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

[Kevin Venzke]

Hmm.

Well, looking at the RangeVoting list, it looks like this is important
to what you're saying:

>You cannot test FB and ICC together, because one requires constant
>voting from all voters and the other requires a change of one vote.

It's true that you probably can't take just two scenarios and make a
point about both FBC and ICC. But nothing stops you from saying that
in the change from some scenario 1 to scenario 2, FBC requires something
of the result, and that in a further change from scenario 2 to scenario
3, ICC requires something about the result. Of course this is only
meaningful if you declare in advance that you're talking about a method
that is supposed to satisfy both criteria.

--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> This is getting frustrating.
> 
> At 11:11 PM 1/25/2007, Kevin Venzke wrote:
> >Hi,
> >
> >--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> >Forest is saying that at least one of properties 6, 7, and 8 would be
> >violated if B didn't win.
> 
> Hello? Anybody home?
> 
> Seriously, I wasn't questioning what Forest is 
> saying. I'm questioning the specific application 
> of ICC to an FB election.

What is an "FB election"? I guess you mean a pair of elections designed
to make a point about what FBC requires.

> ICC involves comparing 
> the results of two elections. *Which two 
> elections?*

In Forest's example (the one you quoted earlier) one of the elections
is 2 B>C>A, 1 C>A>B. The election it is compared to for ICC purposes
would be either 2 B>A, 1 A>B, or else 2 B>C, 1 C>B. The bottom line is
that assuming the properties "majority wins when there are two 
candidates" and ICC, the winner in the first election has to be B.

Actually I'm not sure property 7 (Pareto) has to be invoked here.

> If I'm correct, Forest is following 
> the same line as Warren did previously. Forest's 
> proof is *much* more complex and I'm not sure I 
> followed it all and I don't have time to do so. I 
> just looked to see if he was doing the same thing 
> with regard to ICC, and it seems he is.

Let me give you a simple proof of something that uses both ICC and FBC.
I can prove that there is no method which satisfies these properties:

0. Equal ranking not allowed
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
change)
4. Clone independence

Suppose that a method satisfying all of these properties does exist.
Then given this scenario:

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

There is a tie due to symmetry, so A wins according to property 1.
Change to:

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

Lowered favorite B. By property 3 (FBC) A must still win, since any
other candidate would be an improvement caused by FB.
Change to:

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

We merged clones A and B. By property 4 (ICC) A must still win.
However, by property 2, C must win.

That's it. This method doesn't exist.

I grant that for some other definitions of FBC, this wouldn't cut it.
Having to consider expected utility of a voter given a tie, or saying
that the voter need only have *some* way of voting that doesn't involve
lowering the first preference, creates headaches here without much gain.
To the voter it is not reassuring to know he enjoys a guarantee if he
votes a certain way, without knowing what that way is.

Kevin Venzke


	

	
		
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