Posts made by cfrank

@toby-pereira apparently so, because they left. But honestly in terms of the purpose of a forum, that doesn’t really subtract from anything.

Anyway, this original post was made well before RFK Jr.’s (imo reluctant) alignment with Trump. At least one of RFK Jr.’s predictions was correct, namely that Biden and/or Harris would not beat Trump. His “no spoiler” pledge would have given beating Trump the greatest possible chance, but Democrats refused to cooperate because they are power hungry, greedy, and benefit too much from the duopoly to concede to a third party candidate, even at the cost of Trump winning.

IMO, that’s primarily why RFK Jr. angled against them, in game theory terms it was as a punishment. It was a textbook failed prisoner’s dilemma, and they got a taste of their own medicine in a way that hurt everybody and could have been avoided. But I digress.

@Isocratia I mean, maybe. But if you bail from a conversation just because people are discussing ideas that don’t neatly align with your views, I think that kind of runs counter to productive discussion. Engagement is the whole point of a forum. Why not take the opportunity to make your case? On that point, I don’t think I was being dogmatic, I was just putting a moderate, measured perspective out here. In particular, that if a candidate has comparable support to what Nader did, he should also be on the debate stage.

As for Kennedy, I’m not sure if you followed his campaign directly, but from what I saw, his platform had some surprisingly rational moments. Whatever mistaken views he holds about healthcare, his core message was about dismantling corporate capture of government—which, let’s be honest, is exactly the route that’s brought us to the brink of fascism today. Frankly, he seemed more committed to stopping Trump than the Democrats did.

Like him or not, he was a third-party candidate who genuinely threatened to shake up the duopoly—something we haven’t really seen since Nader. And given how deeply dysfunctional the two-party system has become, that’s not nothing. The political landscape is a real-time disaster, and reform doesn’t just happen on its own. While it wasn’t his main agenda, one thing I appreciated about his run is that he was literally the only candidate to talk about ranked-choice voting and other technical fixes.

That said, I completely checked out when he aligned himself with Trump. At that point, his platform basically collapsed. His current sellout stance disillusioned a lot of his supporters—and honestly, he should’ve just bowed out once it was clear he couldn’t win.

If you see it differently, I’d be interested to hear why. That’s why I brought this topic up in the first place.

@toby-pereira yes definitely. I started trying to actually prove participation last evening, and it got much hairier than I would have liked... lots of branching edge-case conditions. I think an actual proof (or counterexample) of participation for this system would require some nice insights, and/or a larger scale planning and brute-force organized accounting of every relevant case.

For example, I’m quite certain the new participant V can never change the top sorted candidate to somebody they prefer less. So it would have to be an upset via the B2R survivor, who would have to become the new winner, and be preferred less than the old winner by V. But that situation gets complicated in terms of the sorting and the rank tie-breaking authority.

@toby-pereira yes it’s a bit particular, that’s the part that’s designed to preserve participation. The +1 advantage plus tie break conferred to the adversary is essentially to prevent any single vote from changing the outcome of the participation criterion-satisfying method. It still needs proof or more auditing and adjustment. But it was motivated empirically by finding examples of participation failure without introducing the advantage and some other aspects.

I think voters could have an anonymous ID given to them upon voting, it would have to be done with encryption. You’re right that in this case we would have to preclude latecomers, which I think would be fine. I think it could be done securely without an extra trip. This whole situation really makes “recounts” potential difficult though.

Having the “sincere” rank be attached to the original ballot might also be an option, but voters would somehow need to know that the second ballot would not be used in the first election, for instance. The only way they can know for sure is if they don’t provide it until after the first election winners are revealed. That could also be done with encryption.

In terms of preserving participation, the final runoff may not even be necessary. I’m trying to combine two things that can be looked at separately.

Also thanks for reading and your thoughts! I’m starting to wonder about how to guarantee the Condorcet loser criterion while still preserving participation. As of now though I think the method is essentially approval but with significantly stronger Condorcet-like guarantees.

EDIT: I just learned that there is an impossibility theorem about participation, independence of clones, and Condorcet loser. My guess is that the context and proof are similar to Arrow’s theorem, but the details I don’t know.

DISCLAIMER: This is conjectural and needs adjustments, but at least participation compliance can be improved. I have had LLMs run thousands of examples of this iteration, and have found no examples of participation failure. Still, it needs to be audited closely.

To understand this method, you should also know about Bottom Two Runoff (B2R), which you can learn about here for example:

https://www.votingtheory.org/forum/topic/564/bottom-n-and-bottom-2-runoffs-are-equivalent

The method is a development of this concept:

https://www.votingtheory.org/forum/topic/563/direct-independent-condorcet-validation/20

For those perhaps not familiar with some voting theory, there are often desirable criteria of a voting system that are, unfortunately, mathematically impossible to reconcile with each other. An important example of this is the contention between the Condorcet criterion and the participation criterion. It turns out that satisfying the Condorcet criterion in every case guarantees some instances where participation fails, and vice versa.

Consider the following method, which I’m just calling Approval vs. B2R, similar to Approval-seeded Llull as per @Jack-Waugh ---it's a bit technical, but there's a reason for that, so bear with me:

DEFINITION

(1) Voters submit rank ballots with approval cutoffs.
(2) Candidates are sorted by approval rates, with rank-based head-to-head breaking ties if possible—by this, I mean that within a tied score group, we recursively pull Condorcet winners and losers to the top and bottom of the ranking until none remain, and then defer to an authoritative and consistent rank-order over the candidates to break Condorcet cycles and ties in the final sorting.
(3) According to this sorting order, run B2R and identify the B2R survivor, with head-to-head ties broken by the sorting order.
(4) Consider the top-sorted candidate. If this candidate is the same as the B2R survivor, elect them.
(5) Otherwise, the top-sorted candidate will be an adversary to the B2R survivor. Run an independent head-to-head election between the B2R survivor and their adversary, with the following caveats:

--> Voters are not tied down in any way to their original preference between the B2R survivor and the adversary, and can freely vote for either in the independent head-to-head. Also, barring logistical prohibitions, voters who did not participate in the first round should fully be allowed to participate in the final round. By default, voters' original ballots will be used to determine the preference between the B2R survivor and their adversary, but voters may opt in to swap their preference either 0 or 1 times, whichever is necessary for their final ballot to confer an advantage that they wish to disclose.
--> However, based on these swaps, we can count the net number of swaps that are advantageous to the adversary over the B2R survivor compared with the original ballots. If this number is positive, the election proceeds as you would expect, with ties broken by the sort order. However, if the number is not positive, if the original head-to-head was in favor of the B2R survivor, and if a material difference would be incurred, then the adversary will be conferred an automatic +1 head-to-head advantage, and will also automatically win ties.

I conjecture that these caveats about the runoff—including setting aside the B2R survivor, increasing the adversary’s advantage by +1, and giving them ties—together restore the participation criterion under fully sincere ballots, at the expense of the full Condorcet criterion. However, when the margins are not close, the Condorcet criterion is still satisfied if ballots are sincere.

Here’s what else is fascinating about this: we can directly control the tradeoff between participation and Condorcet—in particular, when applicable, we can choose to increase the B2R adversary's margin over the B2R survivor by +1 with probability P. Then the method satisfies participation under sincere ballots with probability at least P, and simultaneously satisfies the Condorcet criterion under sincere ballots with probability at least 1-P.

IN SUMMARY:

This method I believe is participation compliant, which it is supposed to be by intentional design. This still needs to be proved. But as a consequence of this intention, it was also designed to fail Condorcet compliant in a controlled way. As per my comprehension, it will fail Condorcet compliance under these exact conditions:

(1) The Condorcet winner C exists (and will therefore be the B2R survivor);
(2) The approval winner Y is different from C;
(3) The head-to-head rank-based margin of C over Y is either 0 or +1; and
(4) The +1 boost to Y is applied, with ties going to Y.

More generally, it will fail the Smith criterion under these exact conditions:

(1) The approval winner Y is not in the Smith set;
(2) The head-to-head rank-based margin of the B2R survivor over Y is either 0 or +1; and
(3) The +1 boost to Y is applied, with ties going to Y.

In all other cases, it satisfies the Smith criterion. Thus in a sense, this method tries to be as close to Smith compliant as possible while enforcing participation compliance as non-negotiable. It unconditionally satisfies a weakened version of the Condorcet criterion: If the Condorcet winner exists and its weakest margin of victory is at least 2, then the Condorcet winner is elected. It also unconditionally satisfies a more significantly weakened version of the Smith criterion: If every member of the Smith set has a weakest margin of victory against non-Smith set members of at least 2, then the election winner will belong to the Smith set.

However, when the method fails the Smith criterion, it will necessarily fail the Condorcet loser criterion in the following circumstance:

(1) The Condorcet loser is the only available adversary to the B2R survivor, which is only possible if the Condorcet loser alone attains the maximum approval rate.
(2) The B2R survivor beats the Condorcet loser with a margin <=+1.

While this is unlikely to occur in realistic elections, it is possible, and pathological examples can be generated. Addressing this is hard, because independence of clones, participation, and the Condorcet loser criterion are a “cursed triangle,” where any two is incompatible with the third. In our case, we satisfy independence of clones and participation, so examples of failing the Condorcet loser criterion inevitably must exist.

Not that it's necessary to frame the properties in terms of Smith or Condorcet criterion-sounding conditions, but possibly those weak conditions can be strengthened. It is what it is.

Future work will be to prove participation and to refine the mechanisms for guaranteeing participation.

@jack-waugh but that’s always the tradeoff with score systems, which is why people bullet vote and/or min-max (translate to approval). I totally understand what you’re saying though.

I’m not saying the “Condorcet adversary” should be the score winner, just that they should be a strong alternative. Your approval-seeded Llull ballot could accommodate an approval winner as the Condorcet/Smith-method’s adversary, for example.

I think a “rank with approval cutoff” ballot makes sense. Then there could be the approval winner, and, say, the B2R (Smith compliant) winner, followed by an independent head-to-head.

@jack-waugh I see to an extent, but I would argue that your collapse of rankings is incompatible with the distinction by score. You prefer one to the other, even if they are both horrible, right?

@jack-waugh I’m not sure, I think coupling encourages consistency, which is a prerequisite to honesty. The coupled structure is also simpler and more efficient. The score winner is not necessarily the Condorcet winner for instance, and need not be the winner of a Condorcet compliant method when the Condorcet winner doesn’t exist (e.g. B2R).

The same validation logic also works in a homologous sense for Smith compliance. I think it’s less flexible if for instance two divergent Smith compliant methods were pitted against each other in the absence of a Condorcet winner. In that instance though, the Condorcet winner could not be tested with the final head-to-head. I’m not sure about extending to particular special subsets of the Smith set in the same way, depending on the subset.

@toby-pereira I would say they are declared the winner if the methods coincide. Otherwise there may be potential clone issues. It’s also simpler that way. Although if we did have an unambiguous runner up, that would be preferred to validate any potential Condorcet winner status. For example, the second highest scorer.

But if we wanted the decision process to be consistent, that has the potential to cause a recursion of successive head to heads, say, if the two methods repeatedly coincide. Unlikely but still. That would probably be the “right” way to do it in this context.

@jack-waugh exactly, that's the kind of thing I'm considering. Now I'm kind of puzzled though about whether a head-to-head between two winners of clone-independent methods can fail independence of clones. I actually don't think it's possible that the composite system fails independence of clones. FYI my preference for a Condorcet method would be Tideman's Bottom 2 Runoff (B2R), which I'm now certain is equivalent to BNR. For example, we could even pit the Approval winner against the B2R winner head-to-head. That would be a form of Approval-seeded Llull with a final independent head-to-head.

@jack-waugh I agree it isn’t too complex to understand, it could also be made more convenient by setting defaults. I’m actually more concerned with failing independence of clones. But I’m not sure—is it even true that this kind of runoff actually would fail independence of clones? Only the Condorcet and score winners could be pitted against each other, and those are both (roughly) clone independent methods.

I think a real second runoff would only matter in cases where voters were employing tactics. Otherwise, probably we should elect the Condorcet winner (IMO).

@jack-waugh I think I may have proved that the recursive bottom N runoff and ordinary bottom 2 runoff are equivalent: https://www.votingtheory.org/forum/topic/564/bottom-n-and-bottom-2-runoffs-are-equivalent

So actually it may make no difference, which would nice to know.

But I think maybe the concept of a second independent runoff is too complicated… I just got a bit hung up thinking about how the Condorcet winner as computed might not actually be the “real” Condorcet winner.

Bottom 2 Runoff (B2R) is a Smith-compliant cardinal-Condorcet method. We can describe it generally by considering two independent "score" and "head-to-head" functions over a candidate set.

To be precise, if Z+ is the set of natural numbers, let s:Z+ to R ("score") and hth:Z+xZ+ to R (“head-to-head”) be fixed real-valued functions, where s is injective, and where hth(y,x)=-hth(x,y) and hth(x,y)=0 if and only if x=y (“no ties”). In our case we consider a set C of “candidates” to be a finite subset of Z+.

Let B2R(C) denote the “Bottom 2 Runoff” winner in C, determined by repeatedly pitting the bottom two scoring candidates against each other in a head to head, and eliminating the loser until one candidate (the winner) remains.

Also, let RCLE(C) denote the “Recursive Condorcet Loser Elimination” winner in C, determined by repeatedly looking at the set of N lowest-scoring candidates, starting with the full set C, and eliminating any Condorcet loser among them (if none, update N↦N−1 and retry, resetting to the full remaining set with each elimination), until one candidate (the winner) remains.

Both B2R and RCLE are Condorcet compliant and Condorcet loser compliant methods. It also turns out that for fixed score and head-to-head functions with no ties, they are equivalent.

PROOF:

We can proceed by induction on the size of C. With 1 or 2 candidates, the equivalence is trivial. So we consider |C|>=3.

Without loss of generality, we can assume that there is no Condorcet winner in C—with a Condorcet winner, B2R automatically coincides with RCLE, since both methods are Condorcet compliant.

We can also assume that there is no Condorcet loser in C, because that Condorcet loser will immediately be eliminated in RCLE, and will have no material effect on B2R. More formally, if L is the Condorcet loser in C, then B2R(C)= B2R(C omitting L), and RCLE(C) = RCLE(C omitting L).

Now, consider the first candidate eliminated by each method, and call them B2R1 and RCLE1, respectively. If B2R1=RCLE1, then we will be done by the inductive hypothesis. Thus, we have to examine the case where B2R1 differs from RCLE1. In that case, the only option is that RCLE1 is the Condorcet loser in the set of bottom-N scorers for some N >= 3. That is, RCLE1 loses head-to-head against every other candidate in that bottom-N scoring set.

In particular, RCLE1 loses head-to-head against every candidate whose score is lower than its own score. This means that in B2R, no matter which of the candidates whose scores are lower than RCLE1's emerges to face RCLE1, that candidate is guaranteed to beat out RCLE1 head-to-head. Therefore, the presence of RCLE1 in the candidate pool is immaterial to the outcome of B2R.

Formally, it follows that B2R(C)=B2R(C omitting RCLE1). But we also know that by definition, RCLE(C)=RCLE(C omitting RCLE1) as well. By the inductive hypothesis, B2R(C omitting RCLE1) = RCLE(C omitting RCLE1), and this implies that B2R(C)=RCLE(C).

QED

This is true no matter which scoring or head-to-head functions are used, as long as they are static functions and no ties are involved.

This means that, while RCLE superficially appears to be more conservative than B2R, this is actually not the case. B2R is always just as conservative as RCLE, since they are equivalent. B2R is actually just a more efficient implementation of RCLE.

@jack-waugh hm I quite like this in theory. I’m also recollecting that multi round systems tend to fail independence of clones. In principle, approval-seeded Llull could be made more robust by considering larger sets of the lowest-approved candidates and iteratively removing Condorcet losers.

@jack-waugh can you link me to that? I think the only way to address this concern fully would be to do a second, final, independent vote between two candidates, one of whom is the computed Condorcet winner if possible. The rationale is that this is the only possible way to actually test the status of the computed Condorcet winner. Unfortunately it isn’t a complete test, in that it only has a chance of refuting the computed Condorcet winner’s status and cannot confirm it, but at least it’s a test.

If the computed Condorcet winner wins the runoff as expected, that’s evidence supporting their computed status. Otherwise, if the computed Condorcet winner loses the runoff, that’s irrefutable evidence that they were not actually the honest Condorcet winner. For example, if this kind of head-to-head had been performed in Alaska 2022, it’s possible that the IRV winner could have actually beaten out the purported Condorcet winner head-to-head, and that would dismantle all arguments as to his status.

It doesn’t have to be score or approval per se either, any non-Condorcet method would work. More generally, we could have a Condorcet compliant method pitted against a Condorcet non-compliant method to select two front runners, and then do a second independent head-to-head vote.

Furthermore, secondary voting could be on a totally opt-in basis, where voters can choose to keep their preference as stated on their original ballot by default. That way, honest voters will never actually need to worry about the second round. Voters would be disallowed from revoking their ballots altogether, preventing strategic abstention.

A final interesting note is that any voters who choose to change their ballot essentially admit to having voted tactically.

This is just an illustration of a concept and nothing new perhaps.

If we have a score system, for example, we can compute both the score winner and the Condorcet winner (if one exists). In principle, the Condorcet winner should win in a head-to-head match up against any other candidate.

However, the Condorcet winner may have been computed from ballots that included tactical positions such as burial. At the same time, the score winner may have won due to tactics such as bullet voting.

By pitting these two winners against each other in a secondary majority vote, we should in principle have a more robust validation of the Condorcet winner’s position, or an unarguable refutation of it. In fact, I think a secondary majority vote of this kind has the potential to strongly validate and legitimize the outcome of an election, as well as to give us information about the presence of tactical voting.

I wonder what people think about this. I think it’s overlooked that the Condorcet winner by computation may not actually be the Condorcet winner in spirit.

In fact, the only honest validation we can have of the Condorcet winner is if we actually pit that winner against another strong candidate in a final decision election. So even if the score winner is the Condorcet winner, then if we want “steel man” validation, we should STILL have a direct, independent head-to-head runoff with some kind of runner up.

@kodos hm interesting, even as a moderator I wasn’t cognizant of reputation requirements for posting comments, but that would make sense. You’re right that it would be helpful for that information to be transparent. Possibly posting this and introducing yourself will work, I’ll give an upvote and possibly check out what else can be done.

Do you remember which thread was of interest to you?