Look, you can’t complain about this after giving us so many scenarios involving N locked chests and M unlabeled keys.

https://explainxkcd.com/3015/

  • NerdyPopRocks@lemmy.world
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    17 days ago

    Everyone out here using a dice calculator, but can someone answer

    1. Given a value x in (0,1) is there a set of dice and AC such that sum(roll{n_iDv_i}) > AC has probability x?
    2. Is there a sequence of finite set of dice and sequence of some AC for which we can get arbitrarily close to any probability?
    • hangonasecond@lemmy.world
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      17 days ago

      irrational numbers will always be out of reach for a finite set of dice. I think if you restrict the set to the rationals, you will still run into trouble because there’s a finite number of dice and an infinite number of primes, so there will always be a big enough prime whose value you will be unable to get on the denominator. E.g. if you restrict to only a d2 and d3, you can’t get a denominator of 5 for your probability. So add a d5. Now you can’t get a denominator of 7, and so forth.

      But all the primes would suffice especially as you’ve excluded 1 in the set. Otherwise, include a D1 and you’re golden.

      To be extra clear, if you have an infinite set of dice, one for each prime, then you can attain a given probability using a finite subset of those dice. If you allow for the use of infinite dice and infinite rolls, my intuition says you can get the whole interval but let’s think about it.

      It’s true that every real can be expressed as a convergent sequence of rationals, and that between any two real numbers there is at least one rational number. You can use this to construct a sequence of rationals that approaches the real numbers we want in the interval, and because we have all the prime dice and I have (not rigorously) proved this is enough to get any rational, we can roll any probability in our sequence. So we can get as close as we like to the real number.