# [EM] Improved Copeland

Ted Stern dodecatheon at gmail.com
Wed Jun 12 11:29:25 PDT 2019

```As I just wrote,

What's wrong with using Equal-Rated-Zero for pairwise votes, and then
> minimizing the sum of defeating scores against a candidate?  In other
> words, add up all the defeating scores in a candidate's column in the
> pairwise array.  I'm sure that has a name already. In this example,  Abby
> would win with a total of 463 votes Brad>Abby, which is less than the total
> defeating scores for any other candidate.  It also has the advantage of
> being summable with no other information than the pairwise array required.

twice.  That opens up too many strategic options.  So consider the S score
S(Y) = total number of ballots on which Y is ranked strictly below any
candidate who pairwise defeats Y.  The candidate with the minimum S score
is the winner.  When a candidate has only a single defeat, their S score is
the pairwise winning votes of the candidate who defeated them.

This is not easily summable (at first glance), but it has a certain kind of
intuitive sense to it, similar to MMPO.

On Wed, Jun 12, 2019 at 11:14 AM Ted Stern <dodecatheon at gmail.com> wrote:

> Hi Forest,
>
> Try this example:
>
>  98: Abby >  Cora >  Erin >  Dave > Brad
>  64: Brad >  Abby >  Erin >  Cora > Dave
>  12: Brad >  Abby >  Erin >  Dave > Cora
>  98: Brad >  Erin >  Abby >  Cora > Dave
>  13: Brad >  Erin >  Abby >  Dave > Cora
> 125: Brad >  Erin >  Dave >  Abby > Cora
> 124: Cora >  Abby >  Erin >  Dave > Brad
>  76: Cora >  Erin >  Abby >  Dave > Brad
>  21: Dave >  Abby >  Brad >  Erin > Cora
>  30: Dave >  Brad >  Abby >  Erin > Cora
>  98: Dave >  Brad >  Erin >  Cora > Abby
> 139: Dave >  Cora >  Abby >  Brad > Erin
>  23: Dave >  Cora >  Brad >  Abby > Erin
>
>
> Abby defeats all candidates except Brad, and Brad defeats all candidates
> except Dave.  So S(Abby) is the total number of ballots on which Brad is
> ranked above all other candidates except possibly Dave.  So T(Brad) = Brad
> > Abby votes , 463 minus the 23 ballot where Cora > Brad.  So S(Abby) = 440.
>
> Similarly, S(Brad) = T(Dave).  Dave defeats all candidates except Abby and
> Erin.  So T(Dave) = total number of ballots on which Dave is ranked higher
> than all candidates except possibly Abby or Erin.  So T(Dave) = 21 + 30 +
> 98 + 139 + 23 = 311.
>
> I haven't worked out the rest, but I believe they are all higher.  So Brad
> would beat Abby.  I think in this case I would prefer Abby to Brad, so I'm
> not entirely happy.
>
> I am also not seeing an obvious way to make this summable.
>
> What's wrong with using Equal-Rated-Zero for pairwise votes, and then
> minimizing the sum of defeating scores against a candidate?  In other
> words, add up all the defeating scores in a candidate's column in the
> pairwise array.  I'm sure that has a name already. In this example,  Abby
> would win with a total of 463 votes Brad>Abby, which is less than the total
> defeating scores for any other candidate.  It also has the advantage of
> being summable with no other information than the pairwise array required.
>
> On Tue, Jun 11, 2019 at 5:02 PM Forest Simmons <fsimmons at pcc.edu> wrote:
>
>> 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
>>
>> 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
>>
>>
>>
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
>>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20190612/e9828c50/attachment-0001.html>
```