[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
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:
B>A: 51 wins
A>C: 49 wins
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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