[EM] Does MAM use the Copeland method?

[Paul Kislanko]

This is what is intuitively distatefull to us non-specialists about these
methods.

Why should the 33 percent of the voting population who most dislike "A" be
the cause of A to win, whereas 56 pecent rank C>A ? 

> -----Original Message-----
> From: election-methods-electorama.com-bounces at electorama.com 
> [mailto:election-methods-electorama.com-bounces at electorama.com
> ] On Behalf Of Steve Eppley
> Sent: Wednesday, October 06, 2004 12:05 PM
> To: election-methods at electorama.com
> Subject: Re: [EM] Does MAM use the Copeland method?
> 
> Hi,
> 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. 
> 
> The question itself isn't very clear, since some people use 
> the name Ranked Pairs as a synonym for MAM, while others, 
> including me, use the name Ranked Pairs to refer only to 
> Tideman's 1987 method and Zavist-Tideman's 1989 method.  
> Those two methods differ from MAM in several ways, the most 
> important difference being that MAM measures the size of each 
> majority primarily by the number of votes that rank the 
> majority's more-preferred candidate over their less-preferred 
> candidate, whereas Tideman and Zavist-Tideman subtract from 
> that the size of the opposing minority (that is, the number 
> of votes that rank the majority's less-preferred candidate 
> over their more-preferred candidate).  Because of this, MAM 
> satisfies some criteria that Ranked Pairs does not.
> 
> MAM constructs the order of finish of the candidates a piece 
> at a time, by considering the majorities one at a time from 
> largest to smallest, adopting into the finish order each 
> majority's preference that does not conflict
> (cycle) with the partial order of finish already constructed. 
>  The winner is the candidate that tops the order of finish so 
> constructed. (In a multiwinner context, the N winners are the 
> N candidates that top the order of finish.)
> 
> Here's a simple example to illustrate:
> 
>    9 voters, 3 candidates
> 
>    The votes:
>       4: A>B>C
>       3: B>C>A
>       2: C>A>B
> 
>    There are three majorities:
>       7: B>C
>       6: A>B
>       5: C>A
> 
>    MAM first adopts the preference of the largest majority,
>    for B over C.  Thus B finishes over C.  Then MAM adopts
>    the preference of the second largest majority, for 
>    A over B.  Thus A finishes over B.  Since MAM is
>    constructing an ordering and A finishes over B and 
>    B finishes over C, this implies A finishes over C.  
>    So when MAM considers the preference of the third
>    largest majority, for C over A, it is not adopted 
>    since it has already been established that A finishes
>    over C.  Thus the complete order of finish is A>B>C. 
>    Since A tops the finish order, A is elected.
> 
> I've oversimplified a bit.  When the number of voters is 
> small, it's not implausible that two or more majorities will 
> be the same size and it's not implausible that one or more 
> pairings won't have a majority (in other words, a pairing may 
> be a "tie.")  For more details about MAM, see my web pages at 
> <www.alumni.caltech.edu/~seppley>.
> 
> --Steve
> 
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