[EM] RE: A class of ballot set with "unbeaten in mean lotteries."

[Simmons, Forest]

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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