[EM] Does MAM use the Copeland method?

[Gervase Lam]

> Date: Wed, 06 Oct 2004 10:05:24 -0700
> From: "Steve Eppley" <seppley at alumni.caltech.edu>
> Subject: Re: [EM] Does MAM use the Copeland method?

> Gervase L asked:
> > Just a quick question that should clear up my understanding
> > of MAM.  Is it the same as Copeland (i.e. count each
> > candidate's number of wins) except that any pairwise wins
> > that are inconsistent with the Rank Pairs ranking are
> > dropped before the Copeland score is tallied up?
>
> No, that's not MAM.

OK.  I think I was partially confused by the explanation on your web page.

Looking at your web page again, I don't think the Copeland score fits into 
what I was trying to say above.  I'll explain below using one of your 
examples on the web page:

34% x > y > z
10% y > x > z
10% y > z > x
46% z > x = y

This gives the following pairwise matrix:

   x  y  z
x  - 34 44
y 20  - 54
z 56 46  -

> By inspection of the votes, three majorities can be identified:
>
> 34% of the voters ranked x over y.
> 56% of the voters ranked z over x.
> 54% of the voters ranked y over z.
>
> MAM considers the majorities from largest to smallest:
>
> First MAM affirms the "z over x" majority.  
>
> Next, since "y over z" is consistent with "z over x", MAM affirms the
> "y over z" majority.
>
> Next, since "x over y" is inconsistent with "y over z and z over x", MAM
> does not affirm the "x over y" majority.  (The reason "x over y" is
> inconsistent is that "y over z" and "z over x" together imply
> "y over x".)

I understood the above.  This is similar to Ranked Pairs as you now have a 
ranking y > z > x.

However, the above together with what is below got me confused.  What was 
an affirmed majority?

> Alternative x is less preferred by one affirmed majority, "z over x."  
> Alternative y is not less preferred by any affirmed majority.  
> Alternative z is less preferred by one affirmed majority, "y over z."
> Since y is not less preferred by any affirmed majority, MAM elects y.

The only thing I could think of that was like this was keep the "affirmed 
pairwise wins" in the matrix and drop the "unaffirmed pairwise wins" (i.e. 
drop the pairwise wins that are inconsistent with the "Ranked Pairs" 
ranking of y > z > x).  That is, the pairwise matrix turns into:

   x  y  z
x  -  - 44
y  -  - 54
z 56 46  -

That is, the x > y pairwise win is dropped.  So let's count the number of 
pairwise loses each candidate has (i.e. the opposite of Copeland) in this 
new matrix.

x 1
y 0
z 1

y wins because it has the fewest pairwise loses in the matrix.

May be a 4 candidate example would have better illustrated things.

Thanks,
Gervase.