[EM] Ranked Robin

[Forest Simmons]

Better yet use de-cloned Copeland, which has a statistically negligible
chance of ties.

The offensive score of candidate X is the sum of last place votes of the
candidates pairwise defeated by X.

The defensive score of candidate X is the sum of first place votes of the
candidates that pairwise defeat .

In the unlikely case that the offensive and defensive champions are are not
the same, elect the pairwise winner of the two.

El mié., 19 de ene. de 2022 1:58 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> What you are proposing is equuvalent to Borda restricted to the Copeland
> Set ... at least in the case of complete rankings ... in general the Borda
> winner is the candidate with the greatest difference between its row
> average and its column average. With complete rankings the column average
> is 100 percent minus the row average.
>
> Since Borda is well known and can be defined with or without reference to
> the pairwise score matrix, we can say .. Elect the candidate that pairwise
> defeats the most other candidates. In case of ties, eliminate those not
> tied, and elect the candidate with the highest Borda Count relative to the
> other remaining candidates.
>
> If you really want to emphasize the Round Robin Tournament analogy...
> here's the simplest language that most robustly takes both tied games into
> account and tied Copeland scores into account:
>
> The team with the greatest difference between wins and losses is the
> tournament winner. In case of ties, the tied team with the greatest
> difference between total points scored and total points given up, is the
> tournament winner.
>
> Note that the tied team that scores most against opponents could be
> considered the offensive champ, while the team that gives up the fewest
> points to its opponents is the defensive champ.
>
> The team with the greatest difference should be considered the all around
> winner.
>
>
>
> El mié., 19 de ene. de 2022 12:52 p. m., Daniel Carrera <
> dcarrera at gmail.com> escribió:
>
>> Hi guys,
>>
>> Last night I was looking for strategy-resistant Condorcet methods not
>> based on IRV. I went to the list of Condorcet methods on the wiki and I
>> stumbled upon Ranked Robin.
>>
>> https://electowiki.org/wiki/Ranked_Robin
>>
>> Proposed by Sass on VotingTheory.org and Reddit just a couple of months
>> ago. It's Copeland but pairwise ties get a score of 0, plus a tiebreaking
>> mechanism. Sass summarizes the method thusly:
>>
>> "Among the candidates who tie for winning the most head-to-head matchups,
>> elect the candidate with the best average rank."
>>
>> His goal is to get IRV supporters to ditch IRV and pick an "RCV" method
>> that is actually good, while being simple enough that they can say "this is
>> also RCV, it's another one, and it's easier".
>>
>> Now, I personally think that the "average rank" is confusing and the
>> tiebreaking mechanism described on the Wiki is a fabulously complicated "4
>> degree" monstrosity that takes up half the page. And I don't think any of
>> that if needed. If you read this explanation on Reddit you see that what he
>> is trying to do is really simple. I would suggest the following revision:
>>
>> "The candidate that wins the most head-to-head matchups is elected. If
>> there is a tie, grab every finalist and give them a score equal to the sum
>> of all the votes in their favor in every matchup against every other
>> finalist. The finalist with the largest such score is elected. If there's
>> still a tie, conduct a runoff election."
>>
>> Here is an example:
>>
>> 6 votes: D>A>B>C
>> 5 votes: B>C>A>D
>> 4 votes: C>A>B>D
>>
>> This is just a trivial example of a Concorcet cycle to see how to break
>> it.
>>
>> A beats B, 10 vs 5
>> B beats C, 11 vs 4
>> C beats A, 9 vs 6
>> everyone beats D
>>
>> So A,B,C win 1 matchup each, D wins none. ABC are the finalists. Now the
>> tiebreaker round:
>>
>> Score(A) = 10 + 6 = 16
>> Score(B) = 5 + 11 = 17
>> Score(C) = 9 + 4 = 13
>>
>> So B is elected. I'm pretty sure that this should be equivalent to the
>> "1st Degree" tiebreaker described in the wiki, but I think my version is
>> far easier to understand. I would ditch all the other tiebreakers "Degrees"
>> and replace "1st Degree" with my version. That should now produce a system
>> that is as easy to understand as Sass intended.
>>
>> Along those lines, I hope Rob will comment here on whether this system
>> could be used in Burlington. Here's a first stab at legal language:
>>
>> -------------------------------------------------------------------
>> ...
>>   (2) If a candidate receives a majority (over 50 percent of all ballots)
>> of first preferences, that candidate is elected.
>>   (3) If no candidate receives a majority of first preferences, then each
>> candidate is compared in turn to every other candidate in a head-to-head
>> match. The candidate that defeats, by a simple majority of voter
>> preferences, the greatest number of other candidates in a head-to-head
>> match, is elected.
>>   (3) If there is more than one such candidate, a tiebreaking tabulation
>> is conducted among that group of candidates (from here on, called
>> "finalists"). Each finalist is assigned a vote count equal to the sum of
>> the number of ballots that rank said candidate above the other, for every
>> head-to-head match against another finalist. The finalist with the highest
>> vote count is elected.
>> ...
>> -------------------------------------------------------------------
>>
>> Cheers,
>> --
>> Dr. Daniel Carrera
>> Postdoctoral Research Associate
>> Iowa State University
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
>>
>
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