[EM] RE: lotteries unbeaten in mean

[Paul Kislanko]

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: election-methods-electorama.com-bounces at electorama.com
[mailto:election-methods-electorama.com-bounces at electorama.com] On Behalf Of
Simmons, Forest 
Sent: Wednesday, August 24, 2005 5:16 PM
To: election-methods-electorama.com at electorama.com
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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