[EM] question re: converting ballots into a matrix

[Dan Bishop]

rob brown wrote:
> On 12/5/05, *Kevin Venzke* <stepjak at yahoo.fr <mailto:stepjak at yahoo.fr>> 
> wrote:
> 
>      > Has anyone ever considered also adding one-half a "point" to each
>     of the
>      > equals?
> 
>     Yes. It's equivalent to using margins.
 >
> Not sure I understand what you mean. "Using margins" in what sense?

Several Condorcet-completion methods have the idea of "defeat strength". 
When defeats form a cycle, defeat strength is used to decide which 
defeats to keep and which to discard.

Kevin was referring to the use of the pairwise margin of victory as the 
measure of defeat strength.  Other measures are winning votes (i.e., the 
number of votes for the pairwise winner) and relative margins (margin of 
victory as a proportion of the number of preferences in the pairwise 
contest).

For example, given that A pairwise beats B 54 to 12, then the defeat 
strength of A over B is 54-12=42 if measured by margins, 54 if measured 
by winning votes, and (54-12)/(54+12)=63.64% if measured by relative 
margins.

If Margins is used as the measure of defeat strength, then an A=B ballot 
is equivalent to the combination ½ A>B + ½ B>A, as they both have zero 
net contribution to the defeat strength.  This is what Kevin meant.

This list has had lots of discussion on defeat strength; check the archives.

> I guess my question was, for all of the above, doesn't it make more 
> sense to convert the ballots into a matrix as I described?

Nope.  Even if you're using Margins, in which case it doesn't make a 
difference, your proposal, well, doesn't make a difference, so why spend 
the effort of adding half-votes to your matrix?  And if you do have a 
reason for half-votes, you can always create a half-vote matrix from a 
traditional matrix, whereas the converse is not true.