[EM] majority judgement question

Andy Jennings elections at jenningsstory.com
Wed Oct 19 10:23:00 PDT 2011


Good points, Ross and Jameson.

Section 4.3 of my dissertation (http://ajennings.net/dissertation.pdf) talks
about this very thing.

The Ac-Bc rule was proposed by David Gale (
http://en.wikipedia.org/wiki/David_Gale) before his passing, to Balinski and
Laraki directly.  I proposed a rule very similar to Jameson's (Ac - Bc)/(Mc
+ |Ac - Bc|).  Mine was (Ac - Bc)/(2*Mc).  Both are continuous everywhere
(assuming fractional voters), even at the junction where the median changes
from one grade to another.  (See graphs in the pdf.)  Also in my
dissertation, I gave examples for the other two rules where a small change
in the voter profile could cause a candidate to fall multiple rankings.

On the other hand, Balinski and Laraki's rule is constant with respect to
either Ac or Bc almost everywhere.  I think this might make it a little more
resistant to strategic voting.  Plus, the remove-one-median-rating-at-a-time
method has a certain simplicity and elegance to it, especially for very
small electorates, even if it gets a little convoluted for large ones.

~ Andy



On Wed, Oct 19, 2011 at 9:12 AM, Jameson Quinn <jameson.quinn at gmail.com>wrote:

> Great suggestion. I've been thinking along those lines, but I hadn't
> expressed it as clearly.
>
> And now that Ross has given me this idea, I can make it even simpler.
>  Ross's suggested process is of course equivalent to, and harder to explain
> than, using (number above median grade)-(number below median grade) as a
> score. The only disadvantage of my version is that it could give negative
> numbers. But almost all people over the age of 10 (and a lot of people under
> that age) can handle negative numbers just fine, so I think that's OK.
>
> This tiebreaker process is good. It will also tend to agree with the MJ
> one, as long as the tied candidates have approximately the same number of
> votes at the median grade - which will generally be true for two candidates
> whose strengths are similar enough to tie the median grade in the first
> place.
>
> Here's another "tiebreaker" which I've developed. The advantage is that it
> gives a single real-number grade to each candidate, thus avoiding the issue
> of "ties" in the first place. I call it "Continuous Majority Judgment" or
> CMJ.
>
> Rc= Median rating for candidate c (expressed numerically; thus, letter
> grades would be converted to grade-point-average numbers, etc.)
> Mc= Number of median ratings for candidate c
> Ac= Number of ratings above median for candidate c
> Bc= Number of ratings below median for candidate c
> |x| = standard notation for absolute value of x
>
> CMJ rating for c = Rc + ((Ac - Bc)/(Mc + |Ac - Bc|))
>
> For approval (that is, binary ratings), the CMJ rating works out to be
> equal to the fraction of 1s, as you'd expect. Note that the adjustment
> factor is always in the range of -0.5 to 0.5, because the difference |Ac -
> Bc| can never be greater than Mc or it wouldn't be the median.
>
> I prefer either of these methods to the MJ method - not for results, but
> for simplicity. (Ac - Bc) is simplest to explain, while CMJ is simplest to
> compare candidates / post results. All three of them should give the same
> results in almost all cases. But Balinski and Laraki preferred the
> remove-one-median-rating-at-a-time method because they could prove more
> theorems about it, and they wrote the MJ book, so until I write my own book
> about it I'm fine with promoting their method.
>
> JQ
>
>
> 2011/10/19 Ross Hyman <rahyman at sbcglobal.net>
>
>>  It seems to me that there is a simpler way to compare candidates with
>> the same median grade in majority judgement voting than the method described
>> in the Wikipedia page for majority judgement.  Why isn't this simpler way
>> used?
>>
>> Every voter grades every candidate.  Elect the candidate with the highest
>> median grade (the highest grade for which more than 50% of voters grade the
>> candidate equal to or higher than that grade.)  If there are two or more
>> candidates with the same highest median grade, elect the candidate with the
>> highest score of those with the highest median grade.  A candidate's score
>> is equal to the the number of voters that grade the candidate higher than
>> the median grade plus the number of voters that grade to candidate equal to
>> or higher than the median grade.  This is equivalent to giving one point to
>> each candidate for each voter who grades the candidate its median grade and
>> two points for each voter who grades the candidate higher than its median
>> grade.  Motivation:  voters who vote median grade instead of something lower
>> should increase the score for the candidate by the same amount as voters who
>> vote above the median grade instead of equal to the median grade.  With this
>> scoring, going from less than median to median increases the candidate score
>> by one point and going from median to higher than median also increases the
>> candidate score by one point.
>>
>> Example using same example from Wikipedia's majority judgement entry:
>> 26% of voters grade Nashville as Excellent and 42% of voters grade
>> Nashville as Good.  Nashville's median grade is Good and its score is
>> 26+26+42 = 94
>> 15% of voters grade Chattanooga as Excellent and 43% of voters grade
>> Chattanooga as Good.  Chattanooga's median grade is Good and its score is
>> 15+15+43 = 73.
>> Nashville wins.
>>
>>
>>
>>
>>
>>
>>
>>
>> ----
>> Election-Methods mailing list - see http://electorama.com/em for list
>> info
>>
>>
>
> ----
> Election-Methods mailing list - see http://electorama.com/em for list info
>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20111019/25769e4d/attachment-0004.htm>


More information about the Election-Methods mailing list