[EM] Re: Condorcet completed by IRV

[Chris Benham]

James G-A,
You recently wrote:

>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?
>  
>
My answer to your first question is that it seems to me that it does, 
and it also seems to meet  Minimal Defense.
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.

>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.
>
>  
>
CB: In his example, in both the "before" and  "after" cases all  the 
candidates have a  "full majority" pairwise loss.
(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












>
>  
>