[EM] MMV and resolvability

[Michael Ossipoff]

 Two typos in my immediately-recent posting:

1. I'd said:

I'd been referring to MMV as the Drop-Equal-Defeats (DED) version or
Ranked-Pairs, for when a tied result is acceptable, and the use of
randomization in the count is unacceptable (as could be the case in
some Internet votes).

I meant:

"...version _of_ Ranked-Pairs....".

"of", not "or"

So, it should say:

"I'd been referring to MMV as the Drop-Equal-Defeats (DED) version of
Ranked-Pairs, for when a tied result is acceptable, and the use of
randomization in the count is unacceptable (as could be the case in
some Internet votes).":

2. Toward the end of my posting, here is what I meant to say:

D>A is disqualified because it contradicts a set of kept stronger defeats.

(In my posting, I'd left out the word "kept")

I like MMV, as I interpret it. Especially when mid-vote randomization
is unavailable or unaccptable, and a tied outcome is acceptable.

I can't say for sure whether MMV or MAM is better for voting when
mid-vote randomization is available and accepable, and when tied
outcomes aren't acceptable.

I emphasize that I regard Ranked-Pairs (in the MAM or MMV versions) as
the ideal majoritrarian voting-systems, _to be recommended for ideal
majoritarian conditions_, which seem to obtain in polls. But, due to
chicken-dilemma, resulting from intense rivalry among similar parties
and factions, I wouldn't count on ideal majoritarian conditions in
official public political elections in the U.S.

Michael Ossipoff



:



On Sat, Dec 7, 2013 at 6:34 AM, Anders Kaseorg <andersk at mit.edu> wrote:
> Maximum Majority Voting exhibits a silly failure of Tideman’s
> resolvability criterion in the following four-candidate election:
>   3: A > B > C > D
>   1: B > D > A > C
>   2: C > D > B > A
>   1: D > A > B > C
>
> According to my reading of
> http://radicalcentrism.org/resources/maximum-majority-voting/, and also
> according to Eric Gorr’s calculator at http://condorcet.ericgorr.net/, MMV
> proceeds as follows.  The defeats are sorted:
>   5/2: A > C, B > C, C > D
>   4/3: A > B, B > D, D > A
> In the first step, MMV affirms the three 5/2 defeats.  In the second step,
> it ignores all three 4/3 defeats because they form a cycle.  The final
> result is A = B > C > D.
>
> However, there’s no way to add another ballot to make B the unique winner,
> in violation of resolvability.
>
> This failure is silly because, if the three 4/3 defeats had been sorted in
> _any_ strict order, D > A would have been ignored for already forming a
> cycle with the 5/2 defeats, and the final result would unambiguously be A
>> B > C > D.
>
> This feels like an oversight in the MMV definition, and although it’s
> highly unlikely to matter in practice, fixing it is simple enough.  At
> each step, when considering a set of equally strong defeats, we should
> immediately discard any defeats that would complete a cycle with strictly
> stronger affirmed defeats, so that they are not considered for use in
> cycles with equally strong defeats.
>
> Anders