6

I've started to learn some probabilty and it made think about this question: let us assume we randomize virtually any number between 0 and 1. What is the probability for this number to be irrational?

catch22
  • 3,057

1 Answers1

7

Modern probability uses measure theory. In particular the Lebesgue measure.

This means that any countable set has probability zero to be chosen from. In particular this means that the probability for choosing an irrational number is $1$.

In fact not just irrational, but also transcendental and normal, and any other property which occurs outside a countable set.

Asaf Karagila
  • 393,674
  • But its still possible to hit a rational. Just very(?) unlikely. – hjg Oct 22 '12 at 14:24
  • 2
    Yes. It is possible. But it is less possible than the possibility that you will be killed by a tiny black hole that was created due to quantum fluctuation within your heart and sucked dry your blood before collapsing into itself. – Asaf Karagila Oct 22 '12 at 14:26
  • Note that in this context probability 0 is not the same as impossible. There is also the problem what constitutes a (single) random number. If som random generator demon spit out a number digit by digit and would start "0.33333333...", you may need to wait forever to know whether this is $\frac13$ or not. Also, you may question its randomness if it keeps spitting out "3"s. The model is about obtaining a number (not its decimal representation) at once; or when using measure theoretic arguments, it is not about actually obtaining a random number at all, only measures. :) – Hagen von Eitzen Oct 22 '12 at 14:54
  • When the random number generator of your PC outputs consecutive binary digits of a "random" real between $0$ and $1$ then it will surely print the binary expansion of a rational, because in the long run the output will be periodic. – Christian Blatter Oct 22 '12 at 15:25