[EM] Steve's reply:: Deterministic Epsilon Consensus Idea stimulated by a question of Steve Bosworth
steve bosworth
stevebosworth at hotmail.com
Thu Oct 7 19:58:07 PDT 2021
TO: Forest
FROM: Steve
Currently, I understand your pursuit of the optimal "Epsilon" to be an interesting mathematical challenge for you. However, I do not yet see it as having the prospect of improving the use MJ. Partly, my question follows from you previously agreeing that voting using grades is more meaningful than using numbers. You answered: "Yes, I fully agree on that point; nobody I know likes it when the nurse asks how much it hurts on a scale of one to ten!"
If so, the highest median-grade of any MJ election must be either Excellent, Very Good, Good, Acceptable, Poor, or Reject. How could an election benefit from replacing these judgments with numbers?
In any case, what portion of the electorate is going to understand an explanation of a new voting system that includes one of your phrases below like: "... the winning votes consist of fractions greater than (1 - 1/k) of the ballots."
I look forward to our dialogue.
Steve
________________________________
From: Forest Simmons <forest.simmons21 at gmail.com>
Sent: Thursday, October 7, 2021 3:50 PM
To: steve bosworth <stevebosworth at hotmail.com>
Cc: EM <Election-methods at lists.electorama.com>
Subject: Re: Deterministic Epsilon Consensus Idea stimulated by a question of Steve Bosworth
Note that if epsilon is one, the method is just Range Voting on a scale of zero to five ... which gives a clear incentive for concentrating ratings to the extremes of zero and five, or the extremes of "reject" and "excellent" in the MJ terminology.
On the other hand, if epsilon were infinitesimal, the grades above "poor" would serve merely as tie breakers ... not a bad idea for a general purpose tie breaker method!
But as a stand alone method, it might be best to limit the choice of epsilon to a value between zero and 1/6.
Why 1/6? Because of the six levels of grades, a five-sixths majority is a reasonable partial consensus quota. In particular, an initial step that disqualified every alternative pairwise defeated by more than a five-sixths majority of the ballots could never disqualify all candidates when there are only six levels.
[In general, if there are only k levels, there cannot be a beat cycle where all of the winning votes consist of fractions greater than (1 - 1/k) of the ballots.]
This and other considerations make me think that in general when there are k levels of approval, 1/k might be a good nominal value for epsilon.
This "method" is still in the brainstorming stage, so all suggestions are worth considering!
El mié., 6 de oct. de 2021 7:38 p. m., Forest Simmons <forest.simmons21 at gmail.com<mailto:forest.simmons21 at gmail.com>> escribió:
Steve's query about Chiastic Approval included the following ....
Also, correct me if I'm mistaken that XA does not guarantee that its winner will be elected with the support of a majority of all the votes (ballots) cast.
The short answer is "no" ... no method can guarantee majority voter support for its winner, unless they can guarantee that at least one candidate is ranked, rated, scored, or graded above bottom on more than half of the ballots submitted.
The long answer is, "Why stop at half or two-thirds, as some methods require ... why not go for full 100% consensus?"
But, you may object, full consensus is not always possible. Well, neither is forty percent support always possible, but that doesn't stop the Constitution from requiring two-thirds of the voters' support for certain kinds of amendments, etc.
One expedient that has been suggested is the NOTA option for the case when the quota is not met. This option gives new meaning to the word "approval" ... as Mike Ossipoff used to say, you approve a candidate if you would rather see her elected than have to come back next month to vote for someone else.
I would like to suggest another option based on the standard MJ grade ballot ...
Each candidate X gets a score that is given by the sum ..
S(X) = Sum (over j from zero to five) of the product
a(j)*epsilon^j,
where epsilon is a value to be determined by the voters ... and the respective values of a(j), for j in {0, 1, 2, 3, 4} are the number of ballots on which candidate X is graded strictly above reject, poor, acceptable, good, or very good, respectively.
Also each voter has the option of voting for a value of epsilon in the set {.01, .02, ... .99, 1.00}. The median of the distribution of these votes determines the value of epsilon.
Elect the candidate X with the max value of S(X) (once the epsilon value has been determined).
Note that if, for some j, the coefficient a(j) is the total number of ballots, then we can say candidate X is a full consensus candidate at level j.
If there are several full consensus level j candidates, then the higher degree terms will determine the winner.
Thanks!
FWS
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