[EM] Best IRV Tweak

Richard Lung voting at ukscientists.com
Wed Jul 7 10:47:09 PDT 2021

Dear All,

There is science and there is politics. And I believe the latter is getting in the way of the former. And if the science was clearer, I also believe it might open up a better prospect for the politics.   
The question of which candidates to elect and which candidates to eliminate/exclude applies to all election methods. None of them have a rational exclusion count, and consequently none of them have a fully rational election count. (It is possible to have a symmetrical election count to an exclusion count: that is my system.)
The fatal flaw of all these "tweaks" is that they are tweaks or ad hoc after-thoughts. It is that unscientific thinking which also makes the ad hoc additional member system or Mixed Member Proportional, the HOW NOT TO DO IT of election methods.

Richard Lung.

On 7 Jul 2021, at 3:58 am, Susan Simmons <suzerainsimmons at outlook.com> wrote:

It appears that this IRV Tweak satisfies independence from Smith dominated alternatives.(ISDA)

Sent from my MetroPCS 4G LTE Android Device

-------- Original message --------
From: Susan Simmons <suzerainsimmons at outlook.com>
Date: 7/6/21 4:03 PM (GMT-08:00)
To: Daniel Carrera <dcarrera at gmail.com>, election-methods at lists.electorama.com
Cc: election-methods at lists.electorama.com
Subject: Re: [EM] Best IRV Tweak

We're all amateurs! 

You're right ... a Smith candidate cannot be eliminated without suffering a pairwise defeat from another of the remaining candidates, which perforce must be a Smith candidate since Smith candidates cannot be pairwise defeated by non-Smith candidates. When one door shuts another one opens. 

Good on you! I didn't even think to ask the question .... so we have a method that is conceptually and operationally simple, while also seamlessly Smith compliant.

It fails mono-raise, but no worse than IRV.

It seems to be clone independent ... if X has little support against Y, then the same will be true for a clone of X, etc.

Compare Yee diagrams. IRV's are chaotic, unlike those of Condorcet methods (including this one) which are composed of ideal convex Voronoi polygons.

IRV uses ranked preference ballots to simulate Hare style runoffs ...this tweak uses the same ballots to simulate a thorough round robin style tournament ... much more thorough than would be practical without the ranked preference ballots.

For example without ranked ballots a round robin contest among ten candidates would require 45 trips to the polls. No wonder there is no tradition of round robin tournaments in the context of public elections.

In the context of sports tournaments it's hard to imagine an analog of runoff without head to head contests because runoff has only one head to head contest ... the final round that takes place after all of the other contestants have been eliminated by a more superficial criterion.

Sent from my MetroPCS 4G LTE Android Device

-------- Original message --------
From: Daniel Carrera <dcarrera at gmail.com>
Date: 7/6/21 2:21 PM (GMT-08:00)
To: Susan Simmons <suzerainsimmons at outlook.com>
Cc: election-methods at lists.electorama.com
Subject: Re: [EM] Best IRV Tweak

> On Tue, Jul 6, 2021 at 2:00 PM Susan Simmons <suzerainsimmons at outlook.com> wrote:
> While there remains more than one uneliminated candidate eliminate the remaining one that is weakest relative to the other remaining candidates.
> If "weakest" means least top support (relative to the other remaining candidates) then we have IRV.
> But if we interpret "weakest" to mean the candidate with the fewest votes in any head-to-head contest (among the remaining candidates) then we have a Condorcet Compliant method.
> In fact, the weakest candidate in this sense is a pairwise loser to its opponent in its worst head-to-head contest ... which could not happen to a Condorcet Candidate.

I'm just an amateur, but I think your method is also Smith efficient. During an elimination round it is possible that the candidate with fewest votes in a head-to-head contest is in the Smith set, but only by losing a head-to-head contest against another candidate in the Smith set.

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