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So, my (naive) understanding is that in order for Pi to contain every single possible number combination, Pi has to be a normal number, where each digit in base 10 appears with equal probability. However, I don't really see why this is the case? If you look at the first 50 digits of Pi, you'll see every digit from 0-9. Doesn't that mean that there is a non-zero probability of seeing the digits 0-9 for each place? Since there is a non-zero probability, wouldn't every single sequence show up as Pi is infinite then? I'd appreciate someone showing me what I'm not understanding. Thanks guys!

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I don't understand how is it that you jump from “If you look at the first 50 digits of Pi, you'll see every digit from 0-9” to “there is a non-zero probability of seeing the digits 0-9 for each place”. Consider the sequence of $0$'s and $1$'s such that

  • the first term os $1$;
  • all other terms are $0$.

So, you see a $0$ and a $1$ in the first two terms. However, the probabiliy of having a $1$ in the $n$th term is $0$ for every place other than the first one. And $1$ appers only once, not infinitely often, and therefore I don't see why is it that you ask “wouldn't every single sequence show up as $\pi$ is infinite then?”

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  • $\begingroup$ Whoops. Yeah, my bad. Thanks! $\endgroup$ – jack klompus Feb 7 '18 at 11:48

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