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<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">James and all<br>
</span></p>
<span style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold""></span><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">From
a merely technical point of view, the difference between
conventional STV,
including Meek, and binomial STV is that the former use a rational
election
count and an ordinal exclusion count. Binomial STV uses both a
rational
election count and a rational exclusion count.</span>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">The
rational count is more accurate, dealing in exact proportions.
The ordinal
count is less accurate, dealing in questions of more or less
than.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">In
fact, in studies of scales of measurement, such as by SS Stevens
in Science,
the ordinal scale is recognised to be less powerful, less
progressed in the
logic of measurement, than the ratio scale.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold""> </span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">One
thing I can say right away was that counting abstentions was not
something I
dragged in to binomial STV. Rather, it was forced upon me by the
logic of the
binomial count. Abstentions have to be counted because, where
they appear in
the voters preferences determines the relative importance of the
election count
and the exclusion count.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold""> </span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">Ultimately,
the real significance of including abstentions in the count, is
that we then
have a complete dimension of choice. This is of considerable
scientific
significance.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">I
wanted to say more, anyway with regard to you, James, but I’ve
already gone on long enough.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">Regards,</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">Richard
Lung.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold""><br>
</span></p>
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<div class="moz-cite-prefix">On 04/01/2022 23:35, James Gilmour
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:002301d801c3$c663a820$532af860$@globalnet.co.uk">
<pre class="moz-quote-pre" wrap="">It is not only lower preferences that are contingency choices in STV-PR - all preferences after the first are contingency choices.
1. "I want to be represented by my first choice candidate".
2. "The RO may look at my second preference (and transfer my vote to my second preference) ONLY in the contingency that my first
choice candidate cannot represent me in the elected assembly (either because he does not have the support of enough voters or
because she is already elected by a due proportion of other voters and so cannot represent me as well).
Richard's Binomial STV may well be doing things differently from how we commonly understand STV-PR, but that is a completely
different interpretation and perhaps a completely different philosophy. It certainly would be if you drag non-voters in the
calculation in any way.
James Gilmour
Please note change of e-mail address
<a class="moz-txt-link-freetext" href="mailto:jamesgilmour@f2s.com">mailto:jamesgilmour@f2s.com</a>
<a class="moz-txt-link-freetext" href="mailto:james-gilmour@outlook.com">mailto:james-gilmour@outlook.com</a>
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">-----Original Message-----
From: Richard Lung [<a class="moz-txt-link-freetext" href="mailto:voting@ukscientists.com">mailto:voting@ukscientists.com</a>]
Sent: 01 January 2022 16:25
To: <a class="moz-txt-link-abbreviated" href="mailto:jgilmour@globalnet.co.uk">jgilmour@globalnet.co.uk</a>
Cc: Forest Simmons <a class="moz-txt-link-rfc2396E" href="mailto:forest.simmons21@gmail.com"><forest.simmons21@gmail.com></a>; EM <a class="moz-txt-link-rfc2396E" href="mailto:Election-methods@lists.electorama.com"><Election-methods@lists.electorama.com></a>
Subject: Re: [EM] Tie Breaking Version of Burial Resistant Method
James, good to hear from you again,
Reasons do exist for counting abstentions.
Where the abstentions appear in a voters order of preferences influences the relative importance of votes meant to elect
or exclude in a binomial count. A binomial count is where, say you have 10 candidates seeking 5 seats. Preferences 1 to 5
will more or less help elect 5 candidates. Preferences 6 to 10 will less or more help to exclude 5 candidates.
Where the abstentions occur in voters orders of preference will decide the relative importance voters give to electing or
excluding candidates.
Another reason for counting abstentions, with a binomial count, is that if they add up to a quota, one seat will remain
unfilled. This is an incentive for organisations to nominate better candidates, rather than stooges, so no seats are left
unfilled.
I don't see lower preferences as merely contingent but as also positively wanting several representatives.
Regards,
Richard Lung.
On 1 Jan 2022, at 1:45 pm, James Gilmour <a class="moz-txt-link-rfc2396E" href="mailto:jgilmour@globalnet.co.uk"><jgilmour@globalnet.co.uk></a> wrote:
Richard
I think there is a difference of interpretation here about the "meaning" of incomplete preferences - and about not voting.
"Election method (system) is about representing all voters", indeed. Note your word "voters". So those electors who
choose to opt out by not voting (abstain) are themselves making a conscious decision not to take part is the process of
choosing the representatives. I can see nothing that implies abstentions must be counted.
Similarly, in an STV-PR election I always understood that any second and subsequent preferences indicated contingency
choices that would come into play only if your first choice candidate was excluded (because too few voters supported him)
or was already elected (and so did not need your support, and could not in due proportion also represent you). When a
voter does not mark a preference against every candidates on her ballot paper, she is telling the Returning Officer to give
effect to the preferences she has marked, and then said: "If it comes to a choice among any of the remaining candidates
(whom I have not marked), I am not able or don't wish to express any preference among them and I am happy to leave
any choice that has to be made among those candidates to those voters who do have preferences among those
candidates.
James Gilmour
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">-----Original Message-----
From: Election-Methods
[<a class="moz-txt-link-freetext" href="mailto:election-methods-bounces@lists.electorama.com">mailto:election-methods-bounces@lists.electorama.com</a>] On Behalf Of
Richard Lung
Sent: 30 December 2021 18:44
To: Forest Simmons <a class="moz-txt-link-rfc2396E" href="mailto:forest.simmons21@gmail.com"><forest.simmons21@gmail.com></a>; EM
<a class="moz-txt-link-rfc2396E" href="mailto:Election-methods@lists.electorama.com"><Election-methods@lists.electorama.com></a>
Subject: Re: [EM] Tie Breaking Version of Burial Resistant Method
Election method is about representing all voters, not about beating
other candidates. A frequent objection to the latter approach is that in ruling out candidates, voter information about
</pre>
</blockquote>
<pre class="moz-quote-pre" wrap="">them is being lost.
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">
The key to progress in election method is to count all the
preferential information supplied by the voters. That means not only
counting votes that help to elect, but counting votes that help to exclude. Hence, my binomial count. Which also implies
</pre>
</blockquote>
<pre class="moz-quote-pre" wrap="">the abstentions must be counted. -- To properly balance the books, all preferences must be counted.
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">
Regards,
Richard Lung.
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">On 30/12/2021 00:00, Forest Simmons wrote:
Let's say a candidate is at the "bottom" of a set of candidates if it
is not ranked above even one member of the set.
The basic method is to elect the candidate who on the most ballots
pairwise beats every bottom candidate.
Here's the procedure Proc(Beta) with built in tie breaker:
Beta is the set of ballots serving as input for this election
procedure. Let K be the set of candidates ranked by the ballots of Beta.
For each candidate X in K let n(X) be the number of ballots B in Beta
on which X pairwise beats every bottom candidate of B.
Let T be the set argmax(n(X)), i.e.the tied winning set. If T has
only one member, elect that member. Else if T=K, i.e. all candidates
are in an exact tie, then elect by random ballot from K. Otherwise,
elect Proc(Beta(T)), i.e. apply this procedure recursively to the set
of ballots restricted to T.
Without the tiebreaking extension, this method is about as
susceptible to ties as Implicit Approval.
Simple random ballot would be adequate for this single winner method,
but to get an "order of finish" method based on the function
X---->n(X) the recursive version would be very appropriate .... by
transforming n(X) into a polynomial in powers of epsilon with an
additional term for each recursive pass through the procedure.
</pre>
</blockquote>
</blockquote>
</blockquote>
<pre class="moz-quote-pre" wrap="">
</pre>
</blockquote>
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