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

Kevin Venzke stepjak at yahoo.fr
Fri Jan 26 14:59:26 PST 2007

```Hi,

--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> 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.

I agree. I am not trying to argue that ICC and FBC are incompatible end
of sentence. I'm only trying to defend the notion that it would be
possible to prove it, by positing a method that satisfies both, and then

> 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.

Absolutely this is possible.

> 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.

Yes.

> > > 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.

I am slightly confused when you say that a violation of ICC has not been
shown. In Forest's example, the part you quoted didn't explicitly try
to show an ICC failure, it made a conclusion based on the assumption that
the method doesn't fail ICC. Mine is similar...

> > > 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.

Well, I'm confused again. If you're talking about a criterion that could
possibly have value, you have to be talking about a criterion that can
actually be satisfied. But if so, I don't know what you believe this
criterion is.

> >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.

I merely mean that if you consolidate clones (as voted), or introduce
clones, the probability that the winner comes from this set of candidates
is not changed.

(It is possible to define clones on sincere preferences, but this is not
e.g. how Douglas Woodall interprets criteria.)

> 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.

Yes, this is "neutrality." But it complicates the analysis so I did
without it for demonstration purposes.

> >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?

If it works then this method doesn't fit the definition given.

> >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.

You can define this without referring to a "tie" by specifying just that
the situation be completely symmetrical. It doesn't matter how arbitrary or
limited-in-scope this is; as long as we can identify that A is required to
be the winner in the first election, the rest of the demonstration will
work.

Obviously, if you really want to prove an ICC-FBC incompatibility, you
will not want to assume that the method *fails* neutrality, since no
one is hoping to use methods that fail neutrality.

> 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.

No, not necessarily. The FBC and "A wins" properties allow A to win in the
second election. But at least one of the other properties would have to
be sacrificed. (Which ones? I don't know. But we can tell they are not all
compatible.)

> But we could do better than that, possibly. And it may have been done
> by Forest, in particular.

I think so. However, Forest's is more complicated, and I'm not interested
in proving something about FBC and ICC so much as I am in proving that
this kind of demonstration is viable.

(In particular, not only does Forest assume neutrality, he uses a form
of FBC that is much harder to work with. Instead of saying "no strategy
like this must work," it says "if a strategy like this works, there must
be a different strategy that works too," which is a nightmare for
trying to prove something.)

Kevin Venzke

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