James G-A,
I have given some more thought on the problem of how best to
complete Condorcet by using high-resolution ratings ballots,
and I've now concluded that the best is the method I defined in my
last post (Tues.Jun.22):
"Approval Margins":
High-resolution ratings ballots. Infer
Oops! A little mistake in my last post, a "+1" should have read
"-1". Correction and some elaboration below.
"Approval Margins":
High-resolution ratings ballots. Inferring ranking from rating,
eliminate the non-members of the Schwartz-set.
Of the remaining candidates, each ballot approves
James,
Thanks for taking an interest in the comparison between your new "weighted
pairwise" (magnitudes) and my older "compressing ranks" ideas for completing
Condorcet. Unfortunately, when it comes to using high-resolution ratings
ballots to com
Chris,
I greatly appreciate your feedback about my proposal; I find it very
constructive. Some replies follow.
>CB: I should have written "maximised and scaled". Yes, the
>highest-rated Schwartz-set member is changed to the maximum,
>the lowest-rated member is changed to the minimum,
James,
>the ratings should be scaled and maximised among the members of the
>Schwartz set,
>between steps 2 and 3.
Sorry, Chris, what exactly do you mean by "scaled and maximized"? Is that
like raising the highest candidate to 100 and the lowest to 0?
CB: I should have written "maximis
Chris Benham wrote:
>With a very "high intensity" ratings ballot, it should be possible to do
>without the
>plain rankings ballot. With a handful of candidates, why would a sincere
>voter want to give two candidates the same
>ratings score out of 100, and yet rank one above the other?
