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I think this only sounds like a paradox if it is phrased poorly. The accurate way to state it is "The probability of randomly picking a specific number is 0" an

I think this only sounds like a paradox if it is phrased poorly. The accurate way to state it is "The probability of randomly picking a specific number is 0" and that sounds reasonable. The probability of successfully picking any number is 1.

cdavid 9y ago
That's a different statement: OP is alluding to the fact the measure of Q is 0 when using the "standard" sigma algebra on the real line, while you are saying that the measure of a number of 0. [edit] strictly speaking, you would restrict yourself to a bounded interval, e.g. if you pick a random number from a uniform distribution on [0, 1], the probability that this number is rational is 0.
- philipov 9y ago
  oh, yeah, but that's because although Q is dense, it is not a dense subset of R and locally that's equivalent to saying a single point is not dense in R
  - cdavid 9y ago
    no, it is not. The OP statement is about the probability of the event "{X in Q}" (equal to 0), with X a random variable uniformly distributed on a bounded interval. That even contains many points (infinitely many actually), but has a probability 0. You are talking about the probability of a single point event, which is also always 0 on that same sigma algebra. The OP point is not completely trivial because the event contains an infinite (but countably) number of elements. It is fairly easy to understand though since by its very definition, the P[{X in Q}] = sum P[{x}] taken over every rational number (since Q is countable), and each P[{x}] is 0. A deeper statement is that there exists uncountable sets of probability 0.
Houshalter 9y ago
The paradox is that, after picking a random number, you have just done a thing which has probability zero. Doing a thing that has zero probability shouldn't be possible. Ever.
- dorgo 9y ago
  I would argue that the distribution you used to pick a random number was not uniform. Not all the real number were equaliy likely to be picked by you. Hence the probability for some numbers was > 0.
- marcosdumay 9y ago
  Non-countable sets defy intuition on several ways. The silver lining is that we don't have any evidence a non-countable thing exist on the real world. I don't think anybody even has a procedure for gathering that kind of evidence.
  - dorgo 9y ago
    No. Rational number are countable, but have the same property of zero probability for an item in a uniform distribution.
    - pelario 9y ago
      Why the downvotes? I think he is correct; and if someone think he is not, please elaborate
    - marcosdumay 9y ago
      Yes, no reason for the downvotes. Infinities are problematic too.
- rocqua 9y ago
  You can't pick a random real number between 0 and 1. Heck, almost all reals between 0 and 1 can't ever be constructed let alone picked. The here is the non-constructive nature of the real numbers. That is not to say the reals are useless, but they are not much more than a formalism. It's rather useful though because it's hard to get numbers like pi or e. Its really nice that any real interval is compact, but that too is hard to replicate.
  - bjl 9y ago

cdavid 9y ago

That's a different statement: OP is alluding to the fact the measure of Q is 0 when using the "standard" sigma algebra on the real line, while you are saying that the measure of a number of 0. [edit] strictly speaking, you would restrict yourself to a bounded interval, e.g. if you pick a random number from a uniform distribution on [0, 1], the probability that this number is rational is 0.

philipov 9y ago
oh, yeah, but that's because although Q is dense, it is not a dense subset of R and locally that's equivalent to saying a single point is not dense in R
- cdavid 9y ago
  no, it is not. The OP statement is about the probability of the event "{X in Q}" (equal to 0), with X a random variable uniformly distributed on a bounded interval. That even contains many points (infinitely many actually), but has a probability 0. You are talking about the probability of a single point event, which is also always 0 on that same sigma algebra. The OP point is not completely trivial because the event contains an infinite (but countably) number of elements. It is fairly easy to understand though since by its very definition, the P[{X in Q}] = sum P[{x}] taken over every rational number (since Q is countable), and each P[{x}] is 0. A deeper statement is that there exists uncountable sets of probability 0.

Houshalter 9y ago

The paradox is that, after picking a random number, you have just done a thing which has probability zero. Doing a thing that has zero probability shouldn't be possible. Ever.

dorgo 9y ago
I would argue that the distribution you used to pick a random number was not uniform. Not all the real number were equaliy likely to be picked by you. Hence the probability for some numbers was > 0.
marcosdumay 9y ago
Non-countable sets defy intuition on several ways. The silver lining is that we don't have any evidence a non-countable thing exist on the real world. I don't think anybody even has a procedure for gathering that kind of evidence.
- dorgo 9y ago
  No. Rational number are countable, but have the same property of zero probability for an item in a uniform distribution.
  - pelario 9y ago
    Why the downvotes? I think he is correct; and if someone think he is not, please elaborate
  - marcosdumay 9y ago
    Yes, no reason for the downvotes. Infinities are problematic too.
rocqua 9y ago
You can't pick a random real number between 0 and 1. Heck, almost all reals between 0 and 1 can't ever be constructed let alone picked. The here is the non-constructive nature of the real numbers. That is not to say the reals are useless, but they are not much more than a formalism. It's rather useful though because it's hard to get numbers like pi or e. Its really nice that any real interval is compact, but that too is hard to replicate.
- bjl 9y ago