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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