Circular Stand-Off

[Markus Schulze]

Dear Paul,

you wrote (20 May 1999):
> Markus Schulze wrote:
> > You will have to explain why you think that JITW is an improvement.
> > When you write that "there is no 'correct' candidate to withdraw (or
> > to elect) when there is a sincere circular tie," you admit that JITW
> > doesn't lead to a "better" election result than every other Smith
> > Criterion election methods (since every candidate of the top set is
> > equally qualified). And the claim that JITW guarantees that no
> > candidate will regret having competed, is simply not true.
> > 
> > Example: JITW//Smith//Condorcet[EM] is used.
> > 
> >    A:B=48:52
> >    A:C=53:47
> >    A:D=43:57
> >    A:E=44:56
> >    B:C=45:55
> >    B:D=42:58
> >    B:E=62:38
> >    C:D=54:46
> >    C:E=41:59
> >    D:E=40:60
> >    A, B, and C won't withdraw under any circumstances.
> >    Candidate D prefers B to C to E to A.
> >    Candidate E prefers C to B to D to A.
> > 
> > If no candidate withdraws, candidate A is elected.
>
> How do you get this? When A is compared to every other candidate, it
> loses more than any other. Let me be more rigorous. Let us define a vote
> for candidate A as being an instance where A is ranked higher than
> another candidate. So if we add up all the votes, A gets the least
> amount of votes (188 out of a possible 400).  
>
> > If only candidate D withdraws, candidate B is elected.
> > If only candidate E withdraws, candidate C is elected.
>
> If only E withdraws D gets the most votes (as defined above).
>
> > If both candidates, D and E, withdraw, candidate A is elected.
>
> How is this possible? C gets the most votes (102) and A is second (101).

The example above demonstrates that JITW//Smith//Condorcet[EM] doesn't
meet Steve Eppley's criterion that no candidate will regret having
competed. If you want to demonstrate that a given criterion is violated,
you cannot expect that the same example can be used for every election
method.

Markus Schulze