[EM] Problem solved (for pure ranked ballot

Kevin Venzke stepjak at yahoo.fr
Thu Jan 25 07:40:31 PST 2007


--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> >Proof:
> >
> >Initial election consists of three voters voting a tie, each votes 
> >A>B>C>A, but, of course, with their favorite omitted from it. We 
> >will name the voters A, B, C, after their true favorite.

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

> >If the initial condition is a tie, the voter can change the voter's 
> >vote to vote as did the voters of their second preference.

1 A>B>C
1 B>C>A
1 A>B>C (from C>A>B)

> Indeed, 
> >this creates a clone. The election now has two voters for one 
> >candidate, and one for the remaining candidate. As long as the 
> >method would elect a favorite of a 2/3 majority, it violates ICC.

Say A wins. This is only an ICC violation if when you remove one of 
the clones, the other one wins. But A would still have a two-thirds 
majority in that situation.

My thought is that Warren's proof is not very interesting if he doesn't
allow equal ranking, because it means he's testing the strong form of
FBC, which we already know is almost impossible to satisfy.

I believe a neutral, anonymous, deterministic method can only satisfy
strong FBC by treating top ranks as identical. And then, I'm sure we
can state a simple criterion that shows a basic flaw of this treatment.
(Majority favorite becomes impossible, for one thing. But I couldn't
be surprised if ICC is impossible.)

Kevin Venzke


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