Comment by AnthonyMouse

Comment by AnthonyMouse 7 months ago

12 replies

View on Hacker News

> It is never correct to assign a score between the minimum and the maximum, so why allow it in the first place?

Because it is often correct.

Suppose there are candidates A, B and C. Candidates A and B are each polling around 6/10 and candidate C is polling around 4/10, but candidates A and B are quite similar to each other and share a base of support. According to your strategy, A and B are the two most likely to win, so if you prefer A then even though you still prefer B to C you refuse to express your preference and instead assign 10/10 to A and 1/10 to B and C. The voters who prefer candidate B do the same. The result is that A and B end each up at 3.5/10, C ends up at 4/10 and C wins. In other words, you've devolved back into first past the post and caused your least favored candidate to win because of your erroneous strategizing.

By contrast, if you assign 10/10 to A, 5/10 to B and 1/10 to C, you've given A a significant advantage over B without assigning B such a low score that you could deliver the election to C if C defeats A.

amalcon 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. This is also a mistake, and it is a mistake no matter the voting system or the next step in your strategy.

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

> you've devolved back into first past the post

Correct. The potential for this to happen is one of the drawbacks of rated voting systems. It's also, through a different mechanism, one of the drawbacks of ranked systems. It doesn't mean we shouldn't try, since both alternatives give some ability to hedge against incorrect assessments.

> By contrast, if you assign 10/10 to A, 5/10 to B and 1/10 to C, you've given A a significant advantage over B without assigning B such a low score that you could deliver the election to C if C defeats A.

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. Both give the same odds of winning to A and B (well, mine gives B slightly higher odds because its average is 5.5 -- but you get the point). The only difference is that your method outsources the randomness to the rest of the electorate, rather than generating it yourself.

Reply View 10 replies
  • 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.

    Reply View 9 replies
    • 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.

      Reply View 7 replies
      • 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

        Reply View 0 replies
      • 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...

        Reply View 5 replies