Does the 80-20 Rule apply to the Buffon process? 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?
 A: 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.
