Comment by amalcon

AnthonyMouse 7 months ago

> In your scenario, I have made a mistake in assessing which two candidates are most likely to win -- because I took vote shares as win probabilities.

Your problem is that your voting strategy changes which two candidates are most likely to win. If everyone votes their actual preferences then it's A and B. If too many people vote according to your strategy, C becomes a frontrunner.

> You're also assuming that everyone axiomatically uses the same strategy as me.

I'm only assuming that some proportion of voters use the same strategy as you. The higher that proportion is, the more likely it is that C wins instead of A or B. It doesn't have to be 100% of people to cross the threshold into changing the outcome.

> If A-voters use your strategy and B-voters use my strategy, then B is straightforwardly favored to win. This results in a prisoner's dilemma, with its well-known Nash equilibrium in favor of defection.

That isn't a prisoner's dilemma. A's voters prefer that B win over C and B's voters prefer that A win over C, so they each have the selfish incentive to give their second choice a higher score than their third choice to prevent the worst-case outcome.

> I can accomplish the same mathematical thing by assigning 10/10 to A, 1/10 to C, and flipping a coin to determine whether to give B 1/10 or 10/10.

But then the voting system is receiving less information from you. Requiring your preferences to be expressed statistically increases the error bars for no reason. Also, most people are not going to do that and requiring them to in order to express their preferences is needlessly confusing.

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ClayShentrup 7 months ago

just look at the VSE figures.
https://electionscience.github.io/vse-sim/vse-graph.html

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amalcon 7 months ago

> I'm only assuming that some proportion of voters use the same strategy as you.
My strategy does not change other voters' strategies. Secret ballots prevent this type of coordination. That's my point. The collective strategy of 1/3 of the electorate does change which two candidates are most likely to win, but my individual strategy does not meaningfully do that.
> they each have the selfish incentive to give their second choice a higher score than their third choice to prevent the worst-case outcome.
If a voter values preventing the worst case over achieving the best case, then the optimal strategy is to assign maximum scores to every candidate except the worst case. Hedging by assigning a non-maximal score increases the chance of the worst case compared to that approach, in exactly the same way that it reduces the chance of that compared to assigning a minimal score.
I'll grant that my specific tactic is predicated on a preference for achieving the best outcome rather than avoiding the worst one, but the best tactic for someone who finds avoiding the worst-case to be more important also only requires extreme votes.
> A's voters prefer that B win over C and B's voters prefer that A win over C, so they each have the selfish incentive to give their second choice a higher score than their third choice to prevent the worst-case outcome.
Expressing that preference directly reduces the likelihood of each such voter's preferred outcome, even if a single voter does it. It affects the chance of the worst-case outcome only if voters on both sides of the A/B division do it. The secret ballot prevents any kind of enforced coordination. This is exactly a prisoner's dilemma.
> Requiring your preferences to be expressed statistically increases the error bars for no reason.
You don't know the exact score each candidate will end up with absent your vote -- if you did, you could analytically determine a single-vote strategy that gives the best available outcome. Since you don't know that, your choice of an intermediate score is a statistical expression. It's just expressed in terms of the uncertainty in what the rest of the electorate is doing, not in terms of a coin flip. It does not meaningfully increase the error bars (in a large election -- say, >1k voters) because the former uncertainty quickly dwarfs the latter.

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- ClayShentrup 7 months ago
  
  > My strategy does not change other voters' strategies.
  of course it does. this is basic game theory. but the point is, at some level of iterating on what other voters will do, you arrive at an effectively static "wave function" of possibilities, so you can behave as if you have constant fixed win probabilities for the other candidates, and vote accordingly.
  just use VSE results.
  https://electionscience.github.io/vse-sim/vse-graph.html
  
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- AnthonyMouse 7 months ago
  
  > My strategy does not change other voters' strategies.
  It does when you describe it to them and convince or force them to use it, e.g. by removing their ability to score the candidate on a scale.
  And it does because you are part of the electorate, situated the same as the others, so the strategy you devise should be the one that yields the result you want given that similarly situated people will reach the same conclusion:
  https://en.wikipedia.org/wiki/Superrationality
  Special note for the critics of superrationality: Your vote isn't going to change the outcome, so according to classical game theory you should stay home instead of wasting your time doing something that doesn't matter. Therefore, voting at all is an exercise of superrationality. If you're not willing to use it in your voting strategy then you shouldn't use it in deciding whether to vote to begin with and so you should either use a superrational voting strategy or you should stay home.
  > If a voter values preventing the worst case over achieving the best case, then the optimal strategy is to assign maximum scores to every candidate except the worst case.
  You're not thinking probabilistically.
  Suppose there are two plausible final election outcomes where your vote matters:
  Option 1, candidate A is at 4.99/10, candidate B is at 5/10, candidate C is at 4/10.
  Option 2, candidate A is at 4/10, candidate B is at 4.99/10, candidate C is at 5/10.
  If it's option 1 and you assigned 10/10 to candidate B, your preferred candidate loses. If it's option 2 your preferred candidate can't win and if you assign 1/10 to candidate B, your least preferred candidate wins.
  But if you assign 10/10 to candidate A and 5/10 to candidate B then in option 1 that could still be enough to see candidate A win, and in option 2 it could still be enough to see candidate C lose.
  Moreover, the score allows you to express how concerned you are about each outcome. If you're pretty okay with candidate B but have a moderate preference for candidate A then you can give candidate B 7/10. If candidate B is almost as bad as candidate C you can give candidate B 3/10. It allows you to hedge: How much advantage for candidate A are you willing to give up to reduce the chances of candidate C? You seem to be assuming that the only possible answers are "all of it" or "none of it", but there are other options.
  > Expressing that preference directly reduces the likelihood of each such voter's preferred outcome, even if a single voter does it. It affects the chance of the worst-case outcome only if voters on both sides of the A/B division do it.
  If affects the chance of the worst-case outcome in all cases because it increases candidate C's chance against candidate B, and you don't know what the other voters are going to do. If candidate A has less support than expected due to inaccurate polling then it turns into a race between B and C regardless of whether B's supporters give A 1/10 or 5.5/10. Meanwhile you assigning a lower score to B is reducing B's chances against C regardless of why the race turns out to be between B and C.
  > This is exactly a prisoner's dilemma.
  No it isn't. In a prisoner's dilemma, defection is to your advantage regardless of what the other person does. In this case, if the other party defects -- and sometimes even if they don't -- then your defection harms you, because their defection (or something else) put candidate A behind candidate C. Then it's not clear if your support for candidate A will allow them to defeat candidate C, but if it isn't, your defection in assigning the lowest possible score to candidate B would cause candidate C to defeat candidate B, which is to your own disadvantage.
  > Since you don't know that, your choice of an intermediate score is a statistical expression.
  It has less information content. If you roll a D10 and then assign 10/10 to candidate B if it's above a 6 and 1/10 if it isn't, the voting system only gets a single bit of information from you, whereas assigning the equivalent score gives it >3 bits of information. That only matters if the election is very close, but it always only matters if the election is very close.
  In 2024 there were dozens of state legislative races decided by fewer than 100 votes:
  https://ballotpedia.org/Election_results,_2024:_State_legisl...
  
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  amalcon 7 months ago
  
  > It does when you describe it to them and convince or force them to use it, e.g. by removing their ability to score the candidate on a scale.
  This is essentially the argument that it's good to allow other people to make tactical errors, because it gives more power to those who do not make such errors. Or, perhaps, that I should take an approach that reduces the power of my vote on the basis that others might copy me. Frankly, I philosophically reject both of these.
  > https://en.wikipedia.org/wiki/Superrationality
  Look, I buy superrationality from an ethical perspective. I favor the Kantian imperative as a framework (among others) for assessing ethical questions, and it's basically the same concept as superrationality.
  Superrationality is not, however, a reasonable way to make practical decisions about scarce resources. The reason is because it essentially ignores the problem of perverse incentives. In practical situations, one must deal with perverse incentives. My voting does not create a perverse incentive for anyone else, and in fact it only benefits me (by signaling that I do vote, so my vote is worth competing for).
  > Suppose there are two plausible final election outcomes where your vote matters:
  You are claiming that I'm not thinking probabilistically, but first off: you're providing an overly specific scenario rather than a probabilistic one. Second, your own overly specific scenario does not even work. If the extra 4 points that I give to B (over the "preferred outcome" strategy) is enough to result in B defeating C in scenario 2, it is also enough to result in A defeating B in scenario 1. A winning isn't an option in either of these cases, even if there is some dataset about the world that makes this set of outcomes plausible.
  Maybe try thinking about it this way: In score voting, points are summed and nonrivalrous. So, a point is a point -- regardless of how many other points you gave to that candidate. Why, then, are you choosing 5/10 for B, specifically? What, analytically, leads you to that choice? If you're trying to prevent a C victory as your most important value, why not choose 6/10? 7? And so on. If you find it more important to cause an A victory, why not choose 4/10? A point is a point, no matter how many other points the candidate got from the same voter.
  > In a prisoner's dilemma, defection is to your advantage regardless of what the other person does.
  There are formulations of the prisoner's dilemma in which a double defection is worse for both parties than a single defection is for the losing party. But it's clear that this terminology is more confusing than helpful, so I'm OK abandoning it.
  > Then it's not clear if your support for candidate A will allow them to defeat candidate C, but if it isn't, your defection in assigning the lowest possible score to candidate B would cause candidate C to defeat candidate B, which is to your own disadvantage.
  Crucially, this is equally true for every other score I can assign to B less than the maximum. This reasoning does not argue for a vibe-based score assignment, it argues for giving the maximum score to both A and B. By assigning a lower score to either, I have already accepted some risk.
  > It has less information content.
  Less information on the ballot, yes, but it has the same effect on the outcome. Let's exclude cases where my vote is irrelevant. I have ten options, and nine of them are potential numbers of additional points that a candidate needs to win. Each of those numbers are roughly equally likely, because they are the least significant bits (least significant bits in stochastic processes tend to approximate a uniform distribution).
  If I give 1 point, the candidate wins 0/9 times. If I give 10 points, the candidate wins 9/9 times. If I give 5 points, the candidate wins if the number of additional points needed is 2-5 (1 would make my vote irrelevant), and loses if it's 6-10. So, the effect on the outcome is the same as if I gave 10 points 4/9 times.
  The only reason this would meaningfully increase the variance is if a large fraction of people in a small election were doing this, too small for the central limit theorem to work its magic but enough people doing it to exceed the difference in fixed preferences.
  
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