[EM] MAM definitions

[Steve Eppley]

Hi,

Mike O wrote:
> Of course I don't speak for Steve, but it's my understanding 
> that MAM is used in 2 ways:  As a synonym for Ranked
> Pairs(wv).  And, more specifically, as Ranked-Pairs(wv),
> with equal-defeats dealt with by considering them in
> random order.
> 
> I believe that the 2nd definition, the more specific one, 
> is how Steve defines MAM.

It's somewhat misleading to say MAM considers equal-size 
majorities in random order.  The randomness in MAM's 
tiebreaker is in the order ballots are randomly picked, 
one at a time, as many as necessary to construct a strict 
ordering of the alternatives--only one ballot would be 
picked if the first one picked is a strict ordering 
of the alternatives--which is then used to break ties. 
(See my website at www.alumni.caltech.edu/~seppley for 
the details on how this tiebreaking ordering of the 
alternatives is used by MAM to break ties among equal-
size majorities.)

Also, that strict ordering of the alternatives may be used 
as a tiebreaker at the end, if more than one alternative 
is not the less-preferred alternative of any affirmed 
majority.

It would be simpler to just randomly sort the equal-size 
majorities, but that would sacrifice some criteria 
compliance unnecessarily, for instance complete 
independence of clone alternatives.

> The random solution is probably the best, in terms of 
> criteria compliances, for the overall method consisting
> of RP + equal-defeats solution. 
> 
> To me, if it takes a tie to make a method violate a criterion,
> that doesn't seem as important as other violations of that
> criterion. So I'm satisfied with deterministic solutions 
> for equal defeats. 

I would be satisfied too, except that I expect these voting 
methods will need to thoroughly tested in small groups 
before they will be accepted by the public for large 
elections.  In small groups, ties will be much much 
more common.

I think it's misleading to describe as deterministic the 
voting methods that may require tiebreaking at the end, 
if the tiebreaker used at the end is not deterministic.

(How's it going, Mike?  Sorry I've been so unproductive 
lately.)

--Steve