Asklemmy

My guess would be that - depending on the number of digits you are looking for in the sequence - you could calculate the probability of finding any given group of those digits.

For example, there is a 100% probability of finding any group of two, three or four digits, but that probability decreases as you approach one hundred thousand digits.

Of course, the difficulty in proving this hypothesis rests on the computing power needed to prove it empirically and the number of digits of Pi available. That is, a million digits of Pi is a small number if you are looking for a ten thousand digit sequence

it's actually unknown. It looks like it, but it is not proven

This is what allows pifs to work!

It has not been proven either way but if pi is proven to be normal then yes. https://en.m.wikipedia.org/wiki/Normal_number

No, the fact that a number is infinite and non-repeating doesn't mean that and since in order to disprove something you need only one example here it is: 0.1101001000100001000001... this is a number that goes 1 and then x times 0 with x incrementing. It is infinite and non-repeating, yet doesn't contain a single 2.

It's almost sure to be the case, but nobody has managed to prove it yet.

Simply being infinite and non-repeating doesn't guarantee that all finite sequences will appear. For example, you could have an infinite non-repeating number that doesn't have any 9s in it. But, as far as numbers go, exceptions like that are very rare, and in almost all (infinite, non-repeating) numbers you'll have all finite sequences appearing.