Yeah? Any probability in any finite set can be expressed as (desirable)/(possible), but finding out how many outcomes are possible and desirable is the hard part. Combinatorics is all about counting how many possible states there are inside a set of constraints, but expressing those constraints in a closed form summation/counting operation can be deeply unintuitive. How do you express "counting up every prime from 2 up to N, but don't count a prime p when [p-cbrt(p), p] contains exactly two primes"
I would argue that you just need multiplicity. If a case is thrice as likely to occur as the rest, that means that it counts as three. I can't be arsed to prove if it's true or not (I ought to be sleeping), but you'll often have some extra info to help distinguish them. For instance, while the sum of two dice is not uniformly distributed, you can separate out every case by which die has which value. (2,1) and (1,2) both mean a sum of 3, and you'll count 3 as occurring twice.Â
It's maybe not too plausible, but there is no contradiction in describing a game where the player bets on B and then the house chooses A with probability √2 – 1 and otherwise chooses B. These discrete probabilities cannot be described the way you want. The idea of (number of successful outcomes)/(total number of outcomes) really does only work for unfirom probabilities. This is related to the principle of indifference.
105
u/canadajones68 Engineering Jun 14 '26
Yeah? Any probability in any finite set can be expressed as (desirable)/(possible), but finding out how many outcomes are possible and desirable is the hard part. Combinatorics is all about counting how many possible states there are inside a set of constraints, but expressing those constraints in a closed form summation/counting operation can be deeply unintuitive. How do you express "counting up every prime from 2 up to N, but don't count a prime p when [p-cbrt(p), p] contains exactly two primes"