[EM] A New Spinoff of Our Recent Discussions
Forest Simmons
fsimmons at pcc.edu
Wed Jun 5 19:28:48 PDT 2019
I was mulling over Kristofer's ideas abbout using first place counts in new
ways.
It reminded me about what we called "Borda Done Right" where we decloned
Borda by use of the first place counts.
Then a light bulb turned on: Why not de-clone Copeland in rthe same way?
After all, Copeland always chooses from the Landau set, the set of
uncovered candidates. [I first realized this a few years ago when Marcus
expressed doubt that there was a monotonic method that would always choose
from Landau. After a little thought Copeland was the most obvious example
to clear up the question.]
So here is Improved Copeland:
For each candidate X, let S(X) be the sum of first place votes of the
candidates that do not pairwise beat X.
[Note that this sum includes the number of first place votes received by X,
since X does not pairwise beat X.]
Elect from argmax(S(X)).
Note that in large public elections argmax(S(X)) will almost surely consist
of only one candidate.
That version works best when the ballots have have easily discerned
favorites.
Here's a version that works better for a greater variety of ballots,
especially where equal top votes are allowed:
For each candidate Y let P(Y) be the probability that Y would be chosen by
a random ballot lottery. [Actually, any other decent lottery would work
just as well.]
For each candidate X, let S(X) be the sum (over all candidates Y that do
not pairwise beat X) of P(Y).
Elect from argmax(S(X)).
Note that if Z covers X, then S(Z) is greater than or equal to S(X),
because every P(Y) in the sum for S(X) will also be a term in the sum
defining S(Z).
Therefore the max(S(X)) candidate is uncovered.
Like Copeland the method is also monotone, and unlike Copeland the method
is clone proof.
Since Copeland is one of the most familiar Condorcet methods, and has an
obvious appeal until the clone dependence probblem is pointed out, ithis
new method can be presented as a simple , easily understandable solution to
that problem.
How does it hold up on our favorite examples?
Try
49 C
26 A>B
25 B
S(C)=49+26=75
S(A)=26+25=51
S(B)=25+49=74
Arrgmax(S(X))={C}
The method passes the CD criterion.
Is this too good to be true?
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