Are the digits of irrational/transdental numbers random? 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 $0.1$ ?
 A: 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.
A: 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.
