[EM] New Criterion: The Co-operation/Defection Criterion

[Kevin Venzke]

Hi Mike,

--- En date de : Jeu 27.10.11, MIKE OSSIPOFF <nkklrp at hotmail.com> a écrit :
> I'd said:
>  
> > There is only one candidate B. It would be nice if
> {A,B}
> > could be replaced
> > with a larger set of candidates, but the crierion
> would
> > then probably be
> > unattainable.
> 
> [unquote]
>  
> You wrote:
>  
> By allowing multiple "B" candidates, my criterion is
> stronger than
> yours, and it is attainable.
>  
> [unquote]
>  
> By what method?
>  
> If you're right, that's good news.
>  
> I don't deny that yur criterion is stronger. I merely
> question whether
> it's attainable.

I should say, I am starting to have a doubt as to whether my criterion
does everything I think it does. But it is stronger than yours and
should do what you want.

The simplest method I am proposing that satisfies your criterion is
MDD//Condorcet//Approval, let's say.

> > You had said:
> 
>  
> > As far as methods that will satisfy it, we at least
> have
> > some clumsy 
> > ones. Any Condorcet method used to complete
> > majority-defeat-
> > disqualification or CDTT (i.e. Schwartz set that
> replaces
> > sub-majority
> > wins with ties) will do the trick. These methods would
> also
> > satisfy
> > SDSC fully.
> > 
> > [endquote]
> 
> I'd replied:
>  
> > Worth checking out. But, even if you're correct about
> that,
> > how many of those methods
> > meet FBC?
> 
> You write:
> 
> None of them satisfy FBC, but neither does your version of
> MMPO.
> 
> [endquote]
>  
> As has already been asked, do you have an FBC failure
> example for MMPO?
>
> Do you mean that, by ranking a compromise over your
> favorite, you could get
> the compromise into a tie, by lessening the compromise's
> pairwise opposition and
> increasing your favorite's pairwise opposition--Where
> otherwise your favorite would
> go into the tie and lose?
>  
> Can you show an example?

I will try to think of one but it won't be easy, I don't think...

> I'd said:
>  
> > Can those methods be shown to meet CD? Of course, the
> > burden of proof, in the 
> > form of a failure example, is on the person claiming
> that
> > they don't. ...because it's
> > often easier to show noncompliance than compliance.
> But
> > even so, what method other
> > than MMPO can be shown to meet CD?
> 
> [endquote]
>  
> 
> I am quite confident I can convince you that my
> above-listed methods
> satisfy CD.
> 
> Try this method:
> 1. If possible, eliminate every candidate with a majority
> loss
> 2. If there is a CW among remaining candidates, elect him
> 3. Otherwise elect somebody arbitrarily (doesn't matter)
> 
> Isn't it pretty clear that this meets your criterion?
> 
> [endquote]

Ok, let me try to prove to you tha this method satisfies your 
criterion.

In the scenario you're talking about, on the cast ballots:
A will have at least a simple win over B
A will not have a majority loss to C
B will definitely have a majority win over C

MDD//Condorcet//Approval will disqualify C and will not definitely not
disqualify A.

So the only question is whether B, if he is not disqualified, can beat
A. Impossible, because A has a pairwise win over B and is the CW of
those candidates.

Have I ignored any other factors?

By the way, I am a bit confused on one aspect of what you've told me.
You said it isn't necessarily true that A has a *majority* win over
B. I thought A must have a majority because the B voters will be the
only ones not voting A>B, and the B voters certainly are not a 
majority. If that's not true though, so that some *C* voters are also
not voting A>B, then the method really can't be sure whether the CW
is A or B. In that case, a method can pick the pairwise winner between
A or B, but it can't promise which one is sincere CW, so your criterion
would be totally unattainable. I think you need to require that B
cannot possibly be the sincere CW.

> I feel that FBC is more important than CD. 
>  
> If I had to choose between FBC and CD, I'd choose FBC.
>  
> If solve-its-own-ties MMPO (which is what I mean by MMPO) 
> can be shown to fail FBC, then I'd drop it.
>  
> But, by the same token, do your CD-complying methods meet
> FBC?
>  
> If MMPO fails FBC, can you suggest a method that meets FBC,
> SFC &/or SDSC, and CD?
>  
> Can CD be shown to be incompatible with FBC?

My CD-complying methods do not satisfy FBC. (I can't prove it at the
moment, but it would make no sense to me that they do.)

I strongly suspect that in order to satisfy CD, methods must use 
something that looks or acts like a Condorcet mechanism or beatpath
mechanism, and would as a side effect fail FBC. I don't believe I can
prove that they are not compatible though.

Kevin