[EM] Ranked Robin

[Daniel Carrera]

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
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