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

Abd ul-Rahman Lomax abd at lomaxdesign.com
Fri Jan 26 14:12:00 PST 2007

```At 01:45 PM 1/26/2007, Kevin Venzke wrote:

> > 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 mean that you can't devise a method that satisfies all the properties.

I mean a counterexample to the claim that both are satisfied. As far
as I'm concerned, I haven't seen one yet, that is, in this sequence
of examples.

>It sounds like you have in mind some more fundamental kind of

No. A single counterexample satisfying the assumptions proves the
matter. Technically, it would mean that one cannot satisfy both ICC
and FB in the presence of the other assumptions. That's an important
point to remember. One might be able to show, just as well that ICC
and FB were compatible if one of the other assumptions was removed
and perhaps replaced by another.

In other words, it is an incomplete statement to say that it has been
shown that ICC and FB are incompatible in ranked elections;
nevertheless, the other assumptions may be reasonable enough to be
accepted as axioms, i.e., *fundamental* assumptions the violation of
which is "unimaginable" in a desirable election method.

> > 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.

What I'm saying is that a violation of ICC has not been shown. What
appears to be a violation of ICC is actually an *appearance* of a
violation caused by insufficient caution about the application of ICC
and FB together, in the context of the example given.

> > 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.

Yes. However, what I proposed, in thought, was a criterion. Not an
incompatibility. It happens to be a criterion that must be violated
by ranked systems. But it is *not* one that is necessarily a
desirable characteristic of an election, nor that appears intuitively
obvious to be so, as ICC does.

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

I am humoring you, and appreciate that you are humoring me. If we
keep this up, we may both start laughing.

>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
>change)
>4. Clone independence

Yes. Be careful about what clone independence means. It is in the
definition and application of the criterion that the problem lies.

Further, property 1 is a special property that is not even common in
ranked methods. The names of the candidates or the sequence in which
they are listed should not matter, and I think this is specifically
mentioned in other lists of the assumptions.

>These were the three elections:
>
>1 A>B>C
>1 B>C>A
>1 C>A>B

I called this Election 1.

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

I called this election 2, or the FB election because it is election 1
modified by a voter practicing favorite betrayal in order to improve
the election outcome. Does it work?

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

I called this election 3. If we test this by inserting a clone of A,
that is, B, with the additional condition that A>B for all voters,
(which is possible, certainly), we get election 2

Conversely, we may test election 2 to see if it obeys clone
independence by removing the clone set to create election 3.

>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?

What we have is that in the first election, no winner is specified by
the assumptions, except for the assumption that A wins. If we can
assume that A wins the first election, then, we may indeed have a
problem. But I'm not going far down that road. However, could we get
around this by adjusting the names? That is, what if we *define* A as
being the winner? This might seem to be reasonable. I think it is
not. The whole point of the first election is that it is a tie, *by
the method*. If one candidate is the winner if it is a "tie," there
are no ties. A "tie" is an election that cannot be decided by the
method alone without the introduction of some other element, a
tie-breaker. This is generally perceived as fair and allowable only
if it favors no candidate over another. I think it can be said better
-- much better -- but my strong sense is that calling the winner A,
which is essentially the only way I can make sense of this
rigorous proof of this could be constructed.

If C wins election 3, then C must win election 2 if ICC is not
violated. That's clear.

What has been shown is that if we assume that A wins in a tie, if
that rule is one of the assumptions, then there is an FB violation.

But we could do better than that, possibly. And it may have been done
by Forest, in particular. If a voter can improve the outcome of the
election, and improving the probability of a favorable outcome
(rather than causing such an outcome as stated), then, yes, then we
have a possibility of showing this counterexample without the weird
assumption 1. I suspect that the problem I raised remains, but it
will, for me, take more examination, and my kids need to eat dinner
and guess who makes dinner.

```