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If I were to look at the decimal representation of some irrational or even transdental number,

and if I choose a natural number at random can I expect that it is some digit with probability $0.1$ ?

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Look at normal number en.wikipedia.org/wiki/Normal_number. Almost all numbers are normal. –  user17762 Oct 21 '12 at 19:10
    
Concerning the question from the title see Liouville number. –  Raymond Manzoni Oct 21 '12 at 19:23
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up vote 3 down vote accepted

In general all digits of an irrational number do not have to occur with equal probability. A quick example would be the number $$0.1011011101111011111...$$

Here, the number is composed of only $1$'s and $0$'s, so the probability for digits 2-9 is zero. In fact, depending on how you define the probability of finding a digit, the probability of randomly selecting $0$ from the digits might also be zero. This is the case if you define the probability for the infinite decimal representation to be the limit as $n\rightarrow\infty$ of the probability of finding the digit in a truncated approximation of the number with $n$ digits.

Even with this definition, you could get irrational numbers where this limit doesn't exist for some digits. An example would be something like $$0.101100111111000000...$$ where after each run of $0$'s you keep appending a $1$ until the probability of finding zero in the truncated representation is less than $1/4$, and then you go back to appending $0$'s until the probability of finding a $0$ has gone back up to $1/2$. In this case, our definition for the probability of finding $1$ or $0$ in the infinite decimal representation doesn't give an answer, because the limit doesn't exist.

So I guess the point is that probabilities are a bit tricky to define for infinite sets of numbers, but in general irrational numbers do not have to have equal probabilities for each digit.

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Presumably by "at random", you mean "uniformly at random". There is no uniform distribution on the natural numbers; see Probability of picking a random natural number.

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