Is there any finitely-long sequence of digits which is not found in the digits of pi? 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?

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 k-th 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.
