[EM] Fwd: Two MMV definiions (brief, and ordered-procedure)

Michael Ossipoff email9648742 at gmail.com
Sun Dec 8 18:40:57 PST 2013

 Hi Anders--

Yes, by my definitions, MMT returns a tie between A and B. Neither has
a kept defeat.

You said that the problem is that it isn't possible, by adding one
more ballot, to make B win, and so MMT fails the Resolvability
Criterion. I daresay, MMT, and every other voting-system, fails some
criterion or other. Usually more than one, in fact.

Using a criterion to criticize a voting system, to say that another
voting system would be better, it's necessary to say why that
criterion is so necessary. Why is the method's failure of that
criterion more important than its desirable properties?

For example, say we want to use the best ideal majoritarian method,
for a poll. Should it be RP or Beatpath? RP comes in several forms:
MAM, MMT, and CIVS-RP. (But I don't know if MMV, as I now define it,
is different from CIVS-RP. It probably is.)

Maybe mid-count randomizing is unavailable or unacceptable, and so the
choice is between MMT, CIVS-RP, and Beatpath.

So we compare MMT and Beatpath:

MMT fails Resolvability. It has yet to be expained why Resolvability
is important or necessary.

Steve Eppley's simulations have shown that MAM, and therefore probably
MMT (they're both RP versions), when giving a different result from
Beatpath, will, in the vast majority of instances, give a result that
the public prefer to Beatpath's result.

RP has been found to be more stable (less likely to have its result
changed by one additional ballot), as compared to Beatpath.

Yes, Beatpath always chooses from the Schwarz set, but RP always
chooses from the Smith set. The Schwartz set is more exclusive,
because being in a tie with a Schwartz set member, when beaten by
another member, is enough to disqualify a candidate from the Schwartz
set. Sure, it's nice to be more exclusive. But if the pujblic vote
that a nonmember of the Schwartz set is as good as one of its members,
then how bad could it really be to elect hir?

RP, in all of its versions, builds a complete structure of the
strongest mutually-compatible defeats. Maybe that's why it has the
properties by which it's better than Beatpath.

So, anyway, why is Resolvability essential?

Michael Ossipoff

By the ordered procedure definition:

Order the defeats (each row is a place in the order):




Consider the first row:

AC, BC and CD are kept, because they aren't a cycle with any stronger defeats.

Consider the 2nd row:

DA and AB are discarded because they're in cycles with stronger kept defeats.

By the brief definition, the same things are true, as described above,
and the result is the same.

A and B don't have a kept defeat. The method returns a tie between A and B.

On Sun, Dec 8, 2013 at 7:54 PM, Anders Kaseorg <andersk at mit.edu> wrote:
> On 12/08/2013 06:42 PM, Michael Ossipoff wrote:
>> Brief definition:
>> Keep every defeat that doesn't contradict a set of kept stronger defeats,
>> or a set consisting of defeats equal to it, and kept defeats stronger than
>> it.
>> [end of brief MMV definition]
>> Ordered-Procedure MMV definition:
>> In order of stronger first, consider the defeats one at time, as follows:
>> Keep the considerred defeat if it doesn't contradict a set of stronger
>> kept defeats, or a set consisting of defeats equal to it, and kept defeats
>> stronger than it is.
>> [end of ordered-procedure MMV definition]
> This still doesn’t fix the problem I pointed out yesterday[1].  Are you
> intending to fix it, or have you decided to fail resolvability in favor of
> making the definition shorter?
> Anders
> [1]
> http://lists.electorama.com/pipermail/election-methods-electorama.com/2013-December/032423.html

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