[EM] Condorcet and FBC are incompatible
Kevin Venzke
stepjak at yahoo.fr
Sat Jun 25 17:25:56 PDT 2005
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:
a=b 3
a=c 3
b=c 3
a>c 2
b>a 2
c>b 2
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
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