On the contrary, it always happens for any number and base, at n = 0 ! If on the other hand there is prior knowledge constraining n from below --- as in the case of pi, for which the answer is presumably known when n < 10^12 or whatever --- then the probability for a normal number becomes correspondingly exponentially smaller. WFL On 3/20/14, Dan Asimov <dasimov@earthlink.net> wrote:
Just curious: Can one calculate the exact probability of this (assuming for simplicity that the digits of pi are independent and random)?
Or better: Just take flips of a fair coin (H or T*) indexed by Z+. What is the probability there exists an n such that flips n+1 through 2n are all H ??? Dealing with the overlaps looks complicated.
--Dan _____________________________________________________________________________ * If the coin were an old French franc, would a flip be either a T or a Q ???
On Mar 20, 2014, at 7:20 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
The irrationality measure of pi is at most 7.6063 [Salikhov, "On the Irrationality Measure of $\pi$", 2008]. I think that translates into "There are at most finitely many n for which digits n through k*n are all 9's" for some k, but I don't know what k is. But none of this answers Jim's / Dick's question (whose answer is surely "it never happens but we don't have a proof").
--Michael
On Thu, Mar 20, 2014 at 9:34 AM, James Propp <jamespropp@gmail.com> wrote:
Is it possible that for some n, the n+1st through 2nth digits of pi are all 9's?
Dick Hess just gave a wonderful G4G talk which raised this issue.
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