# [EM] Improved Copeland

Forest Simmons fsimmons at pcc.edu
Tue Jun 11 17:02:38 PDT 2019

```Our first attempts at improved Copeland ended up losing monotonicity
without fully achieving clone independence.

I will repeat that version here for comparison:

Elect the candidate with the fewest top rank ballot totals for the
candidates that beat her pairwise.

It turns out that we have to replace the top rank totals with something
that counts a few additional votes beyond the top tallies:

For each candidate X let T(X) be the number of ballots on which candidate X
is ranked above all of the candidates that she beats pairwise.

This total includes all of the unique top votes of candidate X, but also
includes some others.

So here's the method: elect the candidate Y that minimizes S(Y) defined as
the sum of T(X) (over all X that beat Y pairiwise).

Here's an example:

4 A>B
2 B>C
3 C>A

T totals in the form of top votes plus extras from second ranks:
T(A) = 4 + 3 = 7
T(B) = 2 + 4 = 6
T(C) = 3 +2 = 5

S(A) = T(C) = 5 < 6 = T(B)=S(C) < 7 = T(A)=S(B),

so S(A) < S(C) < S(B)

A is the winner, B is the loser, and C is in the middle of the social order
according to this method.

I would appreciate it being tested on your favorite examples. If it needs
clarification I will use your examples to clarify it, as long as it holds
up to scrutiny.

Thanks,

Forest
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