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When considering 3 first numbers in pi, we get 'fixed numbers' 314. However, three integers can have 999 =10^3-1 versions, of which only one 'belongs' to pi. Thus, for n first digits of pi we have 10^n-2 versions of integers which do not belong to pi ratio being 1/(10^n-2)-> 0.0, when n increases. Thus, for increasing number of n, the probability of finding n random numbers in n 'fixed' numbers of pi approaches zero?

Tapio Ruotoistenmäki
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    This is a famous open problem, it is conjectured to be the case, but only based on the many digits that have been calculated. – Peter Jul 27 '19 at 14:13
  • It has been shown that every digit sequence of length at most $11$ occurs in $\pi$ – Peter Jul 27 '19 at 14:15

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There is no reason to expect randomness in digits of $\pi$

It is a common mistake that an infinite sequence of numbers without being eventually periodic must include all combinations.

For example $$1, 1,2,1,2,3,1,2,3,4,1,2,3,4,5,...$$ which is not eventually periodic and does not contain $55555$

Mohammad Riazi-Kermani
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