[EM] Ranked Robin

[Forest Simmons]

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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