[EM] XA (Andy Jennings)

Tue Nov 1 18:12:51 PDT 2016

I see.  Very good!

On Tue, Nov 1, 2016 at 5:54 PM, Jameson Quinn <jameson.quinn at gmail.com>
wrote:

> I don't think you understand my point. Say candidate A has 60 votes at 0,
> 5 votes at 35, and 35 votes at 100; while candidate B has 63 at 0, 5 at 40,
> and 32 at 100. B wins with 37. Now add 5 to the middle scores of both;
> nothing fundamental changes about how any voter views either candidate, but
> now A wins with 40.
>
> My system of EXA (or percentile chiastic approval, PXA) would build
> ratings scales automatically, so that an electorate that liked numbers
> around 80 could get the same results as one that liked numbers around 20.
>
> 2016-11-01 20:32 GMT-04:00 Forest Simmons <fsimmons at pcc.edu>:
>
>> I guess I was assuming normalized ballots all along, in other words you
>> would have at least one candidate that you approved unconditionally and at
>> least one that you would not approve no matter what the other voters did,
>> at least in an election setting.
>>
>> On the other hand if you are just contributing to a five star rating
>> guide ...
>>
>> 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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>>>>
>>>>
>>>
>>
>
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