PI as an infinite set of integers I just had an interesting conversation with my kid who asked an innocent question about the $\pi$:  

If $\pi$ is infinite - does that mean that somewhere in it there's another $\pi$?

I looked around and found this question and a few other, similar ones but I'm afraid my math knowledge is a bit limited to draw a definitive conclusion to the above question.  
Further conversation yielded a secondary question:  

Is there a place within $\pi$ where the complete previous set of numbers (starting with the $3.14...$) repeats all the way to the beginning of that repetition set?  

This question seems rather tricky to me as my assumption is that such a set should exist (because the set is infinite, so the probability of such set's existence should be non-$0$, right?) but the longer we "wait" for such a repetition to occur the longer that repeated set should be, which makes the "wait" longer... and I'm falling into a recursion here :)  
In addition the fact that $\pi$ is irrational means there are no repeatable sequences of digits in it (if my understanding is correct) which kind of throws off the whole "such a sequence should exist since the series is infinite" logic.
An extension to the second question is:  

Is it possible to calculate the probability of such a subset's existence (the one that repeats all the previously seen numbers in the exact same sequence) and if so - what would that probability be?

 A: Just some initial thoughts while I ponder further.
Let me clarify the question I think is being asked a bit.  Does there exist an integer $n > 0$ such that the first $n$ digits of $\pi$ (including the $3$) are followed by the same $n$ digits?  After that, the digits may continue along arbitrarily.
(I interpret the question this way, because assuming that the entirety of $\pi$ is repeated would indicate that $\pi$ was rational, which is known to be untrue.)
I'm pretty sure that the answer to this clarified question is unknown.  If $\pi$ is indeed normal, as many people expect (but it is not known, either), then we might reason as follows:
After the first $n$ digits, what is the probability that the next $n$ digits duplicate those first digits?  If $\pi$ is normal, this probability is $1/10^n$.  If we denote by $N$ the largest number for which this is known not to be true, then the probability that it eventually happens would have a lower bound of $1/10^{N+1}+1/10^{N+2}+\cdots = 1/(9 \cdot 10^N)$.
It's only a lower bound, because the first $n$ digits may well have a suffix that duplicates some of the needed digits of $\pi$ for the repetition.  I don't have a good feeling at the moment, given a random normal number, how deep such a suffix might go, and how that would affect the probability.  It seems unlikely to me, however, that it would raise it anywhere near $1$.
ETA: Perhaps we can place an upper bound on it, in the following way.  We will assume all kinds of independence that are not actually warranted, yet that might give some insight anyway.
Suppose that we know that the first $2n$ digits do not contain any satisfactory duplication.  What is the probability that the last $n-1$ of these $2n$ digits duplicate the first $n-1$ digits of $\pi$?  It is $1/10^{n-1}$ (assuming independence).  What is the probability, given that the last $n-1$ of these $2n$ digits duplicate the first $n-1$ digits of $\pi$, that the next two digits complete a duplication?  It is $1/100$.  The product of these two is $1/10^{n+1}$.
What is the probability that the last $n-2$ of these $2n$ digits duplicate the first $n-2$ digits of $\pi$?  It is $1/10^{n-2}$.  What is the probability, given that the last $n-2$ of these $2n$ digits duplicate the first $n-1$ digits of $\pi$, that the next four digits complete a duplication?  It is $1/10000$.  The probability of these is $1/10^{n+2}$.
And so on.  We should be able to upper bound the probability of a duplication before the $4n$th digit by $1/10^{n+1}+1/10^{n+2}+\cdots = 1/(9 \cdot 10^n)$.  If the largest prefix of $\pi$ we know contains no such duplication has length $2N$, then our total probability should be no more than the sum
$$
\frac{1}{9 \cdot 10^N}+\frac{1}{9 \cdot 10^{2N}}+\frac{1}{9 \cdot 10^{4N}} + \cdots < \frac{1}{9 \cdot (10^N-1)}
$$
Obviously, I've taken a lot of liberties here.  Any thoughts, anyone, on whether any of them matter?
A: Short answer: If $\pi$ contained a $\pi$, then that $\pi$ would contain a $\pi$, and so on, turning the whole thing into a recurring decimal, and therefore rational. However $\pi$ considered as a sequence does contain a subsequence that would be $\pi$, and that contains another such subsequence, and so on. The problem is due to the rather strict requirement that the subsequence should be a nontrivial tail.
A: Turning my comment into an answer: for finite repetitions of finite sequences since the beginning, it is not known as of 20151001 (I believe we have not (yet?) found one, otherwise that would be viral).
It is not known whether $\pi$ is normal or not. It may well have some magic property (for example, start going $\dotsc\,01001000100001\dotsc$ from the decimal place $10^{10^{10^{100}}}$), such that the answer to your question is definitely no (in general i.e. for big enough sequence), or (from the decimal place $10^{10^{10^{100}}}$, repeat since the beginning) definitely yes. Rationality implies the decimal expansion repeats forever. It is not forbidden for an irrational number to repeat (a finite number of times) some (finite) decimal sequence and then continue to go crazy.
