TITLE: Is it irrational?
QUESTION [25 upvotes]: Suppose I generate a number $0 < x < 1$. In general, after the decimal point, the first digit is $1$, the second is $0$, the third is $1$, etc. However, every digit in position $n$ has a $1/n$ chance of being $2$ instead. The tenths place, for example, must have a $2$. The number might look like this:
$$x = 0.221210101020101010101010101010121010101010101010101010101010...$$
(There are an infinite number of possible values of $x$. My question applies to any number in this set, excluding numbers where the digits begin to repeat interminably at a finite point, which has 0 probability of happening. Ex. not interested in 0.222222...)
Such a number doesn't seem to be rational, because $2$ will continue to pop up, even as it becomes less frequent. However, as the position $n$ of a digit approaches infinity, the probability of that digit being $2$ approaches $0$. It seems that as $n$ goes to infinity (and there are an infinite number of digits), the digits approach rational behavior, for lack of a  better phrase.
Would this number be irrational?
Clarification: What I am asking is whether a number would be considered irrational if its digits "approach repetition"; if they could be said to repeat interminably at infinity, but not at a finite point.

REPLY [12 votes]: This isn't a number, it is a set of numbers. Probabilities don't make sense here, a number is part of your set or it isn't. The set is then $0.1010\ldots$ where some of the digits are replaced by $2$. One of the numbers in your set is $0.2222\ldots$, which is rational, another is $0.101010\ldots$, again rational. But it also contains the number where the $2$ show up only on positions that are perfect squares, i.e., $0.2012101012\ldots$, which isn't a repeating decimal and thus irrational.
Your set of numbers isn't countable, by the way, so it has to include transcendental numbers too (the set of algebraic numbers is countable).