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.
Best Answer
If you get infinitely many 2s, but with arbitrarily large gaps between them, then your number is irrational. This happens with probability one. (See the Borel-Cantelli theorem. I'm assuming your events with probability 1/n are independent, you didn't say that...)
But note if you use probabilities $1/n^2$ instead, then you get (with probability one) finitely many 2s, so your number is rational.