[EM] “Monotonic” Binomial STV

[Forest Simmons]

Richard,

I enjoy all of the connections you make including with quantum mechanics
... understanding is a network of connections anchored in the ideas most
familiar and intuitive to us.

My PR idea for Binomial Bucklin is based on the idea once articulated by
Kevin that Bucklin can be thought of as a procedure for arriving at a
reasonable approval cutoff ... i.e. a DesignatedStrategyVoting version of
approval just as Instant Runoff can be considered a DSV version of
Plurality .... a procedure for finding a reasonable candidate for whom to
cast your one and only vote.

And just as Approval can be adapted for PR in the form of PAV, Sequential
PAV, or even the Martin Harper Lottery, so also should Binomial Bucklin
have similar possible modifications to various PR versions.

Best Wishes,

Forest

El mar., 1 de mar. de 2022 1:09 p. m., Richard Lung <voting at ukscientists.com>
escribió:

> Hello Forest,
>
> Yes, with Binomial STV, a blank ballot paper is the same as None Of The
> Above. That would count as a whole vote against any candidate. But any
> blank preferences go towards a fraction of a vote, counting towards a quota
> for an empty seat, by the usual Gregory method, expressed in keep value
> terms.
>
> I had to introduce this feature, to establish the relative satisfaction or
> dissatisfaction with the candidates. To take an extreme example, an
> extremely disaffected voter might vote preference 10 out of ten candidates,
> against some especially detested candidate, and leave the rest blank.
> Preference 1 would help to elect a non-candidate.
>
> I take it this case would be untypical, and preference entropy would weigh
> in favor of election counts, reinforced by low exclusion counts. Binomial
> STV has formally equal election and exclusion counts, like physics laws
> that are formally time-reversible. But in practise, they go one way, except
> on the quantum (very small) scale.
>
> Bucklin method sounds like a method used once in British Columbia in about
> 1951. Enid Lakeman, in How Democracies Vote, explained how it is not
> proportional representation, as was mistakenly suggested during the BC
> Citizens Assembly referendums. One faction could take all the seats with
> 51% of the votes. And Bucklin seems to still employ ordinal scale (only
> calculating by more or less) displacements or transfers of candidates votes.
>
> Binomial STV is essentially Gregory method PR, expressed in keep values,
> that allow accurate calculating quota-deficit candidates, as well as
> quota-surplus candidates. And also applied to counting exclusions, as well
> as elections. This allows for a keep value order of popularity. All kinds
> of STV have theoretical limitations but transfer well from vote to count.
> Adding a rational exclusion count should be worthy of further
> investigation, including real world examples.
>
> Binomial STV is a uniquely scalable system, capable of consistent
> exponential expansion of the count, according to the binomial theorem,
> offering unlimited representation, perhaps of the exponential growth of
> human knowledge.
>
> Regards,
>
> Richard Lung.
>
> On 01/03/2022 00:58, Forest Simmons wrote:
> > "...It follows that if the abstentions add up to a quota, a seat is >
> not taken...." > > Kind of like NOTA ... none of the above. > > I'm trying
> to think how I would design a method in the spirit of > Binomial STV ....
> elections vs exclusions ... preferences vs reverse > preferences. > >
> Perhaps some variant of Bucklin that gradually collapses ballot > rankings
> inward (ER Whole?) when not enough top or bottom votes exist > to meet
> quotas for further inclusion or exclusion ... taking special > care to
> insure both monotonicity and clone independence in the > process, if
> possible. > > I think collapsing has more potential for monotonicity than
> does > elimination, and I'm glad that Binomial stv keeps all of the players
> > in the game until the final count, like Bucklin does. > > -Forest > > > >
> El dom., 27 de feb. de 2022 4:54 p. m., Richard Lung >
> <voting at ukscientists.com> <voting at ukscientists.com> escribió: > > > On
> 28/02/2022 00:45, Richard Lung wrote: >> >> Thanks for your thoughts,
> Kevin, >> >> In this simple instance, the election and exclusion quotas
> cancel. >> But I would be lost without it, in multi-member PR cases of >>
> involved transferable voting. There are a few examples in my >> e-books,
> (The Super-Vote supercharged..., Elect and Exclude..., FAB >> STV...) free
> from Smashwords, in epub format, and pdf versions free >> from archive. org
> where putting "Richard Lung" in quotes in the >> text box should come up
> with about 19 titles. >> >> The square root may not be strictly necessary,
> which may be why I >> keep forgetting it. But it keeps the average keep
> values on a par >> with the election and exclusion keep values. The square
> root is for >> the correct form of the geometric mean, -- an important
> average. >> >> Yes, you are right, there is some other rule not stated --
> All the >> abstentions are counted. in more complex elections, they have to
> >> be, so as not to distort the relative importnce of the election >> and
> exclusion counts. It follows that if the abstentions add up to >> a quota,
> a seat is not taken. This provides an incentive to >> nominate good
> candidates, who work for the voters rather than their >> nominees. >> >>
> So, a candidate is not necessarily electable. More-over a large >> enough
> quota like Hare, with a small number of seats would also be >> prohibitive
> of election, given the voters free choice. >> >> Regards, >> >> Richard
> Lung. >> >> >> >> On 27/02/2022 19:30, Kevin Venzke wrote: >>> Hi
> Kristofer/Richard, >>> >>> I wonder not just about the square root, but
> also if the quota >>> has some additional role in the method, perhaps when
> there are 4+ >>> candidates. >>> >>> Because this expression: ( quota /
> keep ) * ( exclude / quota ) >>> Appears to simplify to: ( exclude / keep )
> >>> >>> This creates the appearance that the quota has no effect on the >>>
> outcome. >>> >>> Richard stated that final values below unity are
> electable. It >>> looks like there will always be an electable candidate,
> unless >>> it's a complete tie, or perhaps if there is some other rule not
> >>> yet stated here. >>> >>> It seems to me that the 3-candidate 1-winner
> case of this method >>> is monotone. It would help to see a four-candidate
> election >>> resolved, too. >>> >>> Kevin >>> >>> Le dimanche 27 février
> 2022, 07:41:20 UTC−6, Kristofer >>> Munsterhjelm<km_elmet at t-online.de>
> <km_elmet at t-online.de> a écrit : >>>> On 27.02.2022 14:04, Richard Lung
> wrote: >>>>> Thank you, Kristofer, >>>>> >>>>> >>>>> for first example.
> >>>>> >>>>> The quota is 100/(1+1) = 50. >>>>> >>>>> Election keep value is
> quota/(candidates preference votes) >>>>> >>>>> Exclusion keep value equals
> quota/(candidates reverse >>>>> preference vote): >>>>> >>>>> Geometric
> mean keep value ( election keep value multiplied by >>>>> inverse exclusion
> keep value): >>>> Geometric mean: >>>> >>>> A: square root of (50/51 x
> 2/50) ~ 0.198 B: square root of >>>> (50/49 x 1/50) ~ 0.143 C: ~= infinity
> (or very high) >>>> >>>> So B wins, having the lowest keep value. Is this
> correct? >>>> >>>> (You seem to have omitted the square root in your
> calculations, >>>> but it shouldn't make a difference. Without the square
> root, A >>>> and B's values are 0.0392 and 0.0204 respectively.) >>> Hi
> Kristofer/Richard, >>> >>> I wonder not just about the square root, but
> also if the quota >>> has some additional role in the method, perhaps when
> there are 4+ >>> candidates. >>> >>> Because this expression: ( quota /
> keep ) * ( exclude / quota ) >>> Appears to simplify to: ( exclude / keep )
> >>> >>> This creates the appearance that the quota has no effect on the >>>
> outcome. >>> >>> Richard says final values below unity are electable. It
> seems >>> like there will always be an electable candidate, unless it's a
> >>> complete tie, or perhaps if there is some other rule not yet >>> stated
> here. >>> >>> Kevin >>> >>> >>> Le dimanche 27 février 2022, 07:41:20
> UTC−6, Kristofer >>> Munsterhjelm<km_elmet at t-online.de>
> <km_elmet at t-online.de> a écrit : >>>> On 27.02.2022 14:04, Richard Lung
> wrote: >>>>> Thank you, Kristofer, >>>>> >>>>> >>>>> for first example.
> >>>>> >>>>> The quota is 100/(1+1) = 50. >>>>> >>>>> Election keep value is
> quota/(candidates preference votes) >>>>> >>>>> Exclusion keep value equals
> quota/(candidates reverse >>>>> preference vote): >>>>> >>>>> Geometric
> mean keep value ( election keep value multiplied by >>>>> inverse exclusion
> keep value): >>>> Geometric mean: >>>> >>>> A: square root of (50/51 x
> 2/50) ~ 0.198 B: square root of >>>> (50/49 x 1/50) ~ 0.143 C: ~= infinity
> (or very high) >>>> >>>> So B wins, having the lowest keep value. Is this
> correct? >>>> >>>> (You seem to have omitted the square root in your
> calculations, >>>> but it shouldn't make a difference. Without the square
> root, A >>>> and B's values are 0.0392 and 0.0204 respectively.) > ----
> Election-Methods mailing list - see https://electorama.com/em > for list
> info > epresentatio
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20220301/b4a6d90b/attachment-0001.html>