[Condorcet] Re: [EM] RE: (crossposted) Revisiting Copeland

Abd ulRahman Lomax abd at lomaxdesign.com
Sun Sep 11 20:47:05 PDT 2005

I now see the source of the confusion. It was a simple error; I was 
then confused because the error was not noticed, instead it was compounded.

Mr. Venke had posted:
 > > > >49 A
 > > > >24 B
 > > > >27 C>B

And I translated these truncated results to

 > > > 49: A>B=C
 > > > 24: B>C>A
 > > > 27: C>B>A

Which was not correct.

The correct translation would have been

49: A>B=C
24: B>C=A
27: C>B>A

Instead of the equation of C and A by the B voters, I had somehow 
written a rank, >. And then I proceeded to analyze the incorrect 
translation. Garbage in, garbage out.

Mr. Venzke apparently did not understand that I had made a simple 
error, and I was led further astray by his response, which made 
incorrect assumptions about what I was saying....

The pairwise matrix now is:

A>B: 49
B>A: 51 wins

A>C: 49 wins
C>A: 27

B>C: 24
C>B: 27 wins

There is a Condorcet cycle, since B>A>C>B

Which is confirmed by the fact that Copeland yields a tie, each 
candidate has one victory.

Am I correct that Copeland always is a tie with respect to members of 
the Smith set? If so, Copeland is simply a way of stating Condorcet. 
Yet it is somewhat easier to understand, I suspect, which would be 
why it was proposed as worthy of attention at this time.

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