<div dir="auto">Richard,<div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">Best Wishes,</div><div dir="auto"><br></div><div dir="auto">Forest</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mar., 1 de mar. de 2022 1:09 p. m., Richard Lung <<a href="mailto:voting@ukscientists.com">voting@ukscientists.com</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div>
Hello Forest,<br>
<br>
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.<br>
<br>
<p>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.</p>
<p>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.<br>
</p>
<br>
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.<br>
<br>
<p>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. <br>
</p>
<p>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. <br>
</p>
<p>Regards,</p>
<p>Richard Lung.<br>
</p>
<br>
On 01/03/2022 00:58, Forest Simmons wrote:<br>
<span style="white-space:pre-wrap;display:block;width:98vw">> "...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
> <a href="mailto:voting@ukscientists.com" target="_blank" rel="noreferrer"><voting@ukscientists.com></a> 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<a href="mailto:km_elmet@t-online.de" target="_blank" rel="noreferrer"><km_elmet@t-online.de></a> 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<a href="mailto:km_elmet@t-online.de" target="_blank" rel="noreferrer"><km_elmet@t-online.de></a> 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.)
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>
</span>epresentatio<br>
</div>
</blockquote></div>