[EM] Re: Condorcet completed by IRV

[James Green-Armytage]

Hi folks, this is James Green-Armytage replying to Chris Benham.

>>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.

	Personally, I don't find that very disturbing. Silly, maybe, but not a
serious problem. I'm guessing that most methods meeting the truncation
resistance/SFC criteria will have this problem. Do you think that the
truncation resistance criterion is misguided? I sort of go back and forth
on it... sometimes I think it's quite important, sometimes I don't. I do
think that it's more important that the random fill incentive, however. 
>
>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".  

	Er, "completely essential"? "must meet"? "all agree"??? I don't know
about that assumption, Chris...

>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.

	My question is whether it is likely that voters will frequently be able
to exploit this strategically, and whether their efforts to exploit it are
likely to seriously distort their reported preference rankings. If not,
then I suggest that the problem is not very severe.

>>
>>  
>>
>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.)

	I was just thinking more than half of the valid vote, to keep things
simple.
>
>
>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.
>
	Okay, I'll take this method as food for thought. I'm guessing that it
doesn't meet truncation resistance, right? So I guess an underlying issue
is whether truncation resistance is important.
	I don't know. The problem using IRV to complete Condorcet is that, when
the IRV and Condorcet winners differ, there will be an incentive for
supporters of the IRV winner to bury the Condorcet winner. If we are using
a CCIRV version that passes truncation resistance, then these folks will
have to use order reversal to do the job, which may make it harder if you
assume that the voters won't all be too savvy on the system, but they can
still get it done if they put their minds to it.
	So, no hard conclusions on my end yet.

my best,
James
>