[EM] calculating margins in condorcet

[Gervase Lam]

> From: Gervase Lam
> Subject: Re: [EM] calculating margins in condorcet
> Date: Thursday 22 April 2004 00:27 am

> Would MinMax Condorcet (a.k.a. Plain Condorcet) do?  Just draw up the
> pairwise matrix then order the candidates by each of their worst
> pairwise defeats.
>
> Or may be something based on Copeland (i.e. number of pairwise wins)?

I thought of one method of doing this that just depends on the final 
ranking obtained from the election.  I suppose it has very slight 
similarities to Borda and the method Ernest posted.  It uses Copeland 
scores.

I'll use the example that Curt used to start this thread:

A->B  67->33
A->C  54->46
B->C  51->49

Election result is A>B>C.

Start from the bottom ranked candidate, C, and give the Copeland score of 
C to this candidate and the candidates ranked above C.  As C has a 
Copeland score of 0, therefore the other candidates also have a score of 0.

Go to the next candidate up, B, and add the Copeland score of B to this 
candidate and the candidate(s) ranked above B.  The candidates now each 
have a score of A=1, B=1, C=0.

The above step is repeated for the next candidate up, A.  Therefore, the 
final scores are A=3, B=1 and C=0.

Due to the fact that the example given by Curt is transitive, as he 
explained in the second paragraph of his post, the final scores are quite 
boring.

Another method, which is a closer to the method Ernest posted, could be to 
use the "reverse" MinMax scores of each candidate.  That is the (No. of 
ballots) - (MinMax score) of each candidate.

The same method as described above could be used except that the "reverse" 
MinMax score is used instead of the Copeland score.  This results in 
A=182, B=82, C=49.

Like the method Ernest mentioned, they still have the unfortunate 
attribute of the scores of the low ranked candidates being very low in 
comparison to the high ranked candidates.  The sacrifice you make for 
making the scores match the ranking.

As a result, I think I still prefer to use plain old MinMax.  The scoring 
is directly related to the method itself.  However, this does mean you are 
stuck with the unremarkable but nevertheless not bad MinMax.

Thanks,
Gervase.