[EM] Does MAM use the Copeland method?

[Steve Eppley]

Hi, 
Adam H wrote:
> The point of the example was to show the way the method 
> performed in an election where each and every person was
> pairwise beaten by one of the others.

I hate to quibble, but I merely wanted a simple example
to demonstrate how MAM works.

But yes, I basically agree with Adam's response to Paul,
and want to add that the context we are working in
assumes that one and only one of the candidates must 
be elected. (Of course, that's a criterion, and Paul 
doesn't care about any criteria, or so he says.)

> A loses to C in 5/9 votes.
> B loses to A in 6/9 votes.
> C loses to B in 7/9 votes.
> 
> Paul Kislanko <kislanko at airmail.net> wrote:
>> I KNOW most pairwise methods elect A in this example. 
>> But pairwise A loses to C by a majority, so why do 
>> the methods elect A?

Assuming one of the candidates must be elected, which one?
If not A, then Paul either wants to elect B, which loses
to A by a larger majority, or he wants to elect C, which
loses to B by a larger majority.  Size matters.

By the way, as I pointed out earlier, non-pairwise methods 
also elect A.

>From another message, here's Paul's latest wording of 
his question:
> 5 of 9 voters voted C>A. 
> Paul's question is how can anyone justify A's win.

Answer: One of the candidates must win; which would be
a better winner than A?  Paul should clarify his point 
by answering this question.

--Steve