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Puzzling

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  • $\begingroup$ I like this reformulation of the answer to clarify things. In the context of this answer, my point from before is that $p$ will have some prior which is neither concentrated at 0 nor at 0.5 - note we do not need to know what the prior is just that it is not concentrated on a specific value, which seems fair since the intention is not specified. Then averaging over this prior gives A the highest probability of winning. Stef argued that you can't do this but I don't see why not. $\endgroup$
    – hexomino
    Commented Dec 24, 2021 at 22:11
  • $\begingroup$ I don't think we can do that. Consider the following statement: "This box contains a ball with probability $p$, otherwise it's empty." Can you say that it has non-zero probability of containing a ball, since if we put a prior over $p$, it's not concentrated on specific value 0? $\endgroup$
    – justhalf
    Commented Dec 25, 2021 at 4:16
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    $\begingroup$ Hm, yea, I guess our differences is whether we view the process that generates $p$ as part of the probability space or not. For me if $p$ is unspecified, then the process that had generated $p$ doesn't matter, since I consider it to already be in the past. If instead the question specifies how to generate $p$, then I would include that in the probability space. So we differ in when we put our baseline for probability. I start it after $p$ is generated, and you start it before $p$ is generated. So in the multiverse of all values of $p$, you are outside any of them, while I'm inside one. $\endgroup$
    – justhalf
    Commented Dec 27, 2021 at 17:40
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    $\begingroup$ Yea, so you're taking into account the puzzle writer's thought into the prior. As in, "there is a possibility that the puzzle writer intends $p$ to be non-zero, so we can say that there is a non-zero probability for B and C to not be perfectly correlated". However, I would say that this isn't valid here, since we are already given a question, and the puzzle writer already has a fixed intention. So if we answer from the PoV of the question, there is no more assigning probability to the prior of $p$. But if we answer from the PoV of external observer ... $\endgroup$
    – justhalf
    Commented Dec 28, 2021 at 4:10
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    $\begingroup$ Interesting discussion overall. I don't know how to resolve our differences. But to respond to your remark "and then not ask for an answer in terms of $p$", I think usually if not mentioned, then you present the answer in terms of the unknowns. E.g., "I have some dogs and chickens, totaling to 10 animals. Do I own more than 30 animal leg?". Then your answer will be basically "well, that depends on how many dogs you have. I can make a formula that is based on the number of dogs $N$, but I can't do better than that." It's the same here, Stef's answer is saying "this is the formula based on $p$." $\endgroup$
    – justhalf
    Commented Dec 28, 2021 at 4:16