On 12.05.2022 08:36, Forest Simmons wrote:
My most recent Copeland attempt ...
Each voter designates a candidate on her ballot B as "The One To Beat
For My Point." (TOTBFMP)
Lacking a Condorcet Winner, elect the candidate that, on the most
ballots, pairwise defeats the TOTBFMP designated candidate.
In other words, for each ballot B let T(B) be the candidate designated
TOTBFMP on ballot B.
Elect argmax N(X), where N(X) is the number of ballots in the set
{B| X pairwise defeats T(B)}.
So ordinary clone infested Copeland elects the candidate that pairwise
defeats the most other candidates, while this version elects the
candidate that fulfills the defeat requests of the most voters.
I would guess that it fails Smith and Landau, both of which ordinary
Copeland passes.
Suppose we didn't have the Condorcet provision, i.e. that the argmax
winner is automatically elected without a check for a CW first. Then I
think something like the usual LCR example could elect either L or R if
the L and R voters designate C to be TOTBFMP.
The Condorcet provision fixes that problem, but I would think that by
cloning C into enough candidates to make a top cycle of C clones, you
could make the method elect L or R instead.
So its main advantage would be (if I'm right) that for a
clone-independent method, it's fairly easy to describe.
Or I could be mistaken, of course :-)
-km
On 12.05.2022 08:36, Forest Simmons wrote:
> My most recent Copeland attempt ...
>
> Each voter designates a candidate on her ballot B as "The One To Beat
> For My Point." (TOTBFMP)
>
> Lacking a Condorcet Winner, elect the candidate that, on the most
> ballots, pairwise defeats the TOTBFMP designated candidate.
>
> In other words, for each ballot B let T(B) be the candidate designated
> TOTBFMP on ballot B.
>
> Elect argmax N(X), where N(X) is the number of ballots in the set
> {B| X pairwise defeats T(B)}.
>
> So ordinary clone infested Copeland elects the candidate that pairwise
> defeats the most other candidates, while this version elects the
> candidate that fulfills the defeat requests of the most voters.
I would guess that it fails Smith and Landau, both of which ordinary
Copeland passes.
Suppose we didn't have the Condorcet provision, i.e. that the argmax
winner is automatically elected without a check for a CW first. Then I
think something like the usual LCR example could elect either L or R if
the L and R voters designate C to be TOTBFMP.
The Condorcet provision fixes that problem, but I would think that by
cloning C into enough candidates to make a top cycle of C clones, you
could make the method elect L or R instead.
So its main advantage would be (if I'm right) that for a
clone-independent method, it's fairly easy to describe.
Or I could be mistaken, of course :-)
-km
