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

Abd ul-Rahman Lomax abd at lomaxdesign.com
Fri Sep 9 08:56:06 PDT 2005

```At 10:52 AM 9/9/2005, Kevin Venzke wrote:
>I couldn't support Copeland unless you use a tiebreaker that satisfies
>minimal defense. Otherwise:
>
>49 A
>24 B
>27 C>B
>
>A could be elected, for instance with a plurality tiebreaker.

It is one thing to use an example like this for discussion purposes
on the Election Methods list; it is quite enough to use it on a list
which is moving toward (or is) a general public discussion. That
example is so unlikely in a public election that we might as well say
it is impossible.

Take 101 random voters in a public election with percentages like
that . Given that almost fifty percent of this population strongly
supports A, strongly enough not to even rate the other candidates,
which is highly unlikely by itself, what is the probability that none
of the other 51 voters would likewise fail to rate A. What is the
chance that none of the other voters would actually prefer A over B or C?

Such an example may well be used to quickly show a characteristic of
an election method; and the writer doesn't want to take the time to
construct a more plausible scenario -- which is understandable -- but
no election method is perfect, probably, so it may well be enough
that harmful scenarios are very rare.

No election method is going to produce great results with a highly
polarized electorate, as is shown in this example.

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

As I read this example, the pairwise elections are

A: 49
B: 51 wins

A: 49
C: 51 wins

B: 24
C: 27 wins

So the Condorcet order is

C > B > A; C is the Condorcet winner.

With Copeland, the number of pairwise victories are counted first.
This would be

A: 0
B: 75
C: 78

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

```