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A large portion of combinatorics cases have probabilities of $\frac1e$.

Secretary problem is one of such examples. Excluding trivial cases (a variable has a uniform distribution over $(0,\pi)$ - what is the probability for the value to be below $1$?). I can't recall any example where $\frac1\pi$ would be a solution to a probability problem.

Are there "meaningful" probability theory case with probability approaching $\frac1\pi$?

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    $\begingroup$ Not completely related, but still amazing (in my opinion): guess what is the probability that the closest integer to $x/y$ is even, when $x$ and $y$ are uniformly randomly chosen in $]0,1[$. Unrelated to $\pi$, you think? Well, you'll find the answer here. $\endgroup$ – Watson Jan 25 '17 at 15:34
  • $\begingroup$ Possibly related: math.stackexchange.com/questions/1243160 $\endgroup$ – Watson Nov 27 '18 at 19:01
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Yes there is!

Here is an example, called the Buffon's needle.

Launch a match of length $1$ on a floor with parallel lines spaced to each other by $2$ units, then the probability that a match crosses a line is

$$\frac 1\pi.$$

You can have all the details of the proof here if you like.

$\qquad\qquad\qquad\qquad\quad $

More generally, if your match (or needle, it's all the same) have a length $a$, and the lines are spaced to each other by $\ell$ units, then the probability that a match crosses a line is

$$\frac {2a}{\pi \ell}.$$

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    $\begingroup$ I don't see how this is a combinatorics problem. Do I have too narrow a view of combinatorics? $\endgroup$ – user856 Jan 24 '17 at 23:34
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    $\begingroup$ @Rahul, I would not call Buffon's needle a combinatorics problem, but I wouldn't call the secretary problem a combinatorics problem either. Seems the OP had general probability problems in mind. $\endgroup$ – Thanassis Jan 25 '17 at 2:27
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    $\begingroup$ I would say that a combinatoric problem is discrete. $\endgroup$ – leonbloy Jan 25 '17 at 2:51
  • $\begingroup$ Just a bit to add: that probability only holds if the length of needle is smaller than the space between the lines. $\endgroup$ – Bonnaduck Jan 25 '17 at 6:08
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    $\begingroup$ I ment probability. Will change the keyword. $\endgroup$ – Stepan Jan 25 '17 at 14:30
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This doesn't addresses the question as asked, but I think is an interesting example in the spirit of what you asked:

Pick $N$ to be an integer. Now, calculate $p_N$ to be the probability that two random numbers $1 \leq m,n \leq N$ are relatively prime.

Then, $$\lim_{N \to \infty} p_N =\frac{6}{\pi^2}$$

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    $\begingroup$ This is a great example. But what is the name of this theorem? Is it somehow linked to zeta(2)? $\endgroup$ – Stepan Jan 24 '17 at 20:58
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    $\begingroup$ @Stepan I don't think that there is any name for this. I think it appears in Apostol,"Introduction to Analytic Number Theory". $\endgroup$ – N. S. Jan 24 '17 at 20:59
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    $\begingroup$ @Stepan The computation is not too difficult, you just have to realize that two numbers are relatively prime if and only if they are not divisible by a common prime. The probability that two numbers less than $N$ are both divisible by $P$ is roughly $\frac{1}{p^2}$. Therefore $$p_N \sim \prod_{ p< N} (1-\frac{1}{p^2})$$ To make it into a proof, you need to estimate the error of the above approximation, and see that in limit the RHS is just the Euler product which corresponds to the reciprocal of $\sum \frac{1}{n^2}$. $\endgroup$ – N. S. Jan 24 '17 at 21:04
  • $\begingroup$ @Stepan Yes, and it works similarly for interpreting $\zeta(n)$ in terms of the probability of $n$ random numbers being relatively prime. I think it's also on Wikipedia under zeta function. $\endgroup$ – Kimball Jan 24 '17 at 23:23
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There are quite a few geometric probabilities related problems, involving $\pi$. The simplest I can think of, imagine this being a darts board.

enter image description here

The probability of hitting the square, assuming the shot didn't miss the board, is $\frac{2}{\pi}$.

If the radius of the circle is $r$, then one side of the square is $r\sqrt{2}$ and $p=\frac{S_{square}}{S_{circle}}=\frac{2 r^2}{\pi r^2}=\frac{2}{\pi}$

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