[EM] Novel Electoral System
Chris Benham
cbenhamau at yahoo.com.au
Thu May 22 12:25:48 PDT 2025
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 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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