[EM] (3) MJ -- The easiest method to 'tolerate'
jameson.quinn at gmail.com
Wed Sep 21 18:38:58 PDT 2016
2016-09-21 18:59 GMT-04:00 Kristofer Munsterhjelm <km_elmet at t-online.de>:
> On 09/06/2016 01:29 AM, Jameson Quinn wrote:
> > ...
> > If that's true, then there's some kind of threshold of number of
> > below which there aren't enough grades for the ballot format to
> > encourage grading against a common standard. So three-slot methods
> > have to be tested to see if voters would vote grading-style or
> > relative-rank style with three grades, or if three grades are still
> > too few.
> > Personally, I'd guess that three would be enough. I think the four most
> > common voting heuristics, in descending order, would be:
> > First, as I've said: approve of one, disqualify greater evil among the
> > frontrunners and any evils greater still.
> > Second: Approve of one, disqualify both frontrunners.
> > Third: if one side of the left-right divide got used to habitually
> > losing, the centrists among them might begin to extend approval all the
> > way to the most-centrist on the other side.
> > Fourth: Approve of one, disqualify unqualified/unserious candidates.
> > Any of the heuristics above would tend to lead to a "successful"
> > resolution of the Chicken Dilemma.
> That's kind of going into the domain of "manual DSV", though, which I'm
> not as much a fan of. MJ has this going for it that there are ways for
> honest voters to vote as long as the categories are well-defined.
> Approval is a *lot* muddier, because it's much less certain that honest
> voters can divide the candidates into "I like these" and "I definitely
> don't like these". Three-slot methods could work if the heuristics are
> common and intuitive enough, but they're kind of in-between MJ's
> expressiveness and having Approval's manual DSV needs.
Median-based methods, especially those which break ties using above-median
votes, can do well with chicken dilemma scenarios. The flipside of this is
that in order to handle center-squeeze scenarios, they require the voters
who favor the centrist CW to put all other candidates below the winning
median; for safety, at bottom-rating.
I think that's somewhat plausible as a naive strategy ("Since I'm Center, I
think Left and Right are equally bad, so even though I don't hate them as
much as they hate each other, I might as well put both of them at bottom
rank"). I also think that when Center voters don't so strategize, the
outcome is not too bad; generally, that will tend to be in cases where
Center might even have lost a score election, and the candidate who beats
Center might even be the utility winner or at least close to it.
But yes, there is this much "manual" strategy necessary, and I don't see
any way around it. I think a system can't be robust to strategy in both
Chicken and Center Squeeze scenarios, and I'd rather get Chicken right in
spite of some strategy (and thus not encourage strategies which, when
over-applied, can end up electing the CL) instead of getting Center Squeeze
right without any strategy.
> > What's a good scenario to use to illustrate IIA failure in ranked
> > Bucklin? Seems to me any such scenario would be solved by a simple
> > truncation from the non-cloning side, which seems strategically obvious
> > enough.
> Perhaps a crowding or "innocent cloning" example would work. E.g.
> something like:
> n candidates are in the running, representing various positions in issue
> space. Someone in the top area of the issue space decides to withdraw,
> and the winner flips from left to right.
> I haven't checked if Bucklin is vulnerable to crowding, however, but in
> the worst case, you could use a Condorcet 3-cycle example where removing
> A changes the winner from B to C. IIA is a lot stronger than
> independence to clones, after all.
Such scenarios depend on the tiebreaker, but I think it should be easy to
build one for most concrete tiebreakers.
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