[EM] beatpath tiebreaking (question)

[Markus Schulze]

Dear Stephen,

you wrote (2 Sep 2004):
> My question is about tiebreaking in the Schulze method
> (beatpath method).
>
> Suppose the part before the tiebreak has yielded more
> than one potential winner, and at least one candidate
> has been eliminated.  Is there any reason why we can´t
> just iterate the method, the difference this time
> being that we do not allow paths through eliminated
> candidates?

This would violate monotonicity.

Example 1:

   3 ABC
   2 BCA
   2 CAB
   2 CBA

   A:B=5:4
   A:C=3:6
   B:C=5:4

   When the method is iterated then candidate B wins with
   certainty.

Example 2:

   One CAB voter is replaced by ACB.

   3 ABC
   1 ACB
   2 BCA
   1 CAB
   2 CBA

   A:B=5:4
   A:C=4:5
   B:C=5:4

   Now, candidate A is elected with a probability of 4/9;
   candidate B is elected with a probability of 2/9; and
   candidate C is elected with a probability of 1/3.

   Thus, ranking the candidate C lower increased his
   probability of being elected.

Of course, it is arguable whether rather monotonicity or
decisiveness should be sacrificed in such a situation.

Markus Schulze