Forum Council

@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, that’s kind of on you. Engagement is the whole point of a forum. Why not take the opportunity to make your case? I don’t think I was being dogmatic, I was just putting a perspective out here.

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.

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.

@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 by V to the old winner. But that situation gets complicated in terms of the sorting and the rank tie-breaking authority.

There is an online obituary if you want to read it here.

@cfrank While I'm not an expert in how to make methods pass particular criteria, participation seems to be a very hard one to get. Most of the methods that pass it seem to be simple adding up ones (e.g. FPTP, Borda, approval, score), although Descending Solid Coalitions and Descending Acquiescing Coalitions are slightly weird methods that do pass it apparently.

@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.

@cfrank said in Fixing Participation Failure in “Approval vs B2R”:

(6) Run a secondary, independent head-to-head election between the B2R winner and their adversary, with the following caveats:
--> Voters are not tied down in any way to their original preference between the B2R winner and the adversary, and can freely vote for either in the independent head-to-head. Also, voters who did not participate in the first round are fully allowed to participate in the final round. By default, voters' original ballots will be used to determine the preference, but voters may opt in to swap their rating either 0 or 1 times, whichever amount is necessary to indicate 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 winner 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 find this part a bit hard to understand.

Also, if it's an independent head-to-head, do you mean a separate trip to the polling station, or just a separate part of the ballot paper? If it's a separate trip, then it would be impossible to manage the swaps and each voter's default position without losing anonymity.

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.

@kodos I've changed it so new users have to register with an e-mail address. I don't think that's too onerous, and it should make the problem go away.