[EM] question re: converting ballots into a matrix

[Dan Bishop]

rob brown wrote:
> On 12/5/05, *Dan Bishop* <daniel-j-bishop at neo.tamu.edu 
> <mailto:daniel-j-bishop at neo.tamu.edu>> wrote:
> 
>     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. 
> 
> So a method that used margins would be unaffected (as I mentioned in my 
> post, "clearly some methods will be unaffected").  Others *would* be 
> affected, to some degree or another.
> 
> Isn't it reasonable to discuss *how* the methods that would be affected, 
> would be affected?

They're affected in exactly the same way as if the measure of defeat 
strength were changed to Margins.

>     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.
> 
> 
> I'm not sure I buy this.  How would you do that?

For example, consider an election with 12 voters and the Condorcet matrix

   A B C
A 0 4 8
B 4 0 8
C 4 4 0

This matrix tells you the number of people who expressed a preference in 
each pairwise contest:

A-B: 4+4 = 8 votes
A-C: 8+4 = 12 votes
B-C: 8+4 = 12 votes

And thus, you also know the number of voters *without* a pairwise 
preference:

A=B: 12-8 = 4 abstensions
A=C: 12-12 = 0 abstensions
B=C: 12-12 = 0 abstensions

You can treat the 4 A=B nonpreferences as 2 A>B + 2 B>A, thus producing 
the half-vote matrix:

   A B C
A 0 6 8 = 14 Borda points
B 6 0 8 = 14 Borda points
C 4 4 0 =  8 Borda points