Schulze tie-breaker, monotonicity problems
Blake Cretney
bcretney at my-dejanews.com
Tue Nov 3 13:23:30 PST 1998
This is my example of Schulze violating monotonicity.
It's a little different than the normal examples, because
the lower ranking change does not ensure the candidates
victory, but increases it from 0% to 33%. I consider this
a monotonicity violation, although you may not.
This is why I do not use the technique of using an additional
round as an LCM tie-breaker, as is done in Schulze.
3 A B C D
2 D A B C
2 D B C A
2 C B D A
A B C D
A X 5 5 3
B 4 X 7 5
C 4 2 X 5
D 6 4 4 X
VA table
A B C D
A X 5 5 0
B 0 X 7 5
C 0 0 X 5
D 6 0 0 X
D >> A 6:5
B >> C 7:5
Eliminating A and C, we get
B D
B X 5
D 0 X
B wins.
----
3 A B C D
1 D A B C
1 A D B C -- represents a lowering of D by one voter
2 D B C A
2 C B D A
A B C D
A X 5 5 4
B 4 X 7 5
C 4 2 X 5
D 5 4 4 X
VA table
A B C D
A X 5 5 0
B 0 X 7 5
C 0 0 X 5
D 5 0 0 X
B >> C 7:5
Eliminating C, we get
A B D
A X 5 0
B 0 X 5
D 5 0 X
With a % chance of victory of
A 4/9
B 2/9
D 3/9
assuming the random ballot tie-breaker
So reducing D's support has changed it from a 0% chance of victory
to 33%. I consider this a violation of monotonicity.
---
Blake
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