[EM] RE : Re: RE : Re: Problem solved (for pure ranked ballot)

Kevin Venzke stepjak at yahoo.fr
Fri Jan 26 10:45:01 PST 2007


--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> Insert a clone into election 1, the result does not change if ICC is 
> satisfied. This says nothing about what happens if you insert a clone 
> into election 2.

I'm not sure why you mention this. The reason I brought up ICC where I
did, is because that is where I want to draw your attention. It wouldn't
be interesting for me to show scenarios where this hypothetical method
*doesn't* fail ICC. If a method fails anywhere, it fails period.

> Favorite Betrayal with election 1 does not change the result if FB is 
> satisfied.
> What has been done here is to apply FB to one election and then apply 
> ICC to a *different* election. This different election happens to be 
> the FB election (the one with FB incorporated), which makes it seem 
> reasonable to link them. It's a logical error; I've seen various 
> false proofs constructed with such errors, I'm sure Warren has seen them
> too.
> To show a contradiction between ICC and FB, there must be *one* 
> election which shows ICC, either by the insertion or removal of a 
> clone, and FB failure, where the election result improves for a voter who
> FBs.

I'm not sure what you mean by "contradiction" here. By "contradiction"
I mean that you can't devise a method that satisfies all the properties.
It sounds like you have in mind some more fundamental kind of 

> What you've done is to create a new criterion: If FB creates a new 
> election which satisfies ICC, the FB has been successful and thus the 
> FB+SecondElectionICC Criterion fails. You can show that FB can do 
> that, therefore no ranked method (and maybe no method at all) can 
> satisfy the FB+SecondElectionICC Criterion. Or something like that.

I don't see what difference it makes to say that "FB+SecondElectionICC"
is a method that doesn't exist, vs. a criterion that can't be satisfied.
Nor how this is different from saying a method can't satisfy both FBC
and ICC.

> Which we can promptly forget about. I'm certainly not going to add it 
> to Wikipeda or the EM wiki!

Incompatibilities don't usually have their own articles, but are often
mentioned in the articles for criteria.

Please humor me and look at the proof I made in my last post. I have a
specific question for you.

These were the properties I gave:
1. When there's a tie due to symmetry, A wins
2. When there are just two candidates, majority wins
3. You cannot cause a better candidate to be elected by lowering your
first preference (define "better" based on the ballot prior to the
4. Clone independence

These were the three elections:

1 A>B>C
1 B>C>A
1 C>A>B

1 A>B>C
1 C>A>B
1 C>A>B

1 A>C
1 C>A
1 C>A

We're considering a single method. We know its properties but we don't
know what it is.

Clearly in the first election, A must win due to property 1. Clearly in
the third election, C must win due to property 2.

Can you suggest to me which candidate can win in the second election,
so that neither FBC nor ICC will necessarily be violated by this method?

Kevin Venzke


Découvrez une nouvelle façon d'obtenir des réponses à toutes vos questions ! 
Profitez des connaissances, des opinions et des expériences des internautes sur Yahoo! Questions/Réponses 

More information about the Election-Methods mailing list