[EM] median rating / lower quartile

[Daniel Bishop]

Fan de Condorcet wrote:

> James,
>
> You wrote:
>
>> Given sincere votes, this may be interesting, but if votes are not
>> necessarily sincere, it would be quite possible for all candidates to
>> receive a social utility of 0. That is, the lower quartile feature makes
>> the method into a kind of 3/4 supermajority method.
>>
>> However, scoring candidates by the median rather than the mean might 
>> be an
>> improvement on standard cardinal ratings. Has this been discussed 
>> before? 
>
> Yes.  I once happened to stumble across an archived message, 
> originally posted by Rob Lanphier, in which he suggested just this.  
> It wasn't long before he shot down his own idea in a follow-up message.
>
> http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/1998-April/001603.html 
>
>
> http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/1998-April/001607.html 
>

Some other problems with Median Ratings:

* It adds a complication to the vote counting: If there are c candidates 
and r possible ratings, there need to be c*r entries in the summation 
array, rather than just c (as in standard Cardinal Ratings).

* The previous problem is minor if you choose a reasonably small value 
of r.  But by doing so, you introduce another problem:

Ballots based on a 0-100 scale:

   A=74, B=42, ...
   A=61, B=87, ...
   A=61, B=59, ...
   A=23, B=25, ...
   A=97, B=72, ...
   Median ratings: A=61, B=59
  
The same ballots rounded for a 0-10 scale:

   A=7, B=4, ...
   A=6, B=9, ...
   A=6, B=6, ...
   A=2, B=3, ...
   A=10, B=7, ...
   Median ratings: A=6, B=6
  
* It fails Neutrality of Spoiled Ballots.

   A=1, B=4
   A=2, B=4
   A=8, B=4
   A=9, B=4

With these ballots, the median ratings are A=5 and B=4, so A wins.  
However, if the ballot (A=0, B=0) is added, then the median ratings 
become A=2 and B=4, so B wins.