<div dir="ltr">Hi Chris,<div><br></div><div>I actually do not believe that the Borda count infers ratings from rankings. I understand how it appears that way, as higher ranked candidates get more points than lower ranked candidates, so it seems like there is something cardinal going on. But the Borda count is simply awarding one point to a candidate each time they are preferred to any other candidate. <br><br>In an election with n candidates, there will be n(n-1)/2 unique pairings of candidates. Each voter will therefore have n(n-1)/2 pairwise preferences. The Borda count can be understood as selecting the candidate who will honor the maximum number of those preferences (or equivalently, violate the minimum number of those preferences). Obviously, there are many ways of defining a social welfare function, but I believe that the Borda count optimizes for this simplest and most intuitive definition.<br><br>In my example with the 26 candidates, selecting the Condorcet winner, candidate Z, violates 499,999*1,000,000 pairwise preferences. Meanwhile, selecting candidate A violates only 500,001 preferences. This is why I find going with candidate Z so objectionable.<br><br>As you note, this is ignoring strategic voting and slate manipulation. I am working on a proof showing that the K-count is less vulnerable to burying and cloning than the Borda count, but it is certainly not invulnerable.</div><div><br></div><div>In your example election, the K-count would choose C (51.5) > A (50.47) > B (48). Perhaps this is my American bias, where political polarization seems to push us ever closer to civil war, but I would like to see C win in this scenario, as they are the least objectionable candidate. In my view, it is better to have a weak leader who is tepidly endorsed than a leader who is passionately hated by nearly half the population. Obviously a very subjective issue. <br><br>I like your metaphor with the engines and fuels. I think of it as two paradigms: preference aggregation, where we can assume that individuals give honest preferences and the slate is not manipulated, vs. electoral methods, where we must assume voters and parties are strategic. I originally conceived the K-count as a preference aggregation function, but of course the two paradigms are closely related.<br></div><div><br></div><div>-Dan</div></div><br><div class="gmail_quote gmail_quote_container"><div dir="ltr" class="gmail_attr">On Mon, May 19, 2025 at 1:59 PM Chris Benham via Election-Methods <<a href="mailto:election-methods@lists.electorama.com">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"><u></u>
<div>
<p>Daniel,<br>
<br>
In common with the pro-Borda minded, you seem to be making the
completely unjustified assumption that the multiple candidates are
more or less equidistant from each other in "issue space" and so
it's ok infer some maybe-sincere ratings from rankings.<br>
<br>
</p><blockquote type="cite">Imagine an election with 26 candidates, A,
B, C... Z, and 1 million voters. Let us suppose that candidate A
is unanimously preferred to every other candidate, 1,000,000 to
0, except for candidate Z, to whom she loses by 2 votes, 500,001
to 499,999. Meanwhile, candidate Z beats every other candidate
by the same 2 vote margin, and is thus very narrowly a Condorcet
winner. Does it really reflect the will of the majority better
to declare candidate Z the winner because of his
extraordinarily narrow margin over all of the opposition when
candidate A is the unanimous favorite versus everyone but
Candidate Z, to whom she barely loses? <br>
</blockquote>
<br>
Yes. The other (related) thing you seem to have in common with
them is obliviousness to Clones. Like Borda, I am sure your
method fails Clone Independence (but maybe less egregiously). <br>
<br>
With only ranking information, other things being equal, I don't
see any justification for rejecting Condorcet (or Smith). Based
on positional information we can sometimes guess that some other
candidate is higher Social Utility, but probably not in a way that
is reliable or isn't very vulnerable to strategy.<br>
<br>
For example say this is an election for a seat in the House of
Representatives in Australia which uses compulsory full strict
ranking Hare:<br>
<br>
49 A>C>B<br>
48 B>C>A<br>
03 C>A>B<br>
<br>
The A and B supporters very likely only ranked C because they were
forced to fully rank. C is the voted Condorcet winner, but it
would never cross anyone's mind in Australia that C should be the
winner. In some places C might have struggled to get on the
ballot, and in Australia would be in danger of forfeiting his or
her cash deposit (for not getting a high enough percentage of the
"primary vote".)<br>
<br>
If truncation was allowed (as I think it should be) then most
likely the A and B voters would have truncated and A would be the
voted CW (A>B 51-49, A>C 49-3).<br>
<br>
But aside from that by far the main reason why a method that fails
Condorcet might be acceptable is that the Condorcet criterion is
incompatible with other criteria that people like. Because
Condorcet is incompatible with Later-no-Help all Condorcet methods
are vulnerable (to varying degrees) to Burial strategy.<br>
<br>
It's a bit like comparing two engines, one performs perfectly with
clean pure fuel but very badly with dirty impure fuel and another
that doesn't do quite as well with clean fuel but copes quite a
bit better with dirty fuel. I'm thinking of course of the
comparison between a Condorcet method and Hare.<br>
<br>
Chris<br>
<br>
<p></p>
<div>On 19/05/2025 9:16 am, Daniel Kirslis
via Election-Methods wrote:<br>
</div>
<blockquote type="cite">
<div dir="ltr">Hi r b-j,
<div><br>
</div>
<div>Thank you for this response. I want to address both of
your principles.</div>
<div><br>
</div>
<div>First, is "one person, one vote". I of course agree
completely that each individual's vote should be treated
exactly equally, and the K-count does this. You say that "for
any ranked ballot, this means that if Candidate A is ranked
higher than Candidate B then that is a vote for A... It
doesn't matter how many levels A is ranked higher than B, it
counts as exactly one vote for A." This is precisely how the
K-count works - if A is ranked above B on one ballot, then A
advances by one along the 'preferred to B' axis. The number of
rankings between them is immaterial to A's position vis a vis
the B axis. However, if A is ranked above other candidates on
that ballot, A will also advance along those candidates' axes,
so it is perhaps not exactly "one vote". But each voter's vote
has the same potential power.</div>
<div><br>
</div>
<div>To your second principle. You say "I cannot understand why,
*if* a Condorcet winner exists, how *any* other method; Hare,
Borda, Bucklin, or Kirslis is more democratic than Condorcet."
Let me give an example to illustrate, which relates to the
principle of majority rule.</div>
<div><br>
</div>
<div>Imagine an election with 26 candidates, A, B, C... Z, and 1
million voters. Let us suppose that candidate A is unanimously
preferred to every other candidate, 1,000,000 to 0, except for
candidate Z, to whom she loses by 2 votes, 500,001 to 499,999.
Meanwhile, candidate Z beats every other candidate by the same
2 vote margin, and is thus very narrowly a Condorcet winner.
Does it really reflect the will of the majority better to
declare candidate Z the winner because of his
extraordinarily narrow margin over all of the opposition when
candidate A is the unanimous favorite versus everyone but
Candidate Z, to whom she barely loses? Many more preferences
are violated by choosing the Condorcet winner in this case
than choosing candidate A. This is the heart of the issue with
the Condorcet winner criteria - if a Condorcet winner exists,
a Condorcet method <b>must</b> completely ignore the size of
the margins of victory, no matter how large. In my view, this
curtails the meaning of 'majority rule' in a way that feels
undemocratic.</div>
<div><br>
</div>
<div>I am not familiar with the Burlington election that you
reference, and I will look into it when I have a chance. I
don't know what the K-count would decide in that case. But I
can try to answer in principle your question "How *possibly*
can Candidate B be elected without counting those 3476 voters'
individual votes a little more (like 17% more) than how much
the votes were counted from the 4064 voters preferring
Candidate A?" In the K-count, for Candidate B to be elected in
this scenario, there would need to be a 3rd candidate (or
multiple other candidates) to whom B was widely preferred but
A was not. So, B would win because the people who favored A
still preferred B to C, while the people who favored B
preferred C to A. If you only look at the head-to-head votes
of A vs. B, this seems anti-majoritarian, but the point I make
in the paper is that you cannot<b> </b>make valid inferences
by decontextualizing the data like that, as doing so can lead
you into the logical contradiction of a Condorcet cycle. It is
in the very nature of multi-option preference aggregation that
the data cannot be decomposed in this way. Another way of
thinking about this is - suppose that while A is preferred to
B, B is preferred to C, and C is preferred to A, so you have a
classic Condorcet cycle. Then, someone must be declared the
winner, so in your reasoning, someone's votes will be counted
for more than someone else's. And, when resolving this issue,
most Condorcet methods will look at the margins of victory,
even though they are ignored in the case when a Condorcet
winner exists. But if margins matter enough to decide a winner
when no Condorcet winner exists, why is it okay to completely
ignore them when a Condorcet winner does exist?</div>
<br>
The K-count is a way of trying to reconcile Condorcet's
conception of majority rule, which looks for majority in terms
of each head-to-head matchup, with Borda's conception of
majority rule, which seeks to honor the maximum number of
individual pairwise preferences.
<div><br>
</div>
<div>Thanks again for your response, and thank you for looking
over the paper. I appreciate your civil tone and good faith
questions, and I hope it is clear that the discussion here is
made with full respect and in a spirit of friendly
intellectual inquiry. And I welcome your response to these
arguments!</div>
<div><br>
</div>
<div>I am also considering the questions from other folks and am
working on responses to those as well.</div>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">On Sun, May 18, 2025 at
3:30 PM robert bristow-johnson via Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank">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"><br>
Hi Dan,<br>
<br>
I made one pass through your paper, but the interaction with
Chris and Andy was helpful. I understand the definition of
your K-count measure, but still don't understand the
motivation of it, solely from the POV of democratic
principles, which is where I draw my Condorcetist
perspective. Admittedly, I am a hard-core Condorcet advocate,
but I am so because of some basic principles.<br>
<br>
I read your section 9 and re-read it, and I still cannot get
past how it justifies *any* non-Condorcet method (including
your K-count) over Condorcet. The principles of free and fair
elections in a democratic context require, among other things,
that our votes are valued equally:<br>
<br>
1. "One person, one vote": Every enfranchised voter has an
equal influence on<br>
government in elections because of our inherent equality
as citizens and this is<br>
independent of any utilitarian notion of personal
investment in the outcome. If I<br>
enthusiastically prefer Candidate A and you prefer
Candidate B only tepidly, your<br>
vote for Candidate B counts no less (nor more) than my
vote for A. The<br>
effectiveness of one's vote – how much their vote counts –
is not proportional to<br>
their degree of preference but is determined only by their
franchise. A citizen with<br>
franchise has a vote that counts equally as much as any
other citizen with<br>
franchise. For any ranked ballot, this means that if
Candidate A is ranked higher<br>
than Candidate B then that is a vote for A, if only
candidates A and B are<br>
contending (such as in the IRV final round). It doesn't
matter how many levels A<br>
is ranked higher than B, it counts as exactly one vote for
A.<br>
<br>
If our votes are not valued equally, then I want my vote to
count more than yours. If that is unacceptable
(understandably) then we must agree to count our votes
equally. In the U.S., too many people have died over that
inequality. So then, in order for our votes to be valued
equally, we must have Majority Rule in single-winner
elections:<br>
<br>
2. Majority rule: If more voters mark their ballots
preferring Candidate A over<br>
Candidate B than the number of voters marking their
ballots to the contrary,<br>
then Candidate B is not elected. If Candidate B were to be
elected, that would<br>
mean that the fewer voters preferring Candidate B had cast
votes that had greater<br>
value and counted more than those votes from voters of the
larger set preferring<br>
Candidate A.<br>
<br>
Those are two ways of, essentially, expressing the same
principle in single-winner elections. For multi-winner
elections, the way to value our votes equally would be
Proportional Representation, but I don't wanna go there in
this discussion. I would like to stay with single-winner
elections.<br>
<br>
Now, of course this doesn't deal with the problem of cycles
and we can discuss what the best and most democratic way to
deal with cycles is, but I cannot understand why, *if* a
Condorcet winner exists, how *any* other method; Hare, Borda,
Bucklin, or Kirslis is more democratic than Condorcet.<br>
<br>
If a CW exists and we *know* (from the Cast Vote Record having
ranked ballot data) that the CW exists and who that CW is, how
is electing the K-count winner, assuming they're different
from the CW, more democratic? Just like with the IRV
failures, we will *know* that a smaller set of voters have
left that election satisfied than that of a larger set of
voters leaving the election dissatisfied. We will know that
the votes coming from that smaller set of voters were more
effective in electing their preferred candidate than the votes
coming from the larger set of voters that not only preferred
someone else, but they preferred a *specific* candidate over
the one who Kirslis elected and marked their ballots saying
so. For the very same reason that IRV failed in Burlington
Vermont in 2009 or in Alaska in August 2022, the elected
candidate will suffer a sense of loss of legitimacy in the
election.<br>
<br>
In Burlington in 2009, 4064 voters marked their ballots that
Candidate A was a better choice than Candidate B and 3476
voters marked their ballots to the contrary. (There were 1436
voters that didn't like either A or B and didn't rank
either.) How *possibly* can Candidate B be elected without
counting those 3476 voters' individual votes a little more
(like 17% more) than how much the votes were counted from the
4064 voters preferring Candidate A?<br>
<br>
Now this is a failure of Hare (IRV) but I can construct the
very same question for an election decided with Kirslis rules
that failed to elect the CW when such exists. How would you
answer that question? How do you justify satisfying a smaller
set of voters at the expense of a larger set of voters that
preferred, not just anyone else, but a specific candidate over
the Kirslis winner? I couldn't glean an answer to that from
section 9 (or anywhere else) in your paper.<br>
<br>
bestest,<br>
<br>
--<br>
<br>
r b-j . _ . _ . _ . _ <a href="mailto:rbj@audioimagination.com" target="_blank">rbj@audioimagination.com</a><br>
<br>
"Imagination is more important than knowledge."<br>
<br>
.<br>
.<br>
.<br>
<br>
> On 05/18/2025 1:51 PM EDT Daniel Kirslis via
Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.electorama.com</a>>
wrote:<br>
> <br>
> <br>
> Hi all,<br>
> <br>
> Thanks so much for the replies. I’ll respond to everyone
in this thread.<br>
> <br>
> Andy - I really appreciate your feedback. Your summary is
correct, and your framing of it as one-norm vs. two-norm vs.
infinity-norm is a way of thinking about it that I had not
considered. It seems like a potentially fruitful lens for
understanding it. And, as perhaps you have surmised, I may
have been mistaken in the statement about the sincere favorite
criteria, but I am working on an analysis of the issue that I
will share.<br>
> <br>
> Toby, making a short summary is a great suggestion. The
argument in the paper is admittedly a bit convoluted before it
presents the actual method. Here is the simplified way that I
would explain it:<br>
> <br>
> Each voter ranks their preferences, with ties allowed and
unranked candidates treated as last-place preferences. Then,
for each candidate, you make a plot, where each axis is the
total number of times that they were preferred to each of
their opponents. So, if the candidates are A, B, and C,
candidate A’s plot would have “number of times preferred to B”
on one axis and “number of times preferred to C” on the other
axis. Candidate B & C could be plotted similarly in terms
of their opponents. The winner is simply the candidate who is
plotted the farthest up and to the right, or closest to
topmost and rightmost point, which is where a candidate who is
the unanimous first-place choice would be plotted. The
distance from that point is calculated using the Pythagorean
theorem, which is where minimizing the sum of squares that
Andy referenced comes in.<br>
> <br>
> The figures in the paper tell the story better than the
words, as it is essentially a geometric idea. And, sections 4,
5, and 6 can really be skipped - they are more about
justifying the approach than explaining it.<br>
> <br>
> Chris, you asked “Why should we be interested in the
"concerns" of Borda (whatever they are)? And so much that we
should embrace a method that fails the Condorcet criterion?”
Great question. If you look at the Stanford Encyclopedia of
Philosophy’s entry on Social Choice Theory, they list
Condorcet and Borda as the original pioneers of this thinking
(<a href="https://plato.stanford.edu/entries/social-choice/" rel="noreferrer" target="_blank">https://plato.stanford.edu/entries/social-choice/</a>).
Borda thinks about majoritarianism in terms of votes, while
Condorcet thinks about it in terms of voters. Obviously, in
FPP elections, these are the same, but the heart of the
interest in these questions comes from the tension that arises
between them in a ranked-choice setting, where each voter has
multiple votes and ‘majoritarianism’ is no longer simple to
define. Don Saari is a thinker who studies these issues and
has argued most persuasively for Borda’s approach over
Condorcet methods. In section 9 of my paper, I explain some of
my philosophical objections to the Condorcet winner criterion.
<br>
> <br>
> You also asked “Do you propose allowing above-bottom
equal ranking or truncation?” Equal ranking is allowed, and
unranked candidates are treated as last place.<br>
> <br>
> And, I am afraid I may have actually been mistaken about
the sincere favorite property, so will have to disappoint you
there.<br>
> <br>
> You asked “Who does your method elect in this example?<br>
> <br>
> 46 A<br>
> 44 B>C<br>
> 10 C”<br>
> <br>
> If I am understanding your notation correctly, A would
win in this example. The full ranking would be:<br>
> A's K-count = 46 =
100-SQRT((100-46)^2+(100-46)^2)/(SQRT(2))<br>
> B's K-count = 44 =
100-SQRT((100-44)^2+(100-44)^2)/(SQRT(2))<br>
> C's K-count = 28.53 =
100-SQRT((100-54)^2+(100-10)^2)/(SQRT(2))<br>
> <br>
> As you can see, when a candidate only appears as a
first-place or last-place preference, their K-count is simply
equal to the number of voters ranking them first.<br>
> <br>
> Thanks all!<br>
> <br>
> <br>
> On Sun, May 18, 2025 at 12:10 PM Andrew B Jennings
(elections) <<a href="mailto:elections@jenningsstory.com" target="_blank">elections@jenningsstory.com</a>>
wrote:<br>
> > Hi Dan,<br>
> > <br>
> > Great paper. Thank you for posting!<br>
> > <br>
> > 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.<br>
> > <br>
> > 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.<br>
> > <br>
> > <br>
> > 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.<br>
> > <br>
> > I would love to see the proof of
non-favorite-betrayal.<br>
> > <br>
> > Best,<br>
> > <br>
> > ~ Andy<br>
> > On Thursday, May 15th, 2025 at 4:25 PM, Daniel
Kirslis via Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank">election-methods@lists.electorama.com</a>>
wrote:<br>
> > <br>
> > > Hello!<br>
> > > <br>
> > > 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.<br>
> > > <br>
> > > 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:<br>
> > > <a href="https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing" rel="noreferrer" target="_blank">https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing</a><br>
> > > <br>
> > > 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.<br>
> > > <br>
> > > Thank you!<br>
> > > <br>
> > > -Dan<br>
> > <br>
> > <br>
> ----<br>
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