[EM] Symmetry and Condorcet--Correction
Alex Small
asmall at physics.ucsb.edu
Thu Jan 16 15:29:09 PST 2003
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
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