots that didn't rank x (above equal last), now rank x directly below the previously
ranked candidates; then x must still win.
Here is Dr. Douglas Woodall's demonstration that using IRV (aka AV) to complete Condorcet by eliminating and then ignoring
the not-allowed-to-win candidates not in the "top tier" creates a method that fails both Mono-add-plump and Mono-append.
CB: In his example, in both the "before" and "after" cases all the candidates have a "full majority" pairwise loss.abcd 10 bcda 6 c 2 dcab 5
All the candidates are in the top tier, and the AV winner is a. But if you add two extra ballots that plump for a, or append a to the two c ballots, then the CNTT becomes {a,b,c}, and if you delete d from all the ballots before applying AV then c wins.
(You don't spell it out, but I assume "full majority" means more than half those ballots that distinguish between any of the
Schwartz-set members.)
So what do I think is the best way to complete Condorcet using IRV?
Identify the members of the Schwartz set. Then, operating on the original ballots, carry on a normal IRV count until
all-but-one of the Schwartz-set members have been eliminated. That remaining Schwartz-set member wins.
(If equal-preferences are allowed, then use the split-votes version.)
Barring the trivial Smith/Schwartz distinction, this is Woodall's "CNTT,AV". It meets Mono-add-plump and Mono-append.
Without your step 2, in common with IRV it meets No Zero-Information Strategy. Woodall tables it as having a "maximal
set of properties" that include Plurality, Symmetric-Completion, Clone-Winner and Clone-Loser.
Chris Benham
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