I just had an interesting conversation with my kid who asked an innocent question about the $\pi$: 

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> If $\pi$ is infinite - does that mean that somewhere in it there's another $\pi$?
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I looked around and found [this question](https://math.stackexchange.com/questions/1403416/infinite-set-always-has-a-countably-infinite-subset) 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: 

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> 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? 
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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: 

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> 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?
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