[EM] (6) To Kristofer and everyone: MJ best to ‘tolerate’

[Jameson Quinn]

Kristofer suggests using the trimmed mean to have a useful number to report
in MJ. That's exactly the point of GMJ (graduated majority judgment): for
each candidate, you essentially use the trimmed mean using the widest trim
that does not include grades on both sides of the median. So if somebody
got 10% A (4.0), 25% B (3.0), 35% C (2.0), 20% D (1.0), and 10% F (0.0),
then you would see that the median is C, the A+B tally (35%) is more than
D+F (30%), so you trim 30% from each side, leaving (2.0*35 + 3.0*5)/40 =
2.125.

This is very, very similar to MJ in outcome; though it's possible to
construct cases where they disagree, it involves two candidates with the
same median but where one is "more polarizing" (has significantly fewer
grades at that median, thus presumably more at the extremes). Even then it
happens less than half the time (half at the extreme). I think that kind of
situation would be negligibly rare in practice; I'd guess, even less common
than honest Condorcet cycles.

Jameson

2016-09-17 13:20 GMT-04:00 Kristofer Munsterhjelm <km_elmet at t-online.de>:

> On 09/17/2016 03:25 PM, Toby Pereira wrote:
> > Arguably a problem with Majority Judgement is that most voters make
> > no difference to the overall "score" for a candidate. Because it uses
> > the median rather than mean, it is a fairly "robust" measure, and a
> > single vote is unlikely to shift the median. People can use that to
> > argue in its favour - if you like or dislike a candidate you can rate
> > them highly or lowly (above or below their current median), and it
> > doesn't matter whether you use the extreme ratings or not. If a
> > candidate's median is C, then I can give an honest B or an
> > exaggerated A - the candidate's overall median won't be affected by
> > my exaggeration.
> >
> > But the flipside of being resistant to strategy in this way is that
> > arguably it is only this way because it's resistant to voting itself!
> > It just has much more "inertia" than other systems. In normal score
> > voting every vote has an effect on a candidate's average. And because
> > of this there is arguably a sense of "power" for a voter. But this
> > isn't the case for median voting. Well, each vote would provide some
> > sort of shift that could make a difference in a tie-break, but it's
> > not the same.
>
> In a way, that's true for ranked methods as well. Suppose that X is the
> Condorcet winner. A vote that ranks X in some place has no effect unless
> it creates a cycle. If the contest is close, then this may easily
> happen, but if it's not, then the vote has no effect except as a way of
> moving the contests closer for some other ballot to change the result.
>
> On the other hand, if you'd generalize usually ranked methods to provide
> scores rather than rankings as their outputs, then a single vote might
> have an effect. So too with MJ: you could have the method return not
> just the median grades of the candidates, but how many votes they're
> away from getting a higher grade. Any vote that grades a candidate
> higher than the MJ result would decrease this counter, and any vote that
> grades the candidate lower would increase it.
>
> > One slightly crazy idea to ameliorate this would be to use a measure
> > somewhere between the mean and median. The mean of a data set is the
> > number that minimises the sum of the squared differences to the data
> > points. The median is the number that minimises the sum of the
> > absolute differences (so the differences^1). So maybe you could
> > instead minimise the differences^1.1 or something. This would still
> > be close to a median method but presumably each vote would make a
> > difference to a candidate's "average" score using this measure. But
> > then arguably it doesn't make sense to use letter grades with a
> > verbal description in a system like this. It becomes about numbers
> > again. It might also be difficult to calculate and hard for voters to
> > understand what's going on.
>
> A more intuitive middle road, I think, would be to use a trimmed mean.
> For each candidate, take the ratings and throw out the k% lowest and
> highest of them, then that candidate's score is the mean of what's left.
>
> If you let k be 50%-1, then you get MJ (since you throw away everything
> but the middle rating). If you let k be 0, then it's Range.
>
> The problem, however, is that if you set k less than the median point,
> the method can fail the majority criterion. Furthermore, the method only
> gets more vulnerable to Approval-style voting the closer to Range you
> get. So if MJ's strategy resistance isn't good enough to keep the voters
> from all voting Approval style, then no lesser-k method will be good
> enough either.
> ----
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