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<p> </p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">Condorcet pairing</span></p>
<p class="MsoNormal"><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"">Suppose a preferential (non-binary) election, by 100
voters for three candidates, A, B and C, contesting two seats,
produces this result:</span> </p>
<p class="MsoNormal"><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"">A<span style="mso-spacerun:yes"> </span>B<span
style="mso-spacerun:yes"> </span>C</span> </p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">46<span style="mso-spacerun:yes"> </span>34<span
style="mso-spacerun:yes"> </span>20</span></p>
<p class="MsoNormal"><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"">A and B are elected on over one third of the votes
each (by Droop quota).</span> </p>
<p class="MsoNormal"><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"">Suppose, however, that this contest result is
contested by Condorcet pairing:</span> </p>
<p class="MsoNormal"><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"">A<span style="mso-spacerun:yes"> </span>B<span
style="mso-spacerun:yes"> </span>A<span
style="mso-spacerun:yes"> </span>C<span
style="mso-spacerun:yes"> </span>B<span
style="mso-spacerun:yes"> </span>C</span> </p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">66<span style="mso-spacerun:yes"> </span>34<span
style="mso-spacerun:yes"> </span>56<span
style="mso-spacerun:yes"> </span>44<span
style="mso-spacerun:yes"> </span>34<span
style="mso-spacerun:yes"> </span>66</span></p>
<p class="MsoNormal"><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"">A gets all C vote and C gets all A vote. But B is
denied its proportionate share of representation. A and C are
the two-seat Condorcet winners. They are implicitly elected each
on single majorities of over half the votes, 50+ votes each,
from an electorate of 100 voters. At any rate, that is the
minimum democratic threshold (even Kenneth Arrow insists on). –
Binary choice is “The tyranny of the (single) majority.” (John
Stuart Mill; Lani Guinier.)</span> </p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">For the sake of argument, consider the logic behind
alternate pairings. Each pairing is a provisional exclusion
count, removing each candidate, in turn, as a third preference.
Thus the A-B pairing amounts to A>B>C.<span
style="mso-spacerun:yes"> </span>The A-C pairing implies
A>C>B. The B-C pairing implies B>C>A. Condorcet
pairing elects A and C on the contradiction that preferences
A>B>C and C>B>A are both correct.</span></p>
<p class="MsoNormal"><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"">On the basis of an original triple preference
(non-binary) count, A>B>C happens to be correct. </span>
</p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">The question is: how does one arrive at such a
preferential count? (That is what I’ve been trying to explain,
against some personal animosity.) It depends on a multi-majority
(proportional) count, as distinct from a single majority count,
summing a (multi-) preference vote, as distinct from a binary
choice.</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold"">Traditional single transferable vote is sufficiently
robust to achieve this, with its rational election count. STV
does contain a residual irrational element, in a sort of last
past the post exclusion count. But the exclusion count can be
calculated on similar rational grounds, as the election count.
Thus, a binomial STV.</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.<br>
</span></p>
<p class="MsoNormal"><span
style="font-size:16.0pt;font-family:"Arial Rounded MT
Bold""> </span></p>
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<div class="moz-cite-prefix">On 06/09/2022 11:28, Colin Champion
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:1e186b90-8463-f0ee-81c5-d847f75d5753@routemaster.app">
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<font face="Helvetica, Arial, sans-serif">Perhaps some people will
be interested in another conclusion I came to from reading
Condorcet's Essai. He proposed a method of breaking cycles which
generated a lot of confusion until Peyton Young glossed it as a
garbled account of the Kemeny-Young method. His reading has been
widely accepted; Tideman (in his 2006 book) declared that
"Condorcet's intent is decoded to my satisfaction" by Young.<br>
<br>
Condorcet described his method twice: forwards in the
Preliminary Discourse and backwards in the body of the work.
Young only discusses the backwards version. In both cases
Condorcet starts from a list of pairwise comparisons, sorted by
margin. In the backwards version he writes: "We will
successively discard from the contradictory set the preferences
which have the smallest majority, and elect the candidate
preferred by those which remain". Presumably he stops discarding
when the residue is consistent; the flaw is that by this point
there may not be enough comparisons left to determine a unique
winner. Young noticed this and remarked that "It seems more
likely that Condorcet meant to *reverse*, rather than to
*delete* the weakest proposition". This is nonsense: no one
writes "delete" when they mean "swap", and Young's reading
doesn't fit the forwards version. <br>
<br>
The forwards statement is clearer: "We thus obtain the following
general rule, that whenever we are required to elect a
candidate, we must take in turn all the pairwise preferences
which have majority support, starting with the largest
majorities, and make a decision according to these initial
preferences as soon as they imply one, without worrying about
the less probable later preferences." In other words, given a
list sorted in decreasing order of margin, take an initial part
which is small enough to be consistent but large enough to
determine a unique winner. This has a corresponding flaw, which
is that as you work through the list, you may be forced to
include a comparison which contradicts those already present
before reaching the point at which you have a winner. But
there's nothing here which you can interpret as meaning "swap"
rather than something else. <br>
<br>
It seems to me as clear as daylight that Condorcet had an
incomplete grasp of Tideman's Ranked Pairs. Tideman recognised
the risk that a new pair may contradict the ones already in the
list, and he saw what to do about it, namely throw it away.<br>
<br>
CJC<br>
</font> <br>
<fieldset class="moz-mime-attachment-header"></fieldset>
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