Consider an elimination method based on a weighted positional system.
WLOG, when dealing with three candidates, the weighted positional system
can be defined so that the candidate ranked first on a ballot gets one
point, the candidate ranked second gets w points, and the candidate
ranked last gets zero, where 0 <= w <= 1.
Furthermore, define that in the two candidate case, the weighted
positional method reduces to Plurality - it elects the candidate which
is ranked first most often, the candidate with a majority.
Then the following example should work to show nonmonotonicity for all
values of w where 0 <= w <= 0.9999:
45: A>B>C
19: A>C>B
14: B>A>C
30: B>C>A
40: C>A>B
9: C>B>A.
The outcome is that, first, B is eliminated and C wins, but when 14
voters change their opinion from A>B>C to B>A>C (lowering A), C is
eliminated and A wins.
Let's calculate this for the edge values (0 and 0.999). First, 0:
Base situation:
A: 64
B: 44
C: 49
So B is eliminated. Then:
A: 78
C: 79
So C wins.
Then, 14 voters lower A as mentioned above and:
A: 50
B: 58
C: 49
So C is eliminated. Then:
A: 90
B: 67
and A wins.
Second, 0.9999:
Base situation:
A: 117.9950
B: 97.9946
C: 97.9951
So B is eliminated. Then:
A: 78
C: 79
So C wins.
Then, 14 voters lower A as mentioned above and:
A: 117.9930
B: 97.9960
C: 97.9951
So C is eliminated. Then:
A: 90
B: 67
and A wins.
It seems that this example can be used on w = 1-eps as well, for
arbitrarily small eps > 0, but I have not proved that. If that is true,
then all we need to know is to show an example of Coombs failing
monotonicity and all WPS-based loser elimination methods that reduce to
"majority winner" in the two-candidate case will have been shown to fail
monotonicity.
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