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by kubanczyk 1544 days ago
If you want to use the concept of "probability" to deal with reality, you better decide first if the reality offers infinite payoff/risk. If it does, your tool (the naive probability) is inadequate and quite easily broken, as you just have demonstrated.

The tool will sway to 100% and back to 0% on like a broken compass, depending on which infinities you thought about this very minute.
1 comments
3np 1544 days ago
Good point - it should follow that given assumption that with infinitesimal probability of infinite payoff, there is also a corresponding non-zero probability of infinite risk.

Overall I think OP is mostly about them struggling to reduce hypothetical outcomes involving infinities to real numbers in order to be able to rank them. For example:

> Agent-neutrality: If there is a welfare-preserving bijection from the agents in w1 to the agents in w2, then w1 and w2 are equally good. > By “welfare-preserving bijection,” I mean a mapping that pairs each agent in w1 with a single agent in w2, and each agent in w2 with a single agent in w1, such that both members of each pair have the same welfare level.

They then proceed to compare a w1 with a 1/0 of 1:1 vs w2 with a 1:3 ratio, "skipping" 2/3 of 0s in w2, claiming a contradiction. So they're really doing a injective non-surjective mapping, not a bijective one, and them being infinite doesn't really absolve that. I think they could get some insight from algorithmic analysis and limits. Just like (probably sloppy notation here) O(0.25n) is faster than O(0.5n) even if both are O(0.5n).

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penteract 1544 days ago
> They then proceed to compare a w1 with a 1/0 of 1:1 vs w2 with a 1:3 ratio, "skipping" 2/3 of 0s in w2, claiming a contradiction. So they're really doing a injective non-surjective mapping, not a bijective one.

There is in fact a bijection between indices that maps the sequences to each other -

  index: 0 1 2 3 4 5 6 7 ...
  w1:    1 0 1 0 1 0 1 0 ...
  w2:    1 0 0 0 1 0 0 0 ...
bijective map such that w2(f(i))=w1(i) :

  f(n) = if (n is even) then n*2 else floor((n*2+1)/3)
  f_inverse(n) = if (n is a multiple of 4) then n/2 else floor((n*3)/4)*2+1
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