[EM] question re: converting ballots into a matrix

[rob brown]

On 12/5/05, Dan Bishop <daniel-j-bishop at neo.tamu.edu> wrote:
>
> They're affected in exactly the same way as if the measure of defeat
> strength were changed to Margins.


Ok, so what you are saying is that if half-votes are counted, using margins
vs. using winning votes will behave identically, i.e. always pick the same
winner?  (for all methods that consider defeat strength)

I suppose it would also make margins and ratio behave identically as well.
To me that seems elegant that they all work the same, so it's less
ambiguous  But since some prefer using winning votes to margins, well.... I
guess they aren't going to like this.  Fair enough.

>     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
>
>   <snip>
>

Ok, but only if we know how many ballots.  I had assumed that we didn't know
that, since that information cannot be derived from a traditional matrix
alone.  So, as long as you store the number of ballots as a separate
variable, a traditional matrix contains all the information that a
half-vote-counting matrix has.  Plus some.

Cool.
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