I know it's likely that, given a finite sequence of digits, one can find that sequence in the digits of pi, but is there a proof that this is possible for all finite sequences? Or is it just very probable?
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Edit: We seem to have lost Gerald Edgar's comment on another answer, which pointed out that the normality property implies but is not equivalent to the property of containing every finite sequence as a substring. For example, consider the sequence enumerating all possible finite strings, separated by increasingly huge oceans of zeros. 


Numbers with this property belong to the set of normal numbers. All known normal numbers are "constructed"; it is not known whether a single "natural" number (square roots of nonsquares, pi, e, logarithms of integers $> 1$) is normal or not. It is believed, however, that these numbers are in fact normal (even independently from the chosen base), and thus have the desired property. 


This is an expansion for Pi in base 16 numeric system: $$\pi = \sum_{k = 0}^{\infty}\frac{1}{16^k} \left( \frac{4}{8k + 1}  \frac{2}{8k + 4}  \frac{1}{8k + 5}  \frac{1}{8k + 6} \right)$$ So to get kth digit you have to get one term and take account for possible translation from a neighboring digit. Thus to find the number of digit from which starts your arbitrary sequence, you should to solve a system of equations about the particular digits: $$\frac{1}{16^k} \left( \frac{4}{8k + 1}  \frac{2}{8k + 4}  \frac{1}{8k + 5}  \frac{1}{8k + 6}\right)=a_1$$ $$\frac{1}{16^{k+1}} \left( \frac{4}{8(k+1) + 1}  \frac{2}{8(k+1) + 4}  \frac{1}{8(k+1) + 5}  \frac{1}{8(k+1) + 6}\right)=a_2$$ $$\frac{1}{16^{k+2}} \left( \frac{4}{8(k+2) + 1}  \frac{2}{8(k+2) + 4}  \frac{1}{8(k+2) + 5}  \frac{1}{8(k+2) + 6}\right)=a_3$$ etc. The numbers $a_k$ are unique for any sequence you are searching for. If the system has no solution, it is likely that your sequence does not appear in the sequence of digits of Pi. 

