[EM] How to break this tie?

[Forest Simmons]

> From: Markus Schulze <markus.schulze at alumni.tu-berlin.de>
> Subject: Re: [EM] How to break this tie?
>
> Dear Chris,
>
> here is an example to illustrate my reservations
> about the uncovered set.
>
> Suppose the defeats are (sorted according to their
> strengths in a decreasing order):
>
>  D > A
>  A > B
>  B > C
>  C > A
>  C > D
>  B > D
>
> The uncovered set is {A,B,C}. When the MinMax method
> (or any other method that is identical to the MinMax
> method in the 3-candidate case) is applied to the
> uncovered set, then candidate A is the winner.
>
> Now suppose some voters rank candidate A higher
> (without changing the order in which they rank the
> other candidates relatively to each other), so that
> the defeats are (sorted according to their strengths
> in a decreasing order):
>
>  A > C
>  D > A
>  A > B
>  B > C
>  C > D
>  B > D
>
> Now the uncovered set is {A,B,D}. When the MinMax
> method is applied to the uncovered set, then
> candidate D is the winner. Thus by ranking candidate A
> higher, candidate A is changed from a winner to a loser.


It is interesting that "Condorcet Lottery" (in keeping with its 
monotonicity) assigns probability 1/3 to A both before and after the 
change.

However, it seems to me that your example could be elaborated without much 
trouble into an example that show that Random Ballot Uncovered fails 
monotonicity, since when D becomes part of the uncovered set it reduces 
the number of ballots on which A is the highest ranked uncovered 
candidate. [so increased support for A would actually reduce A's winning 
probability]

It might be fairly easy to cook up an example in which A's probability 
goes from positive to zero under Random Ballot Uncovered.  Suppose for 
example, that A were part of a subcycle such that on every ballot where A 
is ranked higher than the other members of the subcycle, candidate D is 
ranked even higher.  Then when D becomes part of the uncovered set, A 
loses all of its probability.  In this kind of example at least one of A's 
clones would retain some positive probability. Is there a way of doing it 
without the clones?


Forest