[EM] Novel Electoral System

Sun May 18 18:17:52 PDT 2025

Hi Dan,

> On 05/18/2025 7:46 PM EDT Daniel Kirslis <dankirslis at gmail.com> wrote:
> 
> 
> Hi r b-j,
> 
> Thank you for this response. I want to address both of your principles.
> 
> 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.

It's not enough that the *rules* are consistent with every voter.  FPTP and IRV or Bucklin or Borda or Score Voting have consistent rules, but do not necessarily value each vote equally.  And when the Kirslis method does not elect the CW, neither does that method value each vote equally.

> 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.

Well, it's a little more sophisticated than that.  It's obscured in your method, but for Kirslis to elect a non-CW to exist, the K-count elected candidate has to pick up more points than the CW and the only way to do that is for that candidate to get some spread by defeating other candidates that have preference over the CW in many ballots.  That means there are ballots with more than one level separating the K-candidate from CW.  It's a little Borda-ish as you've alluded to in your paper.  And we *know* that Borda explicitly counts spread, which means if I enthusiastically prefer A>...>B and you tepidly prefer B>A, my vote for A will count more than your vote for B if it comes down to a dogfight between A and B.  Your method is, I believe, vulnerable to burying.

> 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.

Well, so do voters all have the same potential power with Score Voting, but we *know* that if I score A with a 5 and B with a 0 and you score B with a 5 and A with a 4, my vote for A will count five times more than your vote for B if it comes down to A and B being the principal candidates.  Just because the rules are the same for you and me, does not mean that our votes count the same.

> 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.
> 
> 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? 

To make this more comprehensible, let's say 100 voters and three candidates, A, B, and Z.

    A>B 100
    B>A 0
    A>Z 49
    Z>A 51
    B>Z 49
    Z>B 51

Now, how does this break down to these nine numbers?

    A>B>Z
    A>Z>B
    A only

    B>A>Z
    B>Z>A
    B only

    Z>A>B
    Z>B>A
    Z only

Can you turn the six numbers above into the nine numbers below in a consistent manner?  Then I can get a grip on your scenario.  But I am not sure you can do it.

> 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 must completely ignore the size of the margins of victory, no matter how large.

Well, in close elections, that is always the case.  In FPTP if A gets 500,001 and B gets 499,999 there will be a recount, but A will win if these tallies result.

> In my view, this curtails the meaning of 'majority rule' in a way that feels undemocratic.
> 
> 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?"

Just FYI, in the U.S. there have been over 500 RCV elections that FairVote has tracked and analyzed.  I read from someone else that now the number is more than 800, I just haven't yet seen a good consistent searchable database online.  Maybe one exists, I just haven't seen it yet.

Of the 500 RCV elections over 300 had two or fewer candidates, so RCV does nothing at all in those elections differently than FPTP.  Of the 200 that had three or more candidates, more than half had one of the candidates with more than 50% of the vote in the first round, so again RCV does nothing different.  Of the fewer than 100 left that had more than one round, all but 26 had elected the candidate with the most first-choice votes, so again, RCV did nothing different than FPTP.  The remaining 26 had "come-from-behind" winners in which the RCV winner was the candidate in second place in the semifinal round, and that's when RCV did something different than FPTP.

There were four RCV elections where the RCV winner was not the Condorcet winner.  Two of the four had no Condorcet winner (they had a Rock-Paper-Scissors cycle, a Smith set of size 3).  Those were Minneapolis City Council Ward 2 in 2021 and Oakland School Board District 4 in 2022.  That leaves two RCV elections that *had* a Condorcet winner, but didn't elect the Condorcet winner: Burlington mayoral 2009 and Alaska at-large Representative to Congress in a Special Election in August 2022.  Both of those elections were followed immediately by a serious repeal effort.  In Burlington the repeal succeeded for 12 years (RCV has now returned to Burlington) and in Alaska the repeal effort failed by 0.2% margin when there was a 100-to-1 campaign spending ratio ($15 million spent to defeat repeal and $120,000 to support repeal of RCV).  They are launching another petition drive to put repeal on the ballot again in 2026.  When the CW is not elected, it's a close 3-way race and when it's not a cycle, it's the Center Squeeze effect.

I wrote a paper about the the Burlington election where I simply replaced the hypotheticals of what happens when the CW is not elected with real names of real people and real numbers.  It's not a particularly theoretical paper, more political and advocacy, but I try to lay down the principles and maybe channel the thoughts of Condorcet in the paper.  And I happen to live in Burlington and voted in that election in 2009.  Nic Tideman thought the paper was good enough for publication: https://link.springer.com/journal/10602/volumes-and-issues/34-3 but the edited published version gutted some content (mostly a table and figures with color), so I think the submitted version is better and I can share that to anyone without copyright violation: https://drive.google.com/file/d/1jIhFQfEoxSdyRz5SqEjZotbVDx4xshwM/view .

> 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 cannotmake 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?
> 
> 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.
> 

L8r,

robert

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r b-j . _ . _ . _ . _ rbj at audioimagination.com

"Imagination is more important than knowledge."

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