Forest wrote: >>Suppose that ... 1. there are three candidates A, B, and C,
2. ballot rankings are strict, 3. in each ordinal faction second ranked candidates are distributed uniformly between the other two, and 4. there is a beat cycle A>B>C>A . <<< end of quote To which I reply, "suppose not all of those conditions hold?" I wouldn't consider a method that depends upon #2 in any case. _____ From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Simmons, Forest Sent: Wednesday, August 24, 2005 5:16 PM To: [email protected] Subject: [EM] RE: lotteries unbeaten in mean Suppose that ... 1. there are three candidates A, B, and C, 2. ballot rankings are strict, 3. in each ordinal faction second ranked candidates are distributed uniformly between the other two, and 4. there is a beat cycle A>B>C>A . Let (alpha, beta, gamma) equal (m(B,C),m(C,A),m(A,B))/( m(B,C)+m(C,A)+m(A,B)), where m(X,Y) is the margin of defeat of X over Y. Then (if I am not mistaken) the lottery (alpha, beta, gamma) for picking the respective candidates A, B, and C, is unbeaten in mean. Unfortunately this lottery formula is not monotone. To see this, suppose that m(A,B) is increased without changing any of the other margins. Then the value of alpha decreases because its numerator m(B,C) doesn't change while its denominator (the sum of the cyclic order margins) increases. Is this the death knell for lotteries unbeaten in mean? Forest
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