I don't agree that "Sincere Favorite" is practically equivalent to the FBC.
The FBC is about not having to lower your one favorite candidate; it is not
about not having to pick a single favorite from your favorite set. As a
voter, I'd regard the former as a serious dilemma, and the latter as a
trivial detail.
Jameson
2011/11/23 Chris Benham <cbenha...@yahoo.com.au>
> Forest,
>
>
> "Furthermore something must be wrong with the quoted proof (of the
> incompatibility of the FBC and the
> CC) because the winner of the two slot case can be found entirely on the
> basis of the pairwise matrix."
>
> The likely explanation for some odd remarks by you and Jameson has just
> occurred to me. Could it be
> that you and Jameson have mistaken a set of ballots for a "pairwise
> matrix" ??
>
> Here is that quoted proof again, with the ballots represented in the more
> familiar EM notation:
>
> Hello,
>
> This is an attempt to demonstrate that Condorcet and FBC are incompatible.
> I modified Woodall's proof that Condorcet and LNHarm are incompatible.
> (Douglas R. Woodall, "Monotonicity of single-seat preferential election
> rules",
> Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
>
> I've suggested before that in order to satisfy FBC, it must be the case
> that increasing the votes for A over B in the pairwise matrix can never
> increase the probability that the winner comes from {a,b}; that is, it must
> not move the win from some other candidate C to A. This is necessary
> because
> if sometimes it were possible to move the win from C to A by increasing
> v[a,b], the voter with the preference order B>A>C would have incentive to
> reverse B and A in his ranking (and equal ranking would be inadequate).
>
> I won't presently try to argue that this requirement can't be avoided
> somehow.
> I'm sure it can't be avoided when the method's result is determined solely
> from the pairwise matrix.
>
> Suppose a method satisfies this property, and also Condorcet. Consider
> this
> scenario:
>
> 3: A=B
> 3: A=C
> 3: B=C
> 2: A>C
> 2: B>A
> 2: C>B
>
> There is an A>C>B>A cycle, and the scenario is "symmetrical," as based on
> the submitted rankings, the candidates can't be differentiated. This means
> that an anonymous and neutral method has to elect each candidate with
> 33.33%
> probability.
>
> Now suppose the a=b voters change their vote to a>b (thereby increasing
> v[a,b]).
> This would turn A into the Condorcet winner, who would have to win with
> 100%
> probability due to Condorcet.
>
> But the probability that the winner comes from {a,b} has increased from
> 66.67%
> to 100%, so the first property is violated.
>
> Thus the first property and Condorcet are incompatible, and I contend that
> FBC
> requires the first property.
>
> Thoughts?
>
> Kevin Venzke
>
>
> http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html
>
> It is certainly a clear proof of the incompatibilty of the Condorcet
> criterion and Kevin's later
> suggested "variation" of the FBC, "Sincere Favorite":
>
> *Suppose a subset of the ballots, all identical, rank every candidate in
> S (where S contains at least two candidates) equal to each other, and above
> every other candidate. Then, arbitrarily lowering some candidate X from S
> on these ballots must not increase the probability that the winner comes
> from S.*
> A simpler way to word this would be: *You should never be able to help
> your favorites by lowering one of them.*
>
> http://nodesiege.tripod.com/elections/#critfbc
> I can't see any real difference between this and regular FBC, which
> probably partly explains
> why it didn't catch on.
>
> Chris Benham
>
>
> *From:* "fsimm...@pcc.edu" <fsimm...@pcc.edu>
> *Sent:* Wednesday, 23 November 2011 9:01 AM
>
> *Subject:* Re: An ABE solution
>
> You are right that although the method is defined for any number of slots,
> I suggested three slots as
> most practical.
>
> So my example of two slots was only to disprove the statement the
> assertion that the method cannot be
> FBC compliant, since it is obviously compliant in that case.
>
> Furthermore something must be wrong with the quoted proof (of the
> incompatibility of the FBC and the
> CC) because the winner of the two slot case can be found entirely on the
> basis of the pairwise matrix.
> The other escape hatch is to say that two slots are not enough to satisfy
> anything but the voted ballots
> version of the Condorcet Criterion. But this applies equally well to the
> three slot case.
>
> Either way the cited "therorem" is not good enough to rule out compliance
> with the FBC by this new
> method.
>
> Indeed, the three slot case does appear to satisfy the FBC as well. It is
> an open question. I did not
> assert that it does. But I did say that "IF" it is strategically
> equivalent to Approval (as Range is, for
> example) then for "practical purposes" it satisfies the FBC. Perhaps not
> the letter of the law, but the
> spirit of the law. Indeed, in a non-stratetgical environment nobody
> worries about the FBC, i.e. only
> strategic voters will betray their favorite. If optimal strategy is
> approval strategy, and approval strategy
> requires you to top rate your favorite, then why would you do otherwise?
>
> Forest
>
>
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