[EM] Unifying DMTC and DMTCBR, and method X criterion

[Forest Simmons]

Great criterion as far as it goes ... great at not rewarding buriers, but
bad at electing buried candidates ... if I understand it.

Hence the need for a sincere runoff:

40 A>B(Sincere A>C)
35 B>C
25 C>A

The sincere CW is C, which cannot be elected because 25 is less than 100/3,
if I understand the proposed critersion.

But a top three sincere runoff of the form

A vs (B vsC)

will elect C assuming rational voters informed of the true preferences.

On Sun, Aug 13, 2023, 7:51 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> Here's a way to unify DMTC (elects a DMT candidate) and burial resistance:
>
> Suppose that candidate A has more than 1/3 of the first preferences, and
> beats B pairwise. Then B must not win.
>
> This works because B>A voters can't influence A's first preferences, nor
> can they influence whether A beats B pairwise. A neat trick is also that
> this can never lead to a cycle, because there can't be more than two
> candidates with more than 1/3 first preferences, and A can't both beat B
> and B beat A at the same time.
>
> In that spirit, here's a weak criterion that's passed by method X:
>
> Let an election restricted to set S be the election after everybody but
> the candidates in S are eliminated. Then if, in every restricted
> election involving A and B, A has more than 1/|S| of the first
> preferences, and A also beats B pairwise, then B must not be elected.
> Here |S| is the cardinality of set S, i.e. the number of non-eliminated
> candidates in the restricted election in question.
>
> Method X passes this criterion because when B's finding a maximizing
> elimination path, no matter who's eliminated, A always has more than
> 1/|S| first preferences. Therefore A can never be eliminated, so B is
> either forced to be eliminated (in which case he obviously can't be
> elected) or his final matchup is against A, who beats him pairwise. On
> the other hand, A can do no worse than following the same path, leading
> to a matchup against B, which he wins.
>
> It's weak because we need the criterion to hold for all restricted
> elections, not just one of them. So the DMT set has to be
> well-distributed: all its candidates must have about the same support.
> But in return, if the criterion bars B from being elected, voters may
> flip pairwise preferences between DMT members and non-DMT members all
> they want; it'll never help. Furthermore, a voter lowering his last
> ranked DMT set member under a bunch of non-DMT candidates also never helps.
>
> Weak as it is, it could be useful in method design, just like clone
> independence. I'll take every handhold scaling this cliff :-)
>
> -km
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