I was wrong earlier. Consider the following n1 + x A>B>C x C>B>A n2 + y B>C>A y A>C>B n3 + z C>A>B z B>A>C
where all variables are positive. This situation gives rise to the cycle A>B>C>A if: 1) A>B: n1+x+n3+z+y > n2+y+x+z i.e. n1+n3 > n2 2) B>C: n1+x+n2+y+z > x+n3+z+y i.e. n1+n2 > n3 3) C>A: n2+y+n3+z+x > n1+x+y+z i.e. n2+n3 > n1 In other words, none of n1, n2, n3 are greater than 0.5(n1+n2+n3) If we do the cancellations we get n1 A>B>C n2 B>C>A n3 C>A>B and without loss of generality we can assume n1>n2>n3, eliminate C, and A wins. Is A also the plurality winner? The first-place tallies are: A n1+x+y B n2+y+z C n3+z+x Still assuming that 0.5(n1+n2+n2)>n1>n2>n3, A can lose under plurality if n2+z>n1+x or n1-n2<z-x Example n1=24, n2=23, n3=3, x=1, y=2,z=10 25 A>B>C 1 C>B>A 25 B>C>A 2 A>C>B 13 C>A>B 10 B>A>C First place tallies: A 27 B 35 C 14 So, it is most accurate to say that, for 3 candidate races, cancelling out the reversal symmetry and then the rotational symmetry elects either the Condorcet winner or, if no CW exists, the first choice of the largest faction. This method is clearly inferior to plurality-completed Condorcet, although it is no worse than Borda. Anyway, after exploring what happens when we require election methods to respect symmetry, I'm forced to conclude that symmetry isn't a very useful criterion. Alex ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
