Probability of selecting a number in a repeating decimal series For example in a infinitely repeating series such as $\frac{110}{111}=0.\overline{990}$, what would be the probability of selecting a 0 in the series generated by the infinitely repeating decimals?
I thought that the answer seemed obvious seeing that in each repeating segment, there are two 9s and one 0, so the probability of selecting a 0 would be $\frac{1}{3}$. However, couldn't a bijection be created between each of the 9s and 0s? And so, the probability would be $\frac{1}{2}$? 
This is super counterintuitive, and if this isn't true, what's the difference between this and the proof for the number of natural numbers and natural even numbers being equal?
 A: The short answer is just that infinities are weird sometimes (or a lot of times) and have a lot of counter-intuitive properties.
Let's say we assign labels to the decimals like $d_1, d_2, d_3, \ldots$ As $n$ tends to infinity, then the probability of you selecting a $d_i$ associated with a $0$ out of all $d_i$ such that $1 \leq i \leq n$ will tend to $1/3$. I think that this is the best we can do.
Let's try ignoring the limit and just try directly selecting a label instead out of all of the infinite number of labels. This is the same as trying to select a random natural number. Unfortunately, there is no uniform random distribution on the natural numbers, and so we cannot do this. As a thought experiment, what would a random natural number look like? For example, how large should it be?
Yes, we can create a bijection between the $9$s and $0$s, and this is similar to the bijection between the natural numbers and even natural numbers. However, their cardinalities being the same (hopefully) doesn't contradict with our intuition on the limit we took earlier. The trouble comes when we try to think of the infinite sequence all at once, but as noted earlier, we can't actually take a random natural number directly, nor can we take a random digit from the infinite decimal expansion of $\frac{110}{111}$.
