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<p>Dan,<br>
<br>
Returning to my simple example<br>
<br>
46 A<br>
44 B>C<br>
10 C<br>
<br>
We are now all clear that A is the winner, with C coming last.<br>
<br>
What happens if we replace the winner A with a pair of clones, A1
and A2?<br>
<br>
23 A1>A2<br>
23 A2>A1<br>
44 B<C<br>
10 C<br>
<br>
Now according to my calculations this changes the winner to C, the
candidate that formerly came last.<br>
<br>
A1: x 77 sq. + b 54 sq. + c 54 sq. = 11,761<br>
A2: ditto<br>
B: 56 sq. + 56 sq. + 56 sq. = 9,408<br>
C: 46 sq. + 46 sq. + 56 sq. = 7,368<br>
<br>
This a horrible failure of Clone Independence, specifically
Clone-Winner just like your current plurality (aka FPP) system.<br>
<br>
So how is this supposed to be a big improvement on FPP and better
than Hare (aka IRV)? Or any reasonable clone-proof Condorcet
method?<br>
<br>
You indicated that you are at least partly in sympathy with the
Borda Count. That method is of no positive interest aside from
being a historical curiosity.<br>
<br>
It fails Majority Favourite, Clone Independence (including Rich
Party, meaning factions gain by fielding extra candidates) and is
massively vulnerable to Burial strategy. <br>
<br>
So far you have cheerfully admitted that your method fails
Favorite Betrayal ( or "Sincere Favorite"), Condorcet and Clone
Independence.<br>
<br>
But apart from some explanation of how it might be better than
Borda, you haven't given any examples or positive criterion
compliance claims to show why anyone might think it is better than
any other method.<br>
<br>
Chris<br>
<br>
<br>
<br>
<br>
</p>
<div class="moz-cite-prefix">On 22/05/2025 10:54 pm, Daniel Kirslis
via Election-Methods wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CAFFnmiav6siJCd72jwo9Zw=9iSw37AV6QwSQcoB0ozXxQPCHYw@mail.gmail.com">
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<div dir="ltr">Thanks Paul! That is correct.
<div><br>
</div>
<div>For each candidate, you calculate the number of ballots on
which they were ranked below each other candidate. Then, you
square each of these numbers and add them all together to
obtain a total for each candidate. The candidate with the
lowest total is the winner.</div>
<div><br>
</div>
<div>The algebra here is downstream of the geometry. The
intuition comes from the picture.</div>
<div><br>
</div>
<div>We can imagine a Borda count geometrically as one number
line, where a candidate advances by one each time they are
ranked above another candidate on any voter's ballot, and the
candidate who advances the farthest to the right wins. The
K-count instead breaks this out into a different, orthogonal
number line for each opposition candidate, so we move from a
number line into Cartesian space. Now, the candidate who
advances the closest to the 'far corner' of the space wins. </div>
<div><br>
</div>
<div>By aggregating all of the candidates into one number line,
the Borda count treats each opposition candidate identically,
so there is no conception of 'head-to-head' matchups in the
Borda system. The K-count decomposes the matchups in the
maximally independent way (i.e., orthogonally) without
disaggregating the races entirely, as Condorcet methods do.</div>
</div>
<br>
<div class="gmail_quote gmail_quote_container">
<div dir="ltr" class="gmail_attr">On Wed, May 21, 2025 at
5:41 PM Hahn, Paul via Election-Methods <<a
href="mailto:election-methods@lists.electorama.com"
moz-do-not-send="true" class="moz-txt-link-freetext">election-methods@lists.electorama.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 dir="auto">
No I don’t! I should have said rows, not columns. So the
actual numbers are 54 squared times two for A (5,832), 56
squared times two for B (6,272), and 46 squared plus 90
squared (10,216) for C. A still wins, but I think these are
the correct numbers now.
<div><br
id="m_6795413213360472579lineBreakAtBeginningOfSignature">
<div dir="ltr">--pH</div>
<div dir="ltr"><br>
<blockquote type="cite">On May 21, 2025, at 3:55 PM,
Hahn, Paul <<a href="mailto:manynote@wustl.edu"
target="_blank" moz-do-not-send="true"
class="moz-txt-link-freetext">manynote@wustl.edu</a>>
wrote:<br>
<br>
</blockquote>
</div>
<blockquote type="cite">
<div dir="ltr">
<div>
<p class="MsoNormal">I hope Dan doesn’t mind me
stepping in here. I think the issue is that we
are supposed to count non-victories by the number
of ballots failing to express that preference, not
by pairwise differences. If I understand Dan’s
method correctly, one goes down each column of the
Condorcet matrix, subtracting each number from the
total number of ballots cast, squaring those, and
summing them for each column. In this case A has
56 ballots failing to express a preference for A
over B, and 46 failing to express a preference for
A over C. 56 squared plus 46 squared is 5,252.
B’s column-sum is 56 squared plus 90 squared, or
11,236. C’s column-sum is 54 squared plus 56
squared, or 6,052. A’s sum is lowest, so A wins.</p>
<p class="MsoNormal"> </p>
<p class="MsoNormal">Dan, do I have that right?</p>
<p class="MsoNormal"> </p>
<p class="MsoNormal">--pH</p>
<p class="MsoNormal"> </p>
<div>
<div
style="border-right:none;border-bottom:none;border-left:none;border-top:1pt solid rgb(225,225,225);padding:3pt 0in 0in">
<p class="MsoNormal"><b><span
style="font-size:11pt;font-family:Calibri,sans-serif">From:</span></b><span
style="font-size:11pt;font-family:Calibri,sans-serif"> Election-Methods
<<a
href="mailto:election-methods-bounces@lists.electorama.com"
target="_blank" moz-do-not-send="true"
class="moz-txt-link-freetext">election-methods-bounces@lists.electorama.com</a>>
<b>On Behalf Of </b>Chris Benham via
Election-Methods<br>
<b>Sent:</b> Wednesday, May 21, 2025 8:30 AM<br>
<b>To:</b> <a
href="mailto:election-methods@lists.electorama.com" target="_blank"
moz-do-not-send="true"
class="moz-txt-link-freetext">election-methods@lists.electorama.com</a><br>
<b>Subject:</b> Re: [EM] Novel Electoral
System</span></p>
</div>
</div>
<p class="MsoNormal"> </p>
<p style="margin-bottom:12pt">Dan, <br>
<br>
The new short version of your paper I also find
opaque. Earlier you agreed with Andrew that
</p>
<blockquote style="margin-top:5pt;margin-bottom:5pt">
<p class="MsoNormal">It seems like the short
version is that the winner is the candidate with
the smallest sum of SQUARES of non-victories
(defeats plus ties) against their opponents.</p>
</blockquote>
<p class="MsoNormal" style="margin-bottom:12pt"><br>
And then you told me that in this example<br>
<br>
46 A<br>
44 B>C<br>
10 C<br>
<br>
your K-count method elects A.<br>
<br>
C>A 54-46, A>B 46-44, B<C 44-10<br>
<br>
Each candidate has only one "non-victory". So
then I take it then, using Andrew's version the
winner is C, because squaring the pairwise
non-victory scores of C44, B46, A54 doesn't
change their order and C's is the smallest.<br>
<br>
Obviously one of us has it wrong.<br>
<br>
Chris<br>
<br>
</p>
<div>
<p class="MsoNormal">On 20/05/2025 8:58 am, Daniel
Kirslis via Election-Methods wrote:</p>
</div>
<blockquote style="margin-top:5pt;margin-bottom:5pt">
<div>
<p class="MsoNormal">Hi Chris, </p>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">Yes, that is correct. I
have created a simplified version of the
paper that attempts to explain the method in
the most concise possible way. It's only two
pages: <a
href="https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link"
target="_blank" moz-do-not-send="true"
class="moz-txt-link-freetext">https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link</a></p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">It skips over a lot of
the background that explains why I view this
as a compromise between the Borda count and
Condorcet methods and just focuses on
explaining the method itself. Once you see
how the plotting works, it is like Bocce
Ball - closest to the target ball wins.</p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">Thank you for your
engagement on this. I should have started
with this version of the paper!</p>
</div>
</div>
<p class="MsoNormal"> </p>
<div>
<div>
<p class="MsoNormal">On Mon, May 19, 2025 at
12:32<span
style="font-family:Arial,sans-serif"> </span>PM
Chris Benham via Election-Methods <<a
href="mailto:election-methods@lists.electorama.com" target="_blank"
moz-do-not-send="true"
class="moz-txt-link-freetext">election-methods@lists.electorama.com</a>>
wrote:</p>
</div>
<blockquote
style="border-top:none;border-right:none;border-bottom:none;border-left:1pt solid rgb(204,204,204);padding:0in 0in 0in 6pt;margin-left:4.8pt;margin-right:0in">
<div>
<blockquote
style="margin-top:5pt;margin-bottom:5pt">
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">It seems like the
short version is that the winner is
the candidate with the smallest sum
of SQUARES of non-victories (defeats
plus ties) against their opponents.</span></p>
</div>
</blockquote>
<div>
<p class="MsoNormal"><br>
I take that these numbers you are
squaring are the candidate's opposing
and tying vote scores, and not simply
the number of such results. Is that
right?
<br>
<br>
Because otherwise that would often be
very indecisive, like Copeland.<br>
<br>
<br>
On 19/05/2025 1:40 am, Andrew B Jennings
(elections) via Election-Methods wrote:</p>
</div>
<blockquote
style="margin-top:5pt;margin-bottom:5pt">
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">Hi Dan,</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">Great paper. Thank
you for posting!</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">It seems like the
short version is that the winner is
the candidate with the smallest sum
of SQUARES of non-victories (defeats
plus ties) against their opponents.</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">Taking the square
root and dividing can make it
meaningful by scaling it to [0,1] or
[0,s] (where s is the number of
voters), but doesn't change the
finish order.</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">It does seem like
an interesting attempt to "square
the circle" (great pun) and
compromise between Borda and
Condorcet. I hadn't realized that
Borda and Minimax are minimizing the
one-norm and infinity-norm in the
same geometric space. The two-norm
certainly seems like it should be
explored.</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">I would love to
see the proof of
non-favorite-betrayal.</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">Best,</span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif"> </span></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.5pt;font-family:Arial,sans-serif">~ Andy</span></p>
</div>
<div>
<p class="MsoNormal">On Thursday, May
15th, 2025 at 4:25 PM, Daniel Kirslis
via Election-Methods
<a
href="mailto:election-methods@lists.electorama.com" target="_blank"
moz-do-not-send="true"><election-methods@lists.electorama.com></a>
wrote:<br>
<br>
</p>
<blockquote
style="margin-top:5pt;margin-bottom:5pt">
<div>
<div>
<div>
<p class="MsoNormal">Hello!</p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">I am a
newcomer to this mailing list,
so please forgive me if this
message violates any norms or
protocols that the members of
this list adhere to.
</p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">I have
recently developed a novel
method for tabulating
ranked-choice elections that
attempts to reconcile the
concerns of Borda and
Condorcet. I believe that it
maintains the simplicity and
mathematical elegance of the
Borda count while
incorporating Condorcet's
concern with pairwise
dominance. Intuitively, it can
be understood as ordering
candidates by how close they
come to being unanimously
selected when plotted in
Cartesian coordinate space.
Here is a link to the paper:</p>
</div>
<div>
<p class="MsoNormal"><a
href="https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing"
target="_blank"
moz-do-not-send="true"
class="moz-txt-link-freetext">https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing</a></p>
</div>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">Given its
simplicity, I have been very
surprised to discover that this
method has never been proposed
before. I am hoping that some of
you all will take a look at the
paper and share your comments,
questions, and critiques.
Ultimately, it is my hope that
ranked-choice voting advocates
can arrive at a consensus about
the best method for RCV and thus
strengthen efforts to adopt it
and deliver much needed
democratic improvements. But
even if you don't find the
system itself compelling, you
may find the method of plotting
electoral outcomes elucidated in
the paper to be useful for the
analysis of other electoral
systems.</p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">Thank you!</p>
</div>
<div>
<p class="MsoNormal"> </p>
</div>
<div>
<p class="MsoNormal">-Dan</p>
</div>
</div>
</blockquote>
<p class="MsoNormal"> </p>
</div>
<p class="MsoNormal"><br>
<br>
</p>
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<p class="MsoNormal"><br>
<br>
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</blockquote>
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