Condorcet(x( ))

[Bruce Anderson]

On May 17, 12:37pm, Steve Eppley wrote:
>
> Would there be something wrong with a Condorcet that uses x=.5 for
> two unranked candidates and for equally last-ranked candidates if
> there's no truncation, and x=0 otherwise?  Besides the extra
> complexity in the definition?
> 
> While I'm musing... what about using an x() which varies depending on
> the pair's rank position and the number of rank positions on the
> ballot?  So x() would start at 0 for the favorite twins and
> grow smoothly to .5 (or 1?) for the most despised twins?
> 
> Maybe this is too small a concern to spend time on...
> 
> --Steve
> 
>-- End of excerpt from Steve Eppley

The following is an excerpt from a message I sent Mike in August, 1994, on 
alternative ways of addressing ties on voter's ballots.  I doubt that the 
generality below is likely to be worthwhile in comparison to the complexity 
induces.

<<
Kemeny's method makes logical sense only if ties in individual voter's rankings 
count in this half-and-half way, and that's essentially the way that Kemeny 
originally defined his method.

First you convinced me that this approach for counting ties is not sacrosanct.  
While theoretically there are many different ways that voter's ties could be 
counted, three particular approaches stand out.  For the obvious reason, these 
three approaches are "numbered" by (1), (1/2), and(0) as follows:
(1) Count a tie between A and B in favor of A (i.e., as if the voter had ranked 
A over B) when evaluating A, and in favor of B when evaluating B.  (This is the 
approach taken in defining Ossipoff's method.)
(1/2) Count a tie between A and B as half-and-half (i.e., count ties as 
described just above).
(0) Count a tie between A and B in favor of B (i.e., as if the voter had ranked 
B over A) when evaluating A, and in favor of A when evaluating B.

Next you convinced me that it can be reasonable to allow the approach used to 
count ties in a voter's preference order to depend on where these ties are 
occurring--namely, ties for first could count one way, ties for last could count 
a possibly different way, and all other ties could count yet another way.  Taken 
all together, this gives 27 different possibilities.  Using the numbering just 
above and the ordering:
(ties-for-first;ties-in-the-middle;ties-for-last),
these 27 possibilities are:
(1;1;1),     (1;1;1/2),     (1;1;0),
(1;1/2;1),   (1;1/2;1/2),   (1;1/2;0),
(1;0;1),     (1;0;1/2),     (1;0;0),
(1/2;1;1),   (1/2;1;1/2),   (1/2;1;0),
(1/2;1/2;1), (1/2;1/2;1/2), (1/2;1/2;0),
(1/2;0;1),   (1/2;0;1/2),   (1/2;0;0),
(0;1;1),     (0;1;1/2),     (0;1;0),
(0;1/2;1),   (0;1/2;1/2),   (0;1/2;0),
(0;0;1),     (0;0;1/2),     (0;0;0).

In a 4-way race, for example, a ballot that ranked all 4 candidates as being 
tied with one another would count as a 4-way tie for first; while a ballot that 
ranked 2 of the candidates as being tied with each other and being ahead of the 
other 2 candidates, who were also tied with each other, would count as a 2-way 
tie for first and a 2-way tie for last.

At first, I thought that most of these 27 possibilities were obviously 
implausible.  Then I realized that one of the ones I thought was implausible, 
(1/2;1/2;1), is used all the time in sports, and probably elsewhere too.  So now 
I just think it's better not to jump to conclusions.
>>

Bruce