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<p>Forest,<br>
</p>
<p>My main concern is with the autocratic nature of single member
monopolies, generally little or no better than minimally
democratic, and unaccountable. They have not got beyond the
ancient Greek tyranny, deserving of their reputation in the vulgar
sense of tyranny. </p>
<p>R.L.<br>
</p>
<p><br>
</p>
<div class="moz-cite-prefix">On 07/06/2022 19:06, Forest Simmons
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CANUDvfoHvR+P+_hdaCdZgC3HZoJtPybQsyCZNeV=AsKBvrH6jQ@mail.gmail.com">
<div dir="auto">I wish you could have made Nader win .. informal
polls, including a Time magazine poll that was continually
updated online until it became too embarrassing to the
neoliberal establishment, showed that Nader was the landslide
Sincere Condorcet Winner.</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">El mar., 7 de jun. de 2022
10:16 a. m., Richard Lung <<a
href="mailto:voting@ukscientists.com" moz-do-not-send="true"
class="moz-txt-link-freetext">voting@ukscientists.com</a>>
escribió:<br>
</div>
<blockquote class="gmail_quote">
<div>
<p><br>
</p>
<p> </p>
<p class="MsoNormal"><span>Consider a typical single member
scenario, like the 2000 </span><span>US</span><span>
presidential election. Say George W Bush (label B) gets
101. Al Gore (label A) gets 100. Ralph Nader (label C)
gets 10. (After his vote got squeezed from strategic
voting.)</span></p>
<p class="MsoNormal"><span>Bush is elected on single
preference votes. If there were a second ballot, or
instant run-off vote (IRV), Al Gore wins on second
preferences of Nader, the “spoiler” who is eliminated.</span></p>
<p class="MsoNormal"><span>So, IRV passes on independence of
irrelevant alternatives (IIA). But IRV fails on the </span><span>Laplace</span><span>
law of preference gradation. (Orders of preference vote
gradually fall off in count importance.) This also means
that the IIA criterion is inconsistent with the </span><span>Laplace</span><span>
law.</span></p>
<p class="MsoNormal"><span>Binomial STV avoids the dilemma
of this inconsistency.</span></p>
<p class="MsoNormal"><span>Suppose IRV gives:</span></p>
<p class="MsoNormal"><span> </span><span>100 A C _</span> </p>
<p class="MsoNormal"><span>101 B _ _</span></p>
<p class="MsoNormal"><span><span> </span>10 C A _</span></p>
<p class="MsoNormal"><span> </span><span>With IRV, A picks
up the second preferences of C, and wins with 110.</span>
</p>
<p class="MsoNormal"><span>Binomial STV counts abstentions,
shown by the dash lines. The third preferences are all
abstentions, and they do nothing to change the simple
plurality count. It is conceivable that in a less
contentious world than ours, this could be the case. But
it is assumed that the voters have been informed that
last preferences can be given to count against
candidates. And the full slate of preferences is as
follows:</span></p>
<p class="MsoNormal"><span> </span><span>100 A > C > B</span>
</p>
<p class="MsoNormal"><span>101 B > A > C</span></p>
<p class="MsoNormal"><span><span> </span>10 C > A >
B</span></p>
<p class="MsoNormal"><span> </span><span>The keep value
quotient (kvq), of a candidate, is the election keep
value, divided by the exclusion keep value:</span> </p>
<p class="MsoNormal"><span>the keep value is the quota,
211/2 = 105.5, divided by candidate vote.</span></p>
<span></span><span>kvq A = 0/100. Here, zero means close to
0, giving a very small fraction.</span>
<p class="MsoNormal"><span> </span><span>B = 110/101</span>
</p>
<p class="MsoNormal"><span>C = 101/10</span></p>
<p class="MsoNormal"><span> </span><span>Unity, or less, is
the election (or exclusion) threshold of an election (or
exclusion) keep value. Less than unity passes the
threshold. So, A wins with binomial STV, without
breaking the </span><span>Laplace</span><span> law, or
IIA.</span> </p>
<p class="MsoNormal"><span> </span><span>Suppose, however,
that Bush supporters decide to vote insincerely, by
making Gore their last preference – even tho Nader is
the last person they want to see win, but know he can’t,
anyway. </span> </p>
<p class="MsoNormal"><span>(However, this scenario might be
sincere in </span><span>UK</span><span>, with A as
Labour, B as Tory, and C as Liberal Democrat. In either
case, a change in the vote, sincere or otherwise, will
change the count in a toward manner, with good
book-keeping.)</span></p>
<p class="MsoNormal"><span> </span><span></span><span>The
preference slate becomes:</span> </p>
<p class="MsoNormal"><span> </span><span>100 A > C > B</span>
</p>
<p class="MsoNormal"><span>101 B > C > A</span></p>
<p class="MsoNormal"><span><span> </span>10 C > A >
B</span></p>
<p class="MsoNormal"><span> </span><span>Then, kvq becomes:</span>
</p>
<p class="MsoNormal"><span>A = 101/100</span></p>
<p class="MsoNormal"><span>B = 110/101</span></p>
<p class="MsoNormal"><span>C = 0/10</span></p>
<p class="MsoNormal"><span> </span><span>There is a
contradictory answer. B is closer to the quota but A is
closer to the quotient.</span> </p>
<p class="MsoNormal"><span>This is not a logical
contradiction but a contingent contradiction. Decision,
as to the winner, is not a democratic decision, but an
administrative decision, at present FPTP, based on a
convention, reached by previous agreement. (Single
majority is the least democratic system, in the first
place.)</span></p>
<p class="MsoNormal"><span>A first approximation of an
administrative decision (It would be “spurious accuracy”
to go further) is:</span></p>
<p class="MsoNormal"><span>B quota deficit: 105.5/101 ~
1.0446.</span></p>
<p class="MsoNormal"><span>A quotient deficit: 101/100 =
1.01.</span></p>
<p class="MsoNormal"><span> </span><span>Therefore, an
administrative election is of Gore. (The Supreme Court
also made an administrative decision.) </span> </p>
<p class="MsoNormal"><span>The result would most likely be
similar, if Binomial STV, used more accurate figures. To
say nothing of the investigation, by Greg Palast, of the
electoral roll. (The Best Democracy That Money Can Buy.)
</span></p>
<p class="MsoNormal"><span> </span><span>I repeat that for a
democratic decision, as from those Keltic reports,
prevously cited, I don’t recommend less than 4 or 5
member Andrae/Hare system (at-large STV/PR).</span></p>
<p class="MsoNormal"><span>Regards,</span></p>
<p class="MsoNormal"><span>Richard Lung.</span></p>
<p class="MsoNormal"><span><br>
</span></p>
<p class="MsoNormal"><span><br>
</span> </p>
<p> </p>
</div>
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