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

[Kevin Venzke]

Hi Mike,

--- En date de : Mer 26.10.11, MIKE OSSIPOFF <nkklrp at hotmail.com> a écrit :
> Kevin--
>  
> You wrote:
> 
> 
> I see this criterion in effect:
> 
> If some candidate A has simple pairwise wins over every
> candidate in
> a set of candidates "B"...
>  
> [endquote]
>  
> 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.

By allowing multiple "B" candidates, my criterion is stronger than
yours, and it is attainable.

> You wrote:
>  
> 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]
>  
> Worth checking out. But, even if you're correct about that,
> how many of those methods
> meet FBC?

None of them satisfy FBC, but neither does your version of MMPO.

> 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?

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?

> You wrote:
>  
> I am not sure that MMPO as you propose it actually does the
> trick.
> It seems to depend on the A-B tied score which I'm not sure
> is 
> 
> [endquote]
>  
> The A-B tie is sure thing. It's obviously a sure thing if
> there is only one non {A,B} 
> candidate.
> But, to preserve CD's similarity to the 3-candidate
> Approval bad-example, and CD's
>  ability to disitinguish between methods, when there are
> more than
> one non {A,B} candidates, I said, in the criterion
> premise:
>  
> "A majority prefer A and B to all the other candidates, and
> the rest of the voters prefer all
> the other candidates to A and B."
>  
> So, A and B can't not be tied.

Okay, that's true. But this is a point where my version of the criterion
is stronger than yours.

Kevin