# 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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