3

In this question, I mentioned that, assuming the digits of pi are independently-random, then at some point in pi's expansion there will be a sequence of one million consecutive 0's. I decided to determine exactly how many digits ($d$) of $\pi$ we'd have to look at to have a 50% chance of seeing $n$ consecutive 0's.

I was able to solve for $d$:

$$(1-10^{-n})^d = 0.5$$

$$d = \frac{\log(0.5)}{\log(1-10^{-n})}$$

However, while playing around with this expression, I discovered by accident that it is either equal to or nearly equal to

$$d \stackrel{\hbox{?}}{\hbox{=}} \ln(2) \cdot 10^n$$

I've been unable to reduce the first equation to the second, or to show why they should be nearly equal.

Could anyone shed some light - are they equal? And why does the natural-log come into this? Is this somehow related to other results in probability (which I know uses $e$ extensively)?

Community
  • 1
BlueRaja - Danny Pflughoeft
  • 8,434

3 Answers3

3

They are not equal, but (with $\log$ the natural logarithm) $$\left(1-\frac{1}{10^n}\right)^{10^n}\approx \frac{1}{e}\implies 1-10^{-n}\approx e^{-10^{-n}}$$ $$\implies d=\frac{\log(1/2)}{\log(1-10^{-n})}\approx(\log2)10^n.$$

anon
  • 151,657
3

$\log(0.5)=-\log(2)$.

For $x \ll 1, \ln(1+x)\approx x$, which comes from the Taylor series.

The ratio of logs is independent of the base

Ross Millikan
  • 374,822
3

It's very simple $d = \frac{log(0.5)}{log(1-10^{-n})}=\frac{ln(0.5)}{ln(1-10^{-n})}$, because $\log(x)=\frac{ln(x)}{ln 10}$ and in your expression you may multiply numerator and denumerator by $ln(10)$. Then if N is sufficiently large we can to approximate $\ln(1-10^{-N})$ by $-10^{-N}$ and we know that $\ln(0.5)=-\ln(2)$. So we got it. But this expressions arent equal.

Evgeny Savinov
  • 586