[EM] How to break this tie?

[Markus Schulze]

Dear Nathan,

I have attached a copy of my 8 June 1998 mail to Mike
Ossipoff where I criticize Ossipoff's subcycle rule #2.

Markus Schulze

> Mon Jun 08 16:40:30 1998
> To: Mike Ossipoff
> From: Markus Schulze
> Subject: Re: better letter
>
> Dear Mike,
>
> you wrote (8 Jun 1998):
> > For the initial Condorcet count (by the previously-defined
> > Condorcet(EM) ), ignore defeats within cycles. If the winner
> > is in a cycle, then confine the election to the lowest order
> > cycle that includes it (If cycle B is an element of cycle A,
> > then cycle B is of higher order than cycle A).
> >
> > Do a count among the members of that cycle, disregarding
> > defeats in its subcycles. Again, if the winner is in a higher
> > order cycle contained in that cycle, then confine the election
> > to the winner-containing cycle whose order is one greater than
> > the order of the cycle to which the election was previously
> > confined. And, as before, defeats in cycles of higher order
> > than the one to which the election is confined are ignored.
> >
> > So then, each time, the election is confined to the
> > winner-containing cycle of next higher order, ignoring
> > defeats in its contained cycles of higher order.
> >
> > This continues till we get to the highest order cycle
> > containing the winner so far, & then we solve that cycle
> > to find the winner. I forgot to mention: Condorcet(EM) is
> > used in each of these iterative elections.
>
> If I understand your suggestions correctly, then this
> election method still fails to meet Pareto.
>
> **********************
>
> Example (6 candidates; 100 voters):
>
>   24 voters vote D > A > F > C > B > E.
>   20 voters vote B > E > C > D > A > F.
>   20 voters vote B > E > F > C > D > A.
>   20 voters vote A > C > B > E > F > D.
>   11 voters vote D > F > C > B > A > E.
>    5 voters vote A > F > C > B > E > D.
>
>   The matrix of pairwise defeats looks as follows:
>
>   A:B=49:51
>   A:C=49:51
>   A:D=25:75
>   A:E=60:40
>   A:F=69:31
>   B:C=40:60
>   B:D=65:35
>   B:E=100:0
>   B:F=60:40
>   C:D=65:35
>   C:E=60:40
>   C:F=40:60
>   D:E=35:65
>   D:F=55:45
>   E:F=60:40
>
>   As A > E > D > F > C > B > A, every candidate is in
>   the Smith Set.
>
>   Step 1:
>
>   As
>   (1) A > E > D > A,
>   (2) B > A,
>       B > D,
>       B > E,
>   (3) C > A,
>       C > D,
>       C > E,
>   (4) A > F,
>       D > F, and
>       E > F,
>  
>   A > E > D > A is a subcycle of the Smith Set.
>
>   Thus, I ignore the defeats between the candidates of the
>   Smith Set and use the Condorcet(EM) tiebreaker.
>
>   Thus, I get the following matrix of pairwise defeats:
>
>   A:B=49:51
>   A:C=49:51
>   A:F=69:31
>   B:C=40:60
>   B:D=65:35
>   B:E=100:0
>   B:F=60:40
>   C:D=65:35
>   C:E=60:40
>   C:F=40:60
>   D:F=55:45
>   E:F=60:40
>
>   As the worst defeat of candidate A is the smallest,
>   candidate A wins the tiebreaker.
>
>   Step 2:
>
>   As candidate A is in a subcycle, I have to use the
>   tiebreaker in the subcycle:
>
>   A:D=25:75
>   A:E=60:40
>   D:E=35:65
>
>   As the worst defeat of candidate E is the smallest,
>   candidate E wins the elections.
>
> **********************
>
> In the example above, candidate E is elected although
> every voter prefers candidate B to candidate E.
>
> Of course, it can be questioned, whether Pareto is
> important. But from the example above, you can also see,
> that this election method fails to meet monotonicity.
>
> If those 11 voter, who have voted D > F > C > B > A > E,
> change their opinion to D > F > C > B > E > A, then
> we have:
>
>   A:B=49:51
>   A:C=49:51
>   A:D=25:75
>   A:E=49:51
>   A:F=69:31
>   B:C=40:60
>   B:D=65:35
>   B:E=100:0
>   B:F=60:40
>   C:D=65:35
>   C:E=60:40
>   C:F=40:60
>   D:E=35:65
>   D:F=55:45
>   E:F=60:40
>
> Now, there is no subcycle anymore. Candidate B or candidate
> C wins the election, as their worst defeats are the smallest.
>
> **********************
>
> I hope, that I haven't made any mistakes.
>
> Markus