[EM] Novel Electoral System
Chris Benham
cbenhamau at yahoo.com.au
Thu May 22 07:27:16 PDT 2025
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 at 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 at 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 at lists.electorama.com> *On Behalf Of
>> *Chris Benham via Election-Methods
>> *Sent:* Wednesday, May 21, 2025 8:30 AM
>> *To:* election-methods at 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 at 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 at lists.electorama.com>
>> <mailto:election-methods at 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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