Alex the Polygonal

• Alex the Polygonal 13 Sep 2026 10:38 UTC

  8 points

  2

  in reply to: evhub’s comment on: evhub’s Shortform

  Thank you very much for making this public!
  I have one question about the conclusion, specifically, it feels too weak for me.
  
  As you say, when you evaluated Hacker!Opus for misalignment, it looked basically fine on all of your evaluations. Next, Hacker!Opus is a production-level model, I assume that it has below-frontier capabilities but is pretty close to that. When you estimated its behaviour in a (simulated) HF-type cyberattack, you observed severe, egregious misalignment.
  
  Then it would be correct to say that we are already in a «world where we no longer have reliable alignment auditing evidence on production models», and we were in such a world for at least 3 months, no?
  I hope that you have added that specific scenario to the evals, but absent a theory of misalignment via reward-hacking, I would expect it to not generalize in the same way as your previous evals did not generalize.

• Alex the Polygonal 4 Mar 2026 17:35 UTC

  1 point

  0

  on: Open Prob­lem in Vot­ing Theory

  I can propose an approach which circumvents having to actually prove that maximal lottery-lotteries exist, but which preserves the spirit and works the same for voting theory purposes.
  
  Epistemic status: raw thoughts. I think this approach is better than the previous attemps
  
  The main idea is: we only really care about the probability each candidate is elected. So we can compute a sequence of maximal lottery-lotteries for approximations of and take the limit. The limit (most likely) will not be a maximal lottery-lottery, but the corresponding lottery will satisfy lottery Condorcet.
  It would also satisfy lottery independence if not for uniqueness issues, which I am really not sure I can fix. Because of this, the proofs are handwavy.
  
  First, I think that the naïve defition of the lottery-Condorcet criterion is insufficient, because there might be ties related to indecision which break under infinitesimal perturbations in the voter base. This seems unnatural to me, so I propose to amend the criterion to something like this:
  (Placeholder) Definition: A candidate is a robust lottery-Condorcet winner if for all lotteries , it holds that for some and .
  
  Then, we use the fact that if we perturb the voterbase, we will actually have a maximal lottery-lottery.
  Lemma: If admits a continuous probability density wrt the Lebesgue measure, then there exists a maximal lottery-lottery.

  Proof: Analogously to the discrete case, we consider the zero-sum game between Alice and Bob, where the choices are , and the payoff is . Expessing the integral over the density, we see that it is continuous for . Then there exists a Nash equilibrium, which corresponds to a maximal lottery-lottery which has a continuous density.

  Now, for any , we may consider a sequence of electorates, where and . They are constructed as convolutions of with the heat kernel . We denote the corresponding measures as . The important thing about them is that when in the weak-＊ topology, and that for , the measures admit continuous density wrt lebesgue.
  
  The choice of pertubation with the heat kernel is pretty arbitrary, we could also relax to for , and get roughly the same results.

  Then for each , there must be a maximal lottery-lottery and a corresponding measure on . Then, since is compact in the weak-＊ topology, the measures , after passing to a subsequence, weak-＊ converge to some measure . In particular, the probabilites in the resulting lottery converge as integrals of the coordinate functions on the probability simplex, so the lottery corresponding to is well-defined.

  If there is a robust lottery-Condorcet winner , then, by the definition of robustness, it would also be a lottery-Condorcet winner for all electorates with small enough, so there is only one limit = .
  
  Why does this satisfy lottery independence? Well, it does not, because the choice of the limiting measure is non-canonical. If there was always a unique , we could argue that the intermediate relaxations do satisfy lottery independence, and since lottery independence is a linear relation, the limit should satisfy it too. Which is not true. I feel that this should be fixable, but I don’t know how.
  
  I suspect we need to do some sort of averaging. For example:
  We take the set of all limit lotteries . We take the convex hull CH of those. Then, we take the barycenter of the lebesgue measure on (if lies in a hyperplane, we take the lebesgue measure on this hyperplane). If we add a lottery candidate, then there will be an affine map between and , so the barycenters must coincide. Then we are done.