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<p>Thank you, Filip,</p>
<p>The first order Binomial STV is one election count and one
exclusion count, exactly like it (being symmetrical; an
iteration). For the election count I use Meek method of surplus
transfers. The distinction, of that computer count, over the
traditional hand counts, is that preferences, for an already
elected candidate, with a quota, are still recorded. Meek did that
by updating the candidate keep value (the quota divided by a
candidates total transferable vote).</p>
<p>Unlike Meek method, I do keep values for every candidate, losers
as well as winners. Candidates in deficit of a quota have keep
values of more than unity, signifying they are excluded. The
exclusion count is run exactly like the election count but with
the preferences reversed, so a quota now becomes an exclusion
quota. The rule is simple: an election count elects candidates
reaching the quota. An exclusion count excludes candidates
reaching a quota. One voters preferences is another voters
unpreferences. There is no difference in principle between them.</p>
<p>Binomial STV (symbolised as STV^; first order Binomial STV would
be STV^1. Any order bimomial STV would be STV^n. Preceding forms
of STV, including Meek method, are STV^0. The ballot paper looks
just like any Ranked Choice Vote. But the instructions are
different. Every voting method has voters instructions. <br>
</p>
<p>The instructions are, in the case of your example: There are four
seats available and ten candidates to choose from. Your first four
preferences would more or less help to elect candidates. Your next
6 preferences (if you choose to make them) would less or more help
to exclude those candidates. So, a tenth preference counts as much
against a candidate, as your first preference would count for a
candidate. But you don't have to give any order of preference. A
carte blanche is equivalent to NOTA. If a quota of abstentions is
reached, one of the seats is left empty. This election also gives
voters the rational power to exclude candidates.<br>
</p>
<p>Some candidates may be both popular and unpopular enough to gain
both election and exclusion quotas. They are both "alive" and
"dead" to the electorate. (A case of "Schrodingers candidate"
according to Forest Simmons.) Whether they are elected or excluded
is determined by a Quotient of the exclusion quota divided by the
election quota. If the ratio is one or less, they are elected; if
not, excluded. (The Quotient is the square of a geometric mean.)</p>
<p>When inverted, the exclusion count is like a second-opinion
election. The geometric means of the candidates election keep
values and inverse exclusion keep value establish the over-all
order of popularity of the candidates (from lowest to highest
over-all keep values.<br>
</p>
<p>All the voters abstentions have to be counted, to establish
whether they care more to elect or exclude candidates. This also
means there is no reduction of the quota with abstentions, as in
Meek method. Counting abstentions observes the conservation of
(preferential) information.</p>
<p>I hired a programmer for first order Binomial STV, which, unlike
the higher orders, should be much simpler than Meek method, and
simpler in conception than the hand counts. However I have always
supported them all my adult life, and am now an old man.</p>
<p>Kristofer found the GitHub link to the programmers coding, which
he sent me:</p>
<p><a class="moz-txt-link-freetext"
href="https://github.com/Esrot-Clients/STV_CSV/tree/master">https://github.com/Esrot-Clients/STV_CSV/tree/master</a></p>
<p>The programmer also sent me a "frontend" for the use of voters:</p>
<p><a class="moz-txt-link-freetext" href="https://votingstv.cloud/">https://votingstv.cloud/</a></p>
<p>And he sent me two manuals, which I attach, in case useful to a
technical person, unlike myself.</p>
<p>Regards,</p>
<p>Richard Lung.</p>
<p><br>
</p>
<p><br>
</p>
<p><br>
</p>
<div class="moz-cite-prefix">On 05/04/2024 16:26, Filip Ejlak wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CAGsbvGwPnV7AG3DnGU9eFV26VrXDxX3WQS3tNoYQAHYLQ6emNg@mail.gmail.com">
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<div dir="ltr">
<div>This is a question primarily to Richard Lung, as I am
trying to understand Binomial STV (and perhaps simulate it).
If we want to do Binomial STV with 10 candidates and 4 seats,
do we just do an STV contest for 4 winners with a simultaneous
"inversed" STV contest for 6 losers, with a candidate being
excluded in one sub-election iff they have won a seat in the
other sub-election? If that's right, isn't it unfortunately a
very clone-dependent solution? If that's not right, what's the
actual algorithm?</div>
<div dir="ltr"><br>
</div>
<div dir="ltr"><br>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">Richard Lung <<a
href="mailto:voting@ukscientists.com" target="_blank"
moz-do-not-send="true" class="moz-txt-link-freetext">voting@ukscientists.com</a>>
wrote:<br>
</div>
<blockquote class="gmail_quote"
style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div>
<p><br>
</p>
<p> </p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">“I
DO object to STV’s negative response”</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold""> It
does not matter whether Mr Ossipoff, you or I, or
anyone else objects to an election method. As HG Wells
said over a century ago (The Elements of
Reconstruction, 1916) voting method is not a matter of
opinion, but a matter of demonstration.</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">It
is perfectly possible for STV to use equivalent
proportional counts to Webster/Sainte Lague and the
d’Hondt rule divisor methods. Originally STV used the
Hare quota. The Droop quota is merely the minimum PR,
as the Hare quota is the maximum PR. I have
recommended the average PR, a Harmonic Mean quota,
V/(S+1/2) which is equivalent for proportionality to
the Sainte Lague divisor rule. But I invented it for
other reasons. It just turned out to have that extra
confirmation.</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">As
a matter of fact, I hired the programming of (first
order) Binomial STV and supplied the list with some
links, including to GitHub. The other day I learned
that GitHub suffered a mass malware attack in 2023,
from which they maybe did not completely recover. I
have no technical knowledge myself. So I welcome it
being looked into by admin, but that is why the list
has not received the links.</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">First
order Binomial STV is simpler in principle than
conventional STV. It is a one-truth election method,
which makes it unique, not only to STV but to all the
worlds election methods, which are at least two-truth
methods. That is to say, they are “unscientific” or
inconsistent, because their rules differ as to how
they elect or exclude candidates. In principle,
election and exclusion are the same, because one
voters election is another voters exclusion<br>
</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">Binomial
STV (not only first order) uses the same method for
electing as excluding candidates. In other words, it
is symmetrical as to election and exclusion. First
order STV involves two counts, an election count of
preferences and an exclusion count of reversed
preferences. Both counts use Meek method computer
count of surplus transfer, in exactly the same
procedure, whether to elect candidates or exclude them
(to an election quota or an exclusion quota, otherwise
the same quota). The exclusion count is an iteration
of the election count.<br>
</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">However,
first order STV is simpler than Meek method, in that
it dispenses with its “last past the post” exclusion
method, when election surpluses run out. It also
dispenses with the Meek method policy of reducing the
quota as voters abstain their preferences. On the
contrary, abstentions information is counted, thus
satisfying the principle of the conservation of
(preferential) information, fundamental to science or
organised knowledge.</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">Regards,</span></p>
<p class="MsoNormal"><span
style="font-size:14pt;font-family:"Arial Rounded MT Bold"">Richard
Lung.</span></p>
<p><br>
</p>
<p><br>
</p>
<div>On 03/03/2024 02:21, Michael Ossipoff wrote:<br>
</div>
<blockquote type="cite"> My phone fell off its stand,
resulting in premature sending of the reply. So let me
resume:
<div dir="auto"><br>
</div>
<div dir="auto">As I was saying, I DO object to STV’s
negative response, because Sainte-Lague & d’Hondt
don’t have any negative response. …& STV is a
humongously elaborate complex procedure, requiring new
balloting equipment & software…while list-PR
requires no new balloting equipment & no software
modification. The allocations to parties & their
candidates can be determined at any kitchen table
where there’s a hand-calculator.</div>
<div dir="auto"><br>
</div>
<div dir="auto">To return to the matter of Hare
single-winner:</div>
<div dir="auto"><br>
</div>
<div dir="auto">It’s true that sometimes the CW is an
unliked middle compromise, & it would be better to
have the winner-favoriteness that comes with Hare,
which always chooses the favorite of the largest
faction of the Mutual-Majority when there is once.</div>
<div dir="auto"><br>
</div>
<div dir="auto">But, to best & always & most
reliably eliminated perceived lesser-evil
giveaway-need, it’s necessary to always elect the CW,
however unfavorite. So I propose RP(wv) when
rank-balloting is insisted-on.</div>
<div dir="auto"><br>
</div>
<div dir="auto">… but would support a Hare proposal if
Hare is honestly offered. It currently is not.</div>
<div dir="auto"><br>
</div>
<div dir="auto">There might be other comments in that
post that I’d like to reply to, if I can find it.</div>
<div dir="auto"><br>
</div>
<div dir="auto"><br>
</div>
<br>
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