Probability of 8 or 9 digit sequence colliding in the same place in two 65 digit numbers I have two numbers:
3032643431333337636238613038343231383364303731376566303037663231

3861663464383131656131653461343961343364303737663565356561653361

36430373 repeats from digit 36-45 (I only extracted the 8, but the 9th matched too).
What are the chances that the same 8 digits would repeat in two 65 digit length numbers?
What are the chances that the same 9 digits would repeat in two 65 digit length numbers?
Minor Details: I extracted 8 digits from a hash(md5) that was converted to a binary string and then hex encoded to get the 65 digits.  I found the collision and was curious as to the odds.
Thanks
Edit:  As was commented, the digits must be between the 36th and 45th position.  Not elsewhere in the number.

If you want them to be in a specific set of places, the chance of a match starting at a given place is $10^{-8}$ for $8$ digits and $10^{-9}$ for $9$ digits.  Assuming independence and ignoring the chance of two matches (both excellent approximations), there are $58$ places an $8$ digit run can start, so the chance of an $8$ digit match somewhere (but in the same places in both numbers) is $58 \cdot 10^{-8}$  If you want any pair of $8$ continuous digits to match (not necessarily the same places in both numbers) you have $58^2$ pairs of runs, so approximately $58^2 \cdot 10^{-8}=3.364\cdot 10^{-5}$ chance of a match somewhere.
