Novel Electoral System
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.
Dan, do I have that right?
--pH
From: Election-Methods election-methods-bounces@lists.electorama.com On Behalf Of Chris Benham via Election-Methods
Sent: Wednesday, May 21, 2025 8:30 AM
To: election-methods@lists.electorama.com
Subject: Re: [EM] Novel Electoral System
Dan,
The new short version of your paper I also find opaque. Earlier you agreed with Andrew that
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.
And then you told me that in this example
46 A
44 B>C
10 C
your K-count method elects A.
C>A 54-46, A>B 46-44, B<C 44-10
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.
Obviously one of us has it wrong.
Chris
On 20/05/2025 8:58 am, Daniel Kirslis via Election-Methods wrote:
Hi Chris,
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: https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link
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.
Thank you for your engagement on this. I should have started with this version of the paper!
On Mon, May 19, 2025 at 12:32 PM Chris Benham via Election-Methods <election-methods@lists.electorama.commailto:election-methods@lists.electorama.com> wrote:
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.
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?
Because otherwise that would often be very indecisive, like Copeland.
On 19/05/2025 1:40 am, Andrew B Jennings (elections) via Election-Methods wrote:
Hi Dan,
Great paper. Thank you for posting!
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.
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.
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.
I would love to see the proof of non-favorite-betrayal.
Best,
~ Andy
On Thursday, May 15th, 2025 at 4:25 PM, Daniel Kirslis via Election-Methods election-methods@lists.electorama.commailto:election-methods@lists.electorama.com wrote:
Hello!
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.
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:
https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing
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.
Thank you!
-Dan
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list info
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.
--pH
On May 21, 2025, at 3:55 PM, Hahn, Paul manynote@wustl.edu wrote:
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.
Dan, do I have that right?
--pH
From: Election-Methods election-methods-bounces@lists.electorama.com On Behalf Of Chris Benham via Election-Methods
Sent: Wednesday, May 21, 2025 8:30 AM
To: election-methods@lists.electorama.com
Subject: Re: [EM] Novel Electoral System
Dan,
The new short version of your paper I also find opaque. Earlier you agreed with Andrew that
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.
And then you told me that in this example
46 A
44 B>C
10 C
your K-count method elects A.
C>A 54-46, A>B 46-44, B<C 44-10
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.
Obviously one of us has it wrong.
Chris
On 20/05/2025 8:58 am, Daniel Kirslis via Election-Methods wrote:
Hi Chris,
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: https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link
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.
Thank you for your engagement on this. I should have started with this version of the paper!
On Mon, May 19, 2025 at 12:32 PM Chris Benham via Election-Methods <election-methods@lists.electorama.commailto:election-methods@lists.electorama.com> wrote:
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.
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?
Because otherwise that would often be very indecisive, like Copeland.
On 19/05/2025 1:40 am, Andrew B Jennings (elections) via Election-Methods wrote:
Hi Dan,
Great paper. Thank you for posting!
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.
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.
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.
I would love to see the proof of non-favorite-betrayal.
Best,
~ Andy
On Thursday, May 15th, 2025 at 4:25 PM, Daniel Kirslis via Election-Methods election-methods@lists.electorama.commailto:election-methods@lists.electorama.com wrote:
Hello!
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.
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:
https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing
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.
Thank you!
-Dan
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list info
Thanks Paul! That is correct.
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.
The algebra here is downstream of the geometry. The intuition comes from
the picture.
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.
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.
On Wed, May 21, 2025 at 5:41 PM Hahn, Paul via Election-Methods <
election-methods@lists.electorama.com> wrote:
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.
--pH
On May 21, 2025, at 3:55 PM, Hahn, Paul manynote@wustl.edu wrote:
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.
Dan, do I have that right?
--pH
From: Election-Methods election-methods-bounces@lists.electorama.com *On
Behalf Of *Chris Benham via Election-Methods
Sent: Wednesday, May 21, 2025 8:30 AM
To: election-methods@lists.electorama.com
Subject: Re: [EM] Novel Electoral System
Dan,
The new short version of your paper I also find opaque. Earlier you agreed
with Andrew that
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.
And then you told me that in this example
46 A
44 B>C
10 C
your K-count method elects A.
C>A 54-46, A>B 46-44, B<C 44-10
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.
Obviously one of us has it wrong.
Chris
On 20/05/2025 8:58 am, Daniel Kirslis via Election-Methods wrote:
Hi Chris,
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:
https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link
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.
Thank you for your engagement on this. I should have started with this
version of the paper!
On Mon, May 19, 2025 at 12:32 PM Chris Benham via Election-Methods <
election-methods@lists.electorama.com> wrote:
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.
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?
Because otherwise that would often be very indecisive, like Copeland.
On 19/05/2025 1:40 am, Andrew B Jennings (elections) via Election-Methods
wrote:
Hi Dan,
Great paper. Thank you for posting!
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.
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.
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.
I would love to see the proof of non-favorite-betrayal.
Best,
~ Andy
On Thursday, May 15th, 2025 at 4:25 PM, Daniel Kirslis via
Election-Methods election-methods@lists.electorama.com
election-methods@lists.electorama.com wrote:
Hello!
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.
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:
https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing
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.
Thank you!
-Dan
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list
info
Election-Methods mailing list - see https://electorama.com/em for list info
Election-Methods mailing list - see https://electorama.com/em for list
info
Yes, thank-you Paul for your explanation.
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.
Why didn't Dan simply write that in the first (or even second) place?
Chris
On 22/05/2025 10:54 pm, Daniel Kirslis via Election-Methods wrote:
Thanks Paul! That is correct.
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.
The algebra here is downstream of the geometry. The intuition comes
from the picture.
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.
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.
On Wed, May 21, 2025 at 5:41 PM Hahn, Paul via Election-Methods
election-methods@lists.electorama.com wrote:
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.
--pH
On May 21, 2025, at 3:55 PM, Hahn, Paul <manynote@wustl.edu> wrote:
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.
Dan, do I have that right?
--pH
*From:*Election-Methods
<election-methods-bounces@lists.electorama.com> *On Behalf Of
*Chris Benham via Election-Methods
*Sent:* Wednesday, May 21, 2025 8:30 AM
*To:* election-methods@lists.electorama.com
*Subject:* Re: [EM] Novel Electoral System
Dan,
The new short version of your paper I also find opaque. Earlier
you agreed with Andrew that
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.
And then you told me that in this example
46 A
44 B>C
10 C
your K-count method elects A.
C>A 54-46, A>B 46-44, B<C 44-10
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.
Obviously one of us has it wrong.
Chris
On 20/05/2025 8:58 am, Daniel Kirslis via Election-Methods wrote:
Hi Chris,
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:
https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link
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.
Thank you for your engagement on this. I should have started
with this version of the paper!
On Mon, May 19, 2025 at 12:32 PM Chris Benham via
Election-Methods <election-methods@lists.electorama.com> wrote:
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.
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?
Because otherwise that would often be very indecisive,
like Copeland.
On 19/05/2025 1:40 am, Andrew B Jennings (elections) via
Election-Methods wrote:
Hi Dan,
Great paper. Thank you for posting!
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.
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.
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.
I would love to see the proof of non-favorite-betrayal.
Best,
~ Andy
On Thursday, May 15th, 2025 at 4:25 PM, Daniel
Kirslis via Election-Methods
<election-methods@lists.electorama.com>
<mailto:election-methods@lists.electorama.com> wrote:
Hello!
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.
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:
https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing
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.
Thank you!
-Dan
----
Election-Methods mailing list - seehttps://electorama.com/em for list info
----
Election-Methods mailing list - see
https://electorama.com/em for list info
----
Election-Methods mailing list - seehttps://electorama.com/em for list info
----
Election-Methods mailing list - see https://electorama.com/em for
list info
Election-Methods mailing list - seehttps://electorama.com/em for list info
Dan,
Returning to my simple example
46 A
44 B>C
10 C
We are now all clear that A is the winner, with C coming last.
What happens if we replace the winner A with a pair of clones, A1 and A2?
23 A1>A2
23 A2>A1
44 B<C
10 C
Now according to my calculations this changes the winner to C, the
candidate that formerly came last.
A1: x 77 sq. + b 54 sq. + c 54 sq. = 11,761
A2: ditto
B: 56 sq. + 56 sq. + 56 sq. = 9,408
C: 46 sq. + 46 sq. + 56 sq. = 7,368
This a horrible failure of Clone Independence, specifically
Clone-Winner just like your current plurality (aka FPP) system.
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?
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.
It fails Majority Favourite, Clone Independence (including Rich Party,
meaning factions gain by fielding extra candidates) and is massively
vulnerable to Burial strategy.
So far you have cheerfully admitted that your method fails Favorite
Betrayal ( or "Sincere Favorite"), Condorcet and Clone Independence.
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.
Chris
On 22/05/2025 10:54 pm, Daniel Kirslis via Election-Methods wrote:
Thanks Paul! That is correct.
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.
The algebra here is downstream of the geometry. The intuition comes
from the picture.
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.
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.
On Wed, May 21, 2025 at 5:41 PM Hahn, Paul via Election-Methods
election-methods@lists.electorama.com wrote:
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.
--pH
On May 21, 2025, at 3:55 PM, Hahn, Paul <manynote@wustl.edu> wrote:
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.
Dan, do I have that right?
--pH
*From:*Election-Methods
<election-methods-bounces@lists.electorama.com> *On Behalf Of
*Chris Benham via Election-Methods
*Sent:* Wednesday, May 21, 2025 8:30 AM
*To:* election-methods@lists.electorama.com
*Subject:* Re: [EM] Novel Electoral System
Dan,
The new short version of your paper I also find opaque. Earlier
you agreed with Andrew that
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.
And then you told me that in this example
46 A
44 B>C
10 C
your K-count method elects A.
C>A 54-46, A>B 46-44, B<C 44-10
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.
Obviously one of us has it wrong.
Chris
On 20/05/2025 8:58 am, Daniel Kirslis via Election-Methods wrote:
Hi Chris,
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:
https://drive.google.com/file/d/1F_I2ZBUKXKbmcS-uSvMAf_gNdNO8m0GB/view?usp=drive_link
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.
Thank you for your engagement on this. I should have started
with this version of the paper!
On Mon, May 19, 2025 at 12:32 PM Chris Benham via
Election-Methods <election-methods@lists.electorama.com> wrote:
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.
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?
Because otherwise that would often be very indecisive,
like Copeland.
On 19/05/2025 1:40 am, Andrew B Jennings (elections) via
Election-Methods wrote:
Hi Dan,
Great paper. Thank you for posting!
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.
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.
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.
I would love to see the proof of non-favorite-betrayal.
Best,
~ Andy
On Thursday, May 15th, 2025 at 4:25 PM, Daniel
Kirslis via Election-Methods
<election-methods@lists.electorama.com>
<mailto:election-methods@lists.electorama.com> wrote:
Hello!
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.
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:
https://drive.google.com/file/d/152eNheS2qkLHJbDvG4EwW3jdO4I_NwcX/view?usp=sharing
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.
Thank you!
-Dan
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