Jobst and All:
I did make one
(inconsequential) boo boo. The normalization factor for the weights x+y-z,
y+z-x, and z+x-y should be 1/N, not 1/(2N), since the sum of these weights is
x+y+z=N.
To physically carry out the lottery, draw
from a bag of x+y-z red, y+z-x green, and z+x-y blue marbles.
This basic result can be extended to any
three candidate ballot set that doesn't have equal rankings:
If the ballot set has a Condorcet Winner
Candidate, then give that candidate 100% probability.
If not, cancellation of opposite
ballots like A>B>C and C>B>A will reduce the set to the already
solved case.
To see that this cancellation works,
consider the contributions to p(A), p(B), and p(C) by a set of n pairs of this
type. n:A>B>C and n:C>B>A :
Obviously p(A) and p(C) are increased by n,
which is half the number of ballots, but how about B?
The contribution to p(B) is
gamma*n/(gamma+alpha) + alpha*n/(gamma+alpha), which reduces to n.
I'll write of a practical approach in
another posting.
Forest
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