[EM] RE: (crossposted) Revisiting Copeland

[Abd ulRahman Lomax]

My original post was bounced by the moderator 
because he thought I had made a mistake in my 
numbers. I don't think so, but, of course, I do 
make mistakes and my knowledge is less than 
perfect. He invited me to resubmit if I thought 
it was correct, which may create some confusion 
since I think Mr. Venzke already responded to me 
due to the cross-post on the election-methods 
list. If anyone wants to read my original post, 
which had some other comments not quoted below, 
it is in the election-methods archive. 
Cross-posting between a moderated and an 
unmoderated list can be problematic....

At 04:38 PM 9/9/2005, Kevin Venzke wrote:
>--- Abd ul-Rahman Lomax <abd at lomaxdesign.com> a écrit :
> > At 10:52 AM 9/9/2005, Kevin Venzke wrote:
> > >49 A
> > >24 B
> > >27 C>B
> > >
> > >A could be elected, for instance with a plurality tiebreaker.

[and I replied]
> > As I understand Copeland, truncated votes, as shown, would be
> > considered as equal ratings below the rated candidate(s), so the
> > complete description of the votes would be

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

[Mr. Venzke continued:]

>No, it would be:
>49 A>B=C
>12 B>A>C
>12 B>C>A
>27 C>B>A
>
>This is the margins treatment (that is, pretend that the B voters split half
>and half on A vs. C).

This does not seem to be how Copeland is counted. 
I'd appreciate a reference if it is. If the B 
voters are split, why not also the C voters?

I did not see in descriptions of the Copeland 
method an explicit description of how truncation 
is handled. However, there is an example 
(http://en.wikipedia.org/wiki/Condorcet_method -- 
see under Defeat Strength) which clearly 
considers unranked candidates as rated below all ranked candidates.

Mr. Venzke's method is interesting, but I don't 
see it described anywhere that Copeland is mentioned.

So I continued:

> > With Copeland, the number of pairwise victories are counted first.
> > This would be
> >
> > A: 0
> > B: 75
> > C: 78

The moderator, I think, thought this was in 
error. After all, A had 49% of the vote, and 
first preference for B was only 24% and C only 
27%. But if we consider truncated votes as 
ranking all ranked candidates above all unranked 
candidates (which is how it seems it is done), 
51% of the voters ranked A last. There are no pairwise victories for A.

> >
> > C is likewise the winner. What is this example supposed to show?


>Copeland returns a three-way tie. By the way, did you not have a problem with
>C supposedly winning?? C receives acknowledgment from only 27% of the voters!

That depends. B>C is an "acknowledgement" of C if 
it implies that C>A. If it implies that half of 
the B>C voters rank C>A, as it seems Mr. Venzke 
would analyze the result, then 39% of the voters 
would prefer C. And the way I counted it, 51% preferred C.

Do I have a problem with C winning? Yes. But C is 
the Condorcet winner, if truncations are counted 
as I have stated. Counting them as Mr. Venzke has 
counted them seems unlikely to be what the voters 
would have intended. (However, the rules should 
be explicit; but if half the B voters actually 
preferred A to C as Mr. Venzke would have it, they should have so voted!)

>The example shows that Copeland with a plurality tie-breaker can elect
>candidate A, even when more than half of the voters preferred B to A and
>didn't prefer A to anybody.

Why does Mr. Venzke split the B votes, but 
apparently not the C votes? They are the same, 
both truncated, leaving A out. I'd restate the 
election with both split, but since I don't have 
time and I think the splitting is probably an error, I'm not doing it.

>In my opinion, this is a majority rule issue.

It seems to me that Mr. Venzke is counting 
Copeland truncated votes one way in the example 
and then criticizing it based on an alternate 
count (i.e., the one I actually used). If 
Copeland is counted the way he has claimed, then 
he'd be right, there would be a problem.