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Whether the needle crosses a line depends greatly on the angle the needle makes with the line, 90 degrees being of course the most favorable for a line-crossing. Does the 80-20 Rule apply to these angles? That is, since 20% of 90 degrees is 18 degrees, are about 80% of the line-crossings (for a large number of needle-tosses) accounted for by about 20% instances, in which the angle of the needle relative to the line was between 72 and 90 degrees? Has anyone done a statistical study or simulation of this?

Mike Jones
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  • Reminds me of a paper by Edwin Jaynes on the Bertrand paradox. Short answer: depends how you throw your needle. – Lucas May 25 '13 at 22:44
  • http://bayes.wustl.edu/etj/articles/well.pdf – Lucas May 25 '13 at 22:44
  • Edwin Jaynes once posed this exercise: A very limp piece of string of length $L$ is thrown very unskillfully onto the floor. Find the probability distribution of the distance between its ends. – Michael Hardy May 25 '13 at 22:46

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The angle $\Theta$ is uniformly distributed betwen $0$ and $\pi/2$ radians.

It actually took me a while to figure out that the question is this: If the number $c$ is so chosen that $\Pr(\Theta>c\mid\text{crossing})=0.8$ then is it true that $\Pr(\Theta >c)=0.2$?

I.e. do $80\%$ of crossings occur for only the $20\%$ of values of $\Theta$ most favorable to crossings? Short answer: no.

The prior density of $\Theta$ is $$ f_\Theta(\theta) = 2/\pi\quad\text{for }\theta\text{ between }0\text{ and }\pi/2 $$ and $=0$ for $\theta$ not in that space.

The likelihood function is $$ L(\theta\mid\text{crossing}) = \Pr(\text{crossing}\mid\Theta=\theta) = \sin\theta. $$

Bayes tells us to multiply the prior by the likelihood function and then normalize to get the posterior probability density $f_{\Theta\mid\text{crossing}} (\theta)$.

Thus the posterior is $$ f_{\Theta\mid\text{crossing}} (\theta) = 1\cdot\sin\theta, $$ since $\int_0^{\pi/2}\sin\theta\,d\theta=1$.

Then we have, for example, $\Pr(\Theta>\pi/4)=1/2$ and $$ \Pr(\Theta>\pi/4\mid\text{crossing}) = \int_{pi/4}^{\pi/2}\sin\theta\,d\theta = \cos\frac\pi4=\frac{\sqrt{2}}{2} \approx 0.707\ldots. $$

So the most crossing-productive half of the population produces about $70.7\%$ of all crossings.

Plug in numbers to get other results.

Michael Hardy
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