A common illustration of the nature of infinity is that, given an infinite amount of time, a monkey on a typewriter will, with probability $1$, produce the complete works of Shakespeare.

Consider now a (very clever) monkey that is able to choose, with uniform probability, real numbers from some interval, say $(0,1$). In an infinite time, is it correct to state that the monkey will eventually choose any given number (e.g. $\sqrt{2}/2$) with probability $1$?

Does the fact that the cardinality of $\mathbb{R}$ is larger than that of $\mathbb{N}$ come into the equation?

I have tried understanding it as follows. Consider the probability that the monkey chooses a number in $(\sqrt{2}/2-\epsilon/2, \sqrt{2}/2+\epsilon/2)$, which is $\epsilon$. The probability that in $N$ trials the monkey has not chosen a number in this interval is $(1-\epsilon)^N$. Does the question therefore boil down to evaluating:

$$\lim_{\substack{N\rightarrow\infty\\ \epsilon\rightarrow 0}}(1-\epsilon)^N$$

Does this limit exist? Is it $0$?

Follow up question: what if the monkey is now free to choose from an unbounded interval, like $\mathbb{R}$?

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    $\begingroup$ The rough answer is, no, you would not get all real numbers. Indeed, probably all you can say is that, with probability 1, the set of random numbers will be dense in $(0,1)$. $\endgroup$ – Thomas Andrews Feb 26 '14 at 19:22
  • $\begingroup$ The probability that in $N$ trials (be $N$ as large as you want) the monkey has selected number (say) 1/2 is zero. $\endgroup$ – leonbloy Feb 26 '14 at 19:48
  • $\begingroup$ A related (and perhaps slightly more interesting) question would be the same restricted to a countable set; say, the rationals. But then we cannot have a "uniform" probability math.stackexchange.com/questions/311360/… $\endgroup$ – leonbloy Feb 26 '14 at 19:51

Presumably the monkey only chooses countably many numbers. You can adapt the proof that the measure of $\Bbb Q$ is zero to show that the chance any given real is chosen is still $0$. Choosing from an unbounded interval doesn't change anything. To show that, take your favorite bijection between $(0,1)$ and $\Bbb R$

Specifically for your limit, you need to define how the two limits are taken. If you let $N \to \infty$ first, the limit is zero. If you let $\epsilon \to 0$ first, the limit is $1$. If you let $\epsilon=\frac 1N$ and they go together, you get $\frac 1e$. You can get anything on $[0,1]$ you like.

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    $\begingroup$ Great answer. However, is saying that the "chance of any given real occurring is $0$" not incorrect? Some reals will definitely be chosen. $\endgroup$ – MGA Feb 26 '14 at 19:27
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    $\begingroup$ @MGA: It is correct, despite being counterintuitive. You could see this question $\endgroup$ – Ross Millikan Feb 26 '14 at 19:35
  • $\begingroup$ @MGA: Choosing some real occurs with probability $1$. Fixing a particular real number, and asking what is the probability that this real is chosen, that's zero. Fixing a particular set of reals and asking what is the probability that some/almost all/all reals are chosen from that set depends on the set (whether it is measurable) and its measure. $\endgroup$ – Asaf Karagila Feb 26 '14 at 19:41
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    $\begingroup$ @RossMillikan That's amazing, and I want to learn more. I guess the 'place' to study this formally would be "probability theory"? Do you have a favourite textbook, by any chance? $\endgroup$ – MGA Feb 26 '14 at 19:55

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