4 ms·

Strictly speaking our own existence doesn't even tell us the probability is greater than zero. For an event to have probability zero doesn't imply it can't occu

Strictly speaking our own existence doesn't even tell us the probability is greater than zero. For an event to have probability zero doesn't imply it can't occur. If a number is chosen from a uniform distribution on the reals between zero and one, whatever the result is the probability of that exact result occurring was zero.

dotancohen 5y ago
> For an event to have probability zero doesn't imply it can't occur. The distinction in meaningless. We exist, ergo intelligent life can develop in this universe.
- danbruc 5y ago
  It is correct, if you split 100% across infinitely many possible outcomes, each outcome will have probability zero, still one of the possible outcomes will occur.
  - rsj_hn 5y ago
    Doesn't this require a continuum? I don't think you get there with a measure on a countable set. That would make this an unphysical argument, IMO.
    - omnicognate 5y ago
      Yes, it requires uncountability. It could be that physics is ultimately best modelled with countable sets, but that hasn't been established and our current best physical theories are certainly full of uncountability. The distinction between "surely" and "almost surely" [1] is "just" a curiosity about probability theory though, albeit a rather fundamental one, and I only brought it up as such. It's interesting to think about, and if you do so it quickly brings you up against deep philosophical questions about what probability means. [1] https://en.m.wikipedia.org/wiki/Almost_surely https://en.m.wikipedia.org/wiki/Almost_surely
      - danbruc 5y ago
        Just out of curiosity, is there are short answer why it requires uncountability? Naively something like pick a random natural number would also seem to lead to probability zero. I can see that pick a random natural number might be problematic, how would you do this? Pick on digit and then with some probability either stop or continue and pick another digit, but it is at the very least not obvious that one could make this work without larger numbers just having smaller and smaller probabilities and there might also be issues with termination. On the other hand it is not obvious to me why one could not work with uniform distributions over the set [0, n) and then look at the limit as n goes to infinity.
        rsj_hn 5y ago
        If you have a measure on a countable set, lets number it 0, 1, 2, .. then you must have: m(i) >= 0 (since it's a measure). And must also have 1 = m(0) + m(1) + ... (because it's a measure) so 1 = Lim S(i) Where S(i) is the partial sum going from 0 to i. But if each m(i) = 0, then each partial sum is zero. So 1 = Lim 0 = 0
        danbruc 5y ago
        There is probably something wrong with this, but I was thinking something like the following. Take the sequence of sets M(n) = { 0, 1, 2, ... n - 1 } with measure m(n, i) = 1 / n. The m(n, i) are non-negative and the sum over all m(n, i) for a fixed n is 1. Then take the limit. The set M(n) will seemingly approach the natural numbers but I am not sure that this is valid. The m(n, i) will approach 0, I think that is uncontroversial. But I guess it might not be valid to argue that the sum remains 1 even though it seemingly equals n * 1 / n.
        rsj_hn 5y ago
        No, you can't interchange limits like that. In your case, when you say "sum" of the n identical things, you just mean multiplying n by the integer 1/n. So you have 1 = 1/n *n != lim(1/n)lim(n). The last is an indeterminate form of 0*infinity and so you don't get to conclude that it's one.
  - mensetmanusman 5y ago
    Infinity times zero might be meaningless in a quantized reality
- omnicognate 5y ago
  > The distinction in meaningless. We exist, ergo intelligent life can develop in this universe. I didn't say anything contrary to this. I was just pointing out an interesting detail about probability theory. It's impolite to edit your comment in such a way as to turn its existing replies into non-sequiturs. For the record, this comment initially cast doubt on the claim about zero probability events, hence the reply saying it was correct.

dotancohen 5y ago

> For an event to have probability zero doesn't imply it can't occur. The distinction in meaningless. We exist, ergo intelligent life can develop in this universe.

danbruc 5y ago
It is correct, if you split 100% across infinitely many possible outcomes, each outcome will have probability zero, still one of the possible outcomes will occur.
- rsj_hn 5y ago
  Doesn't this require a continuum? I don't think you get there with a measure on a countable set. That would make this an unphysical argument, IMO.
  - omnicognate 5y ago
    Yes, it requires uncountability. It could be that physics is ultimately best modelled with countable sets, but that hasn't been established and our current best physical theories are certainly full of uncountability. The distinction between "surely" and "almost surely" [1] is "just" a curiosity about probability theory though, albeit a rather fundamental one, and I only brought it up as such. It's interesting to think about, and if you do so it quickly brings you up against deep philosophical questions about what probability means. [1] https://en.m.wikipedia.org/wiki/Almost_surely https://en.m.wikipedia.org/wiki/Almost_surely
    - danbruc 5y ago
      Just out of curiosity, is there are short answer why it requires uncountability? Naively something like pick a random natural number would also seem to lead to probability zero. I can see that pick a random natural number might be problematic, how would you do this? Pick on digit and then with some probability either stop or continue and pick another digit, but it is at the very least not obvious that one could make this work without larger numbers just having smaller and smaller probabilities and there might also be issues with termination. On the other hand it is not obvious to me why one could not work with uniform distributions over the set [0, n) and then look at the limit as n goes to infinity.
      - rsj_hn 5y ago
        If you have a measure on a countable set, lets number it 0, 1, 2, .. then you must have: m(i) >= 0 (since it's a measure). And must also have 1 = m(0) + m(1) + ... (because it's a measure) so 1 = Lim S(i) Where S(i) is the partial sum going from 0 to i. But if each m(i) = 0, then each partial sum is zero. So 1 = Lim 0 = 0
        danbruc 5y ago
        There is probably something wrong with this, but I was thinking something like the following. Take the sequence of sets M(n) = { 0, 1, 2, ... n - 1 } with measure m(n, i) = 1 / n. The m(n, i) are non-negative and the sum over all m(n, i) for a fixed n is 1. Then take the limit. The set M(n) will seemingly approach the natural numbers but I am not sure that this is valid. The m(n, i) will approach 0, I think that is uncontroversial. But I guess it might not be valid to argue that the sum remains 1 even though it seemingly equals n * 1 / n.
        rsj_hn 5y ago
        No, you can't interchange limits like that. In your case, when you say "sum" of the n identical things, you just mean multiplying n by the integer 1/n. So you have 1 = 1/n *n != lim(1/n)lim(n). The last is an indeterminate form of 0*infinity and so you don't get to conclude that it's one.
- mensetmanusman 5y ago
  Infinity times zero might be meaningless in a quantized reality
omnicognate 5y ago
> The distinction in meaningless. We exist, ergo intelligent life can develop in this universe. I didn't say anything contrary to this. I was just pointing out an interesting detail about probability theory. It's impolite to edit your comment in such a way as to turn its existing replies into non-sequiturs. For the record, this comment initially cast doubt on the claim about zero probability events, hence the reply saying it was correct.