[EM] MMV and resolvability

[Michael Ossipoff]

 Again, Markus says that the strongest defeats are all skipped
(discarded) because they're in a cycle. But neither my definitions of
MMV, nor that of Prabhakar, say to do that. Remember that Anders
Kaseorg was specifically referring to MMV as defined by Prabhakar.

Because my brief, non-ordered, definition can sometimes fail to give
any resut at all--which could be called a "tie",but which could be
considered undesirable, I'll refer to my ordered-procedure definition,
which I stated in a recent previous posting here.

Markus' 2nd example, like his 1st one, doesn't show a nonmonotonic
result with MMV, by my ordered procedure definition. MMV starts by
considering the strongest set of defeats, and it keeps them because
they aren't in a cycle consistisng only of them and stronger defeats.
So, at that point, Markus' narrative is in error, with regards to what
MMV (as I define it) does.

And, if you don't consider my definition relevant to this dicussion,
remember that Prabhakar refers to me as the proponent of MMV.

I was referring to my own ordered-procedure definition of MMV. Here is
Prabhakar's posted definition at his website:

[quote]
This list of matchups is used to rank the candidates, starting from
the largest win on down. The order is important, because in rare cases
[5]a later matchup may conflict with an earlier one:

(i) If a matchup later in the list conflicts with the
previously-determined order, the latter matchup is superseded
(ignored).

(ii) In the even unlikelier case where several matchups with same-size
majorities conflict with each other, all such conflicting matchups are
ignored (though any non-conflicting matchups of that size are still
included).
[/quote]

Paragraph (i) doesn't say to ignore the strongest defeats in Markus'
1st or 2nd example, because there is no previously-determined order
for them to conflict with. They aren't later in the list.

Paragraph (ii), admittedly, does say to ignore cycles of same-size
defeats. By a literal interpretation, that would mean ignoring
(discarding) all such cycles in Markes' examples.

I suggest that that isn't the intent of Prabhakar's definitiion. I
suggest that that definition should be modified to speak of keeping
instead of ignoring.

Then, when the strongest, first-considered, set of equal defeats are
considered, they're all kept, because there is no
preiously-determiined order with which for them to conflict.

(ii) could be re-worded to say:

"(ii) When, considered together at the time appropriate to their
strength,  several matchups with same-size majorities conflict with
each other,or eachother and stronger defeats, all such conflicting
matchups are skipped (declared not kept) (though any non-conflicting
matchups of that size are still included)."

In a recently previous posting here, I stated my ordered-procedure MMV
definition.Of course I like my own definition best. I wrote it because
it's what I mean by MMV.

Markus is saying only that, as currently worded, Prabhakar's
definition can be interpreted to discard the strongest defeats, and
then keep a weaker one. But that isn't an inherent property of the
Drop-Equal-Defeats way of dealing with equal defeats.

In Markus' 1st example, after the vote-change, it might, at first,
appear that A is the rightful winner, because it has a defeat that's
weaker than all of the other defeats. But I remind you that A has
another defeat that is among the strongest.

Markus hasn't showed a problem with Ranked-Pairs in general, or with
the Discard-Equal-Defeats version in particular (as defined by me). He
has shown only that Prabhakar's definition needs a small modification.

Either by my ordered-procedure definition, or by a slight modification
of Prabhakar's definition, DED/MMV doesn't have nonmonotonicity in
Markus' examples, and hasn't been shown here to have a problem.

I suggest DED/MMV for voting when mid-count randomization is
unavailable or unacceptable (maybe voters don't trust the
randomization), and a tied outcome is acceptable. (of course a tied
outcome could be solved by a randomizing process, and that could be
done in a way that could be more feasibly-acceptable than mid-count
randomizations).

Randomization, whether mid-count or as a solution to a tied outcome,
should be solved by random-ballot.

Steve Eppley's simulations showed that, when MAM (Eppley's RP version)
and Beatpath give different results, MAM's result is
publicly-preferred to Beatpath's result, in an overwhelming majority
of instances.

In polling, Beatpath seems to have demonstrated less stability than
MAM. With Beatpath, there are more instances of one added ballot
changing the outcome.

Michael Ossipoff







On Sun, Dec 8, 2013 at 12:05 AM, Markus Schulze
<markus.schulze at alumni.tu-berlin.de> wrote:
> Hallo,
>
> here is another example to illustrate MMV's violation
> of monotonicity.
>
> Situation 1:
>
>   A>B, B>C, C>D, D>A, D>E, E>A each have the same
>   strength and are stronger than every other pairwise
>   defeat.
>
>   The other pairwise defeats are (sorted by their strength
>   in a decreasing order):
>
>   A>C
>
>   C>E
>
>   E>B
>
>   B>D
>
>   MMV skips A>B, B>C, C>D, D>A, D>E, and E>A, since they
>   form a directed cycle.
>
>   Then it locks A>C,C>E,E>B, and B>D, so that A is the
>   unique winner.
>
> Situation 2:
>
>   Some voters rank candidate A higher (relatively to
>   candidate D), so that the pairwise defeat D>A becomes
>   weaker.
>
>   A>B, B>C, C>D, D>E, E>A each have the same
>   strength and are stronger than every other pairwise
>   defeat.
>
>   The other pairwise defeats are (sorted by their strength
>   in a decreasing order):
>
>   A>C
>
>   C>E
>
>   E>B
>
>   D>A
>
>   B>D
>
>   MMV skips A>B, B>C, C>D, D>E, and E>A, since they
>   form a directed cycle.
>
>   Then it locks A>C,C>E,E>B, and D>A, so that D is the
>   unique winner.
>
> Thus, by ranking candidate A higher candidate A is changed
> from a winner to a loser.
>
>
> Markus Schulze
>
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