[EM] “Monotonic” Binomial STV
Richard Lung
voting at ukscientists.com
Tue Mar 1 13:09:05 PST 2022
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> 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> 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> 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
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