[EM] MAM vs Schulze

Forest Simmons fsimmons at pcc.edu
Sun Oct 9 14:13:11 PDT 2016

Do I remember correctly that MAM is just Ranked Pairs with a better tie

It seems like MAM and Ranked Pairs put in as many defeats as possible
without creating cycles, but like Juho says do we really need all of those
defeats to decide the winner?

River recognizes that we just need one defeat for each non-winner. It seems
to me that the one defeat for each non-winner should be as strong as
possible, but (having already been eliminated) their other defeats don't

I want to propose a new Condorcet Method below, but first a simple three
slot method inspired by Jameson's MAS, but one that truly satisfies the
Chicken Defense Criterion:

Ballots are scores or ratings on a scale of zero to three.

The Smith candidate with the highest average score wins.

In other words, the method is Smith//Score (or is it Smith\\Score ?).with
three slot ballots.


49 C
27 A>B
24 B (sincere is B>A)

With sincere votes A is elected as the only member of Smith.

Under the B faction defection C wins as the Smith candidate with the
highest Score.

Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.

For each ballot on which a candidate is ranked above bottom but below top
that candidte receives one point.  For each ballot on which the candidate
is ranked top or equal top that candidate receives two points.

The Smith candidate with the greatest number of points wins.

[End of definition]

Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.

How does it do on burial?

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