[EM] XA (Andy Jennings)

[Monkey Puzzle]

Jameson, could you please explain your EXA (or PXA) method in a little more
detail?  I had some trouble following your example.

I am also uncomfortable with the emphasis on making the ratings numerical.

 Frango ut patefaciam -- I break so that I may reveal

On Tue, Nov 1, 2016 at 2:28 PM, Jameson Quinn <jameson.quinn at gmail.com>
wrote:

> One issue I have with XA is that it makes numerical votes inherently
> meaningful; it is entirely possible to change the election outcome by
> adding or subtracting a constant from all ballots. I'm wondering if this is
> fixable.
>
> What if you transformed all ratings into percentiles first? Let's call
> this system empirical chiastic approval, EXA. So if you had something like
> (candidates W-Z and grades A-F)
>
> 3: WA XB YC ZF
> 2: WF XC YD ZA
>
> ... that would be an empirical grade distribution of 5As 3Bs 5Cs 2Ds 5Fs,
> or in percentiles A75 B60 C35 D25 F0. So the EXA score is W60, X60, Y35,
> Z40. (This example gave a tie because I deliberately made it so there would
> be round numbers. In general, a tie would be highly unlikely. Clearly, in
> order to minimize strategy, this tie should be broken in favor of W, the
> candidate who had the most excess goodwill on their weakest
> positive-influence ballot.)
>
> I think there are probably strategies involving manipulating the empirical
> distribution (non-semi-honestly), but I doubt that anyone would have the
> fine-grained info necessary to pull such a strategy off.
>
> If you condition on a given percentile distribution, it satisfies the same
> kind of criteria that XA does, including "individual non-strategy".
>
> Interestingly, though not very usefully, this is a voting system which
> would work just fine with allowing a negative infinity to positive infinity
> ballot scale. (it would work even better than Bucklin in that sense.)
>
> 2016-11-01 16:54 GMT-04:00 Forest Simmons <fsimmons at pcc.edu>:
>
>> It is more likely that two candidates will have the same median score (an
>> MJ tie situation) than having the same XA score.
>>
>> Part of the reason is that the XA scores depend continuously on the
>> distribution of ratings, while the median can be a discontinuous function
>> of the distribution.
>>
>> From another point of view, the graph of y = x is more likely to be
>> perpendicular to the graph of the distribution function F(x) = Probability
>> that on a random ballot candidate X will have a rating of at least x.  An
>> orthogonal intersection minimizes error due to random perturbations.
>>
>> The graph of F stair steps down from some point on the y axis between (0,
>> 0) and (0, 1) to some point on the vertical segment connecting (1, 0) to
>> (1, 1).  If the distribution is uniform, then the graph of F is the
>> diagonal line segment connecting (0, 1) to (1, 0), perpendicular to the
>> line y = x.
>>
>> The median point (used in MJ and other Bucklin variants) is the
>> intersection of the graph of F with the vertical line given by x = 1/2,
>> cutting the square with diagonal corners at (0, 0) and (1, 1) in half.
>>
>> The midrange Approval value is the intersection of the graph of F with
>> the horizontal line y = 1/2.
>>
>> The XA value is the intersection of the graph of F and line y = x, which
>> bisects the right angle formed by  x = 1/2 and y = 1/2 at the intersection
>> (1/2, 1/2).
>>
>> So XA can be thought of as a method half way between midrange Approval
>> and score based Bucklin.
>>
>> More later ...
>>
>> Forest
>>
>>
>>
>>
>>> From: Andy Jennings <elections at jenningsstory.com>
>>> To: Michael Ossipoff <email9648742 at gmail.com>
>>> Cc: "election-methods at electorama.com"
>>>         <election-methods at electorama.com>
>>> Subject: Re: [EM] XA
>>>
>>> On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <
>>> email9648742 at gmail.com>
>>> wrote:
>>>
>>> > What makes XA do that more effectively than MJ? What's the main
>>> advantage
>>> > that distinguishes how XA does that from how MJ does it, or the
>>> results,
>>> > from the voters' strategic standpoint?
>>>
>>>
>>> Michael,
>>>
>>> As Rob said, the median is not terribly robust if the distribution of
>>> votes
>>> is two-peaked:
>>> http://www.rangevoting.org/MedianVrange.html#twopeak
>>> And I'm afraid many of our contentious political elections are
>>> two-peaked,
>>> at least in the current environment.
>>>
>>> With MJ, I like the fact that if the medians for all candidates will fall
>>> between B and D, then I can use the range outside that for honest
>>> expression.  Yet in the back of my head, I know that if everyone tries to
>>> "use the range outside that for honest expression", then the medians
>>> won't
>>> be in that range anymore and it seems like a slippery slope to everyone
>>> using only the two extreme grades.
>>>
>>> XA solves this problem by making the more extreme grades more difficult
>>> to
>>> achieve.  As Rob said, in the case where everyone grades at the extremes,
>>> the XA will match the mean.
>>>
>>> On the other hand, I admit that:
>>> 1) with the median, 50% would have to give the top grade for a candidate
>>> to
>>> receive that grade.  And 50% would have to give the bottom grade for a
>>> candidate to receive that grade.  I consider both of these very unlikely.
>>> 2) MJ is not just "the median", it has a tie-breaking scheme which
>>> mitigates this somewhat.
>>>
>>> ~ Andy
>>>
>>>
>>
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