Random Ballot Tiebreaker

[Markus Schulze]

Dear Blake,

you wrote (28 Aug 1998):
> I suggest that the way to judge the randomness of a method should
> be by outcome.  If given a certain example, method X has
> 5 possible outcomes, while method Y has only 3, I would conclude that
> in that case at least, X has behaved more randomly.
>
> The example you provide does not show this.  You will find that if you
> eliminate D, E will be the winner.  If you eliminate E, one of A, B,
> or C will be chosen randomly.  If you eliminate A, then B, then C, the
> result will be D.  So all results are possible, just as in my method.

I don't agree with you.

If you have a random tiebreaker and a situation with 5 potential winners
and your random tiebreaker guarantees that only 3 of the 5 potential
winners could win (independently on which ballots or candidates you
choose randomly), then you implicitely use an additional deterministic
tiebreaker, that reduces the number of potential winners from 5 to 3.
In other words: It is possible in your situation to get a further
reduction of the set of potential winners in a deterministic way without
violating desired criteria. In other words: You have used a random
tiebreaker in a situation, where a random tiebreaker wasn't yet needed.

To my opinion, the randomness of a random tiebreaker cannot be defined
via the number of possible winners. The reason: A random tiebreaker will
be used _only_ in those situations, in which it is not possible to get a
further reduction of the set of potential winners without violating
some desired criteria. But in these situations every potential winner
has a legitimacy to win the elections.

To my opinion, the randomness of a random tiebreaker should be defined
via the expected number of random steps.

Markus Schulze