RL
Richard Lung
Mon, Feb 28, 2022 12:53 AM
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 Munsterhjelmkm_elmet@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 Munsterhjelmkm_elmet@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.)
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@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@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.)
FS
Forest Simmons
Tue, Mar 1, 2022 12:58 AM
"...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@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
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
will always be an electable candidate, unless it's a complete tie, or
there is some other rule not yet stated here.
It seems to me that the 3-candidate 1-winner case of this method is
It would help to see a four-candidate election resolved, too.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<
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
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
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
always be an electable candidate, unless it's a complete tie, or
is some other rule not yet stated here.
Kevin
Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<
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
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.)
"...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@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@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@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
>
RL
Richard Lung
Tue, Mar 1, 2022 9:09 PM
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@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 >>> Munsterhjelmkm_elmet@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 >>> Munsterhjelmkm_elmet@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
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@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@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@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
FS
Forest Simmons
Tue, Mar 1, 2022 10:25 PM
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@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@ukscientists.com voting@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 >>> Munsterhjelmkm_elmet@t-online.de
km_elmet@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 >>> Munsterhjelmkm_elmet@t-online.de
km_elmet@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
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@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@ukscientists.com> <voting@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@t-online.de>
> <km_elmet@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@t-online.de>
> <km_elmet@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
>
KM
Kristofer Munsterhjelm
Wed, Mar 2, 2022 11:03 PM
On 01.03.2022 01:58, 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.
Symmetric weighted positional methods could be considered to work this
way. The three-candidate single-winner Binomial STV works like fpA/lpA;
you could also imagine fpA-lpA, which is really the weighted positional
method (1, 0, 0, ...., -1): first preferences plus zero times second
preferences plus zero times ... plus -1 times last preferences.
And if the weight vector is symmetric (e.g. x_1, x_2, ..., x_k, -x_k,
-x_2, ..., -x_1), then it could be considered to factor in both
inclusion (the positive weights) and exclusion (the negative ones). Then
Borda could be considered a system of this form, for instance.
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.
I have been considering generalizations of fpA-fpC to multiple
candidates along these lines, although I'm missing significant pieces of
the puzzle. Suppose we want to pass single-candidate DMTBR, no matter
how many candidates run. Then an obvious way is to somehow reduce the
n-candidate election into a three-candidate election.
One way of doing that would be, for a candidate A, to collapse every
other candidate into either B or C. Let the sets c_B and c_C be the
candidates that are meant to be collapsed to B or C respectively. Then
modify a ballot by removing every candidate but the highest ranked
candidate in c_B, and then relabeling that candidate B; and removing
every candidate but the highest rated candidate in c_C, and relabeling
that candidate C.
(This looks like elimination, but it's more like relabeling every
candidate in c_B to B and then removing redundant later ranks, e.g.
X>Y>A>Z>W becomes B>B>A>B>C, and then B>A>C.)
If the assignment of candidates into c_B and c_C are held constant, then
the transformation is monotone, because ranking some candidate in c_B
higher either has no effect or ranks B higher on the transformed ballot
(all else equal), and the same holds for c_C. And since fpA-fpC itself
is monotone for three candidates, it can't produce a monotonicity
failure, either.
If A is a Condorcet winner with >1/3 fpp, then no matter how the other
candidates are partitioned into c_B and c_C, A keeps this property. And
if the burial doesn't change the partition of the other candidates, then
it also passes single-candidate DMTBR, because burying A under some
candidate X is weaker than burying A under B in the three-candidate
election, which three-candidate DMTBR already protects against.
So one could imagine a method where f(A, {B, C, D...}, {E, F, G...}) is
A's score with c_B being {B, C, D, ...} and c_C being {E, F, G, ...},
and then letting A's score be the min (or max, or sum, or some other
operator) of a number of these collapsed-election functions.
But that's the significant piece I lack: I don't know how to compose
these to preserve DMTBR, because the relative values of the various f
functions may change as a consequence of burial; and if f doesn't
contain an f for every possible partition, it's not obvious which
partitions belong in there, and how to make sure the selection thereof
can't be exploited in a way that fails single-candidate DMTBR. And if
the partition depends on the outcome so far, it becomes more like
elimination and may lead to path dependence problems again, just like
ordinary elimination.
-km
On 01.03.2022 01:58, 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.
Symmetric weighted positional methods could be considered to work this
way. The three-candidate single-winner Binomial STV works like fpA/lpA;
you could also imagine fpA-lpA, which is really the weighted positional
method (1, 0, 0, ...., -1): first preferences plus zero times second
preferences plus zero times ... plus -1 times last preferences.
And if the weight vector is symmetric (e.g. x_1, x_2, ..., x_k, -x_k,
-x_2, ..., -x_1), then it could be considered to factor in both
inclusion (the positive weights) and exclusion (the negative ones). Then
Borda could be considered a system of this form, for instance.
> 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.
I have been considering generalizations of fpA-fpC to multiple
candidates along these lines, although I'm missing significant pieces of
the puzzle. Suppose we want to pass single-candidate DMTBR, no matter
how many candidates run. Then an obvious way is to somehow reduce the
n-candidate election into a three-candidate election.
One way of doing that would be, for a candidate A, to collapse every
other candidate into either B or C. Let the sets c_B and c_C be the
candidates that are meant to be collapsed to B or C respectively. Then
modify a ballot by removing every candidate but the highest ranked
candidate in c_B, and then relabeling that candidate B; and removing
every candidate but the highest rated candidate in c_C, and relabeling
that candidate C.
(This looks like elimination, but it's more like relabeling every
candidate in c_B to B and then removing redundant later ranks, e.g.
X>Y>A>Z>W becomes B>B>A>B>C, and then B>A>C.)
If the assignment of candidates into c_B and c_C are held constant, then
the transformation is monotone, because ranking some candidate in c_B
higher either has no effect or ranks B higher on the transformed ballot
(all else equal), and the same holds for c_C. And since fpA-fpC itself
is monotone for three candidates, it can't produce a monotonicity
failure, either.
If A is a Condorcet winner with >1/3 fpp, then no matter how the other
candidates are partitioned into c_B and c_C, A keeps this property. And
if the burial doesn't change the partition of the other candidates, then
it also passes single-candidate DMTBR, because burying A under some
candidate X is weaker than burying A under B in the three-candidate
election, which three-candidate DMTBR already protects against.
So one could imagine a method where f(A, {B, C, D...}, {E, F, G...}) is
A's score with c_B being {B, C, D, ...} and c_C being {E, F, G, ...},
and then letting A's score be the min (or max, or sum, or some other
operator) of a number of these collapsed-election functions.
But that's the significant piece I lack: I don't know how to compose
these to preserve DMTBR, because the relative values of the various f
functions may change as a consequence of burial; and if f doesn't
contain an f for every possible partition, it's not obvious *which*
partitions belong in there, and how to make sure the selection thereof
can't be exploited in a way that fails single-candidate DMTBR. And if
the partition depends on the outcome so far, it becomes more like
elimination and may lead to path dependence problems again, just like
ordinary elimination.
-km
RL
Richard Lung
Fri, Mar 4, 2022 9:41 PM
HelloForest,
All the sciences that share the same structure of measurement are
subject to formal inter-disciplinary comparisons. Evolutionary theory
has been adapted by several disciplines, including speculation on the
multi-verse. Transferable voting especially suits evolution, as Enid
Lakeman observed, in How Democracies Vote.
Perhaps because I was too late to be educated in the New maths, it took
me too long to tumble to the fact that a comparison of election method
or “electics” with physics depends on a complex value election count.
And that depends on an election, in at least two dimensions.
Binomial STV is a complete single dimension of choice. The operative
word is complete, which makes possible its consistent binomial theorem
expansion into exponentially higher orders of count, for unlimited
analysis in depth.
The completeness also makes possible the combination of a second
dimension into a complex number election count.
Binomial STV is a complete dimension, because it accurately rationally
book-keeps, in keep values, all the voters preferential information.
Other electoral systems do not do this. They make arbitrary or expedient
rules to come to a result that does not well follow the voters wishes.
They either exclude preferential information or they do not include it,
in the first place. They use the preferences in ways not requested by
the voters. In the case of traditional STV methods, they do try to
always follow the voters preferences, but do so, less rationally
accurately, only on an ordinal scale, in the exclusion count.
Regards,
Richard Lung.
On 01/03/2022 22:25, Forest Simmons wrote:
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@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@ukscientists.com>
<mailto:voting@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@t-online.de> <mailto:km_elmet@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@t-online.de> <mailto:km_elmet@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
HelloForest,
All the sciences that share the same structure of measurement are
subject to formal inter-disciplinary comparisons. Evolutionary theory
has been adapted by several disciplines, including speculation on the
multi-verse. Transferable voting especially suits evolution, as Enid
Lakeman observed, in How Democracies Vote.
Perhaps because I was too late to be educated in the New maths, it took
me too long to tumble to the fact that a comparison of election method
or “electics” with physics depends on a complex value election count.
And that depends on an election, in at least two dimensions.
Binomial STV is a complete single dimension of choice. The operative
word is complete, which makes possible its consistent binomial theorem
expansion into exponentially higher orders of count, for unlimited
analysis in depth.
The completeness also makes possible the combination of a second
dimension into a complex number election count.
Binomial STV is a complete dimension, because it accurately rationally
book-keeps, in keep values, all the voters preferential information.
Other electoral systems do not do this. They make arbitrary or expedient
rules to come to a result that does not well follow the voters wishes.
They either exclude preferential information or they do not include it,
in the first place. They use the preferences in ways not requested by
the voters. In the case of traditional STV methods, they do try to
always follow the voters preferences, but do so, less rationally
accurately, only on an ordinal scale, in the exclusion count.
Regards,
Richard Lung.
On 01/03/2022 22:25, Forest Simmons wrote:
> 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@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@ukscientists.com>
> <mailto:voting@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@t-online.de> <mailto:km_elmet@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@t-online.de> <mailto:km_elmet@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
>
