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What is this the probability that a random positive integer power of 2 begins with 1?BY a random power I mean a natural number chosen randomly from the set of natural numbers.

AgnostMystic
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    How are the integers 'randomly' chosen? With what distribution? I suspect you really want some kind of asymptotic answer. – Potato Jan 29 '14 at 06:58
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    This question is not meaningful until you specify the probability distribution on the set of powers of 2 under which you want to compute the probability. – Prateek Jan 29 '14 at 06:59
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    The power of 2 is chosen uniformly and randomly from the set {1,2,3,.....,n}.I want the limiting probability as n goes to infinity – AgnostMystic Jan 29 '14 at 07:02
  • There are infinite integers !!!. – Felix Marin Jan 29 '14 at 07:04
  • See also http://math.stackexchange.com/questions/417486/probability-that-the-first-digit-of-2n-is-1 – Chris Culter Jan 29 '14 at 07:13

2 Answers2

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30.102999566398119521373889472449302676819%

Chris Culter
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We want to know when we have $10^n \leq 2^k < 2\cdot 10^n$. Taking $\log$, this becomes $n \leq k \log_{10}(2) < \log_{10}(2) + n$. Modulo $\mathbb{Z}$, our question is now about asymptotics of how often $k\log_{10}(2) \pmod{\mathbb{Z}}$ lands in the interval $[0,\log_{10}(2)] \subseteq [0,1]$. Since $\log_{10}(2)$ is irrational, I believe that the distribution is basically uniform, and so the probability should simply be $\log_{10}(2) \approx 0.301$.

Zach L.
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