[EM] RE: (crossposted) Revisiting Copeland
stepjak at yahoo.fr
Sun Sep 11 12:47:49 PDT 2005
There is some general confusion. In your previous message you used margins
to complete truncated rankings. I replied that your completion was not right
according to margins, since your completion changed C>B votes to C>A votes.
However, Copeland doesn't use margins. So in this message below, you believe
that I am saying Copeland uses margins.
--- Abd ulRahman Lomax <abd at lomaxdesign.com> a écrit :
> 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
Actually, this isn't even complete: The 49 voters have to be split also,
if we use margins.
> 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?
The C voters aren't split because they explicitly gave a second preference,
so that there is no equal ranking to resolve. This has nothing to do with
Copeland only looks at pairwise wins and losses, so it doesn't matter what you
do with equal rankings:
> 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.
Yes, that's right.
> > > 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.
The key word is "victories" not "votes." A has a victory over C, who has
a victory over B, who has a victory over A. Thus a three-way tie.
> >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.
B>C would be an acknowledgment of C if anyone had actually voted that way.
But no one did.
> Do I have a problem with C winning? Yes. But C is
> the Condorcet winner, if truncations are counted
> as I have stated.
No treatment of truncation creates or changes the Condorcet winner. I do not
think the way you counted the ballots in this scenario is self-consistent.
That is, I don't believe you can define a rule which treats truncation as you
propose to in this scenario.
> >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.
The C votes only truncate the last preference. The B voters truncate two
preferences. If you split C then you're counting "C>B" ballots as if they
were "C" ballots.
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