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Earlier results have shown that when $l < d$, the expected number of crossings of a needle of length $l$ with vertical lines spaced $d$ apart is $\frac{2l}{\pi d}$, which is also the expression for the probability that a needle intersects a line. I'm looking for an intuitive explanation for why that is the case (is that even the case...?) when the needle is longer ie. $l > d$ (consider $l = 3, d = 1$ for example).

This does not match the expression for the probability that a needle intersects a line when $l > d$; rather, it matches the expression for the probability that a needle intersects a line when $l < d$. Is this just because the possible numbers of crossings are no longer restricted to $0$ and $1$ (ie. the $0$ term cancels out when computing the expected value)?

And, how would one find the PMF of the number of crossings when $l > d$ (for a simpler case such as $l = 3, d = 1$)? The possible values for the numbers of crossings are $0, 1, 2, 3, 4$ if I'm not mistaken. But I don't know where to go from there.

edit: still looking for the PMF!

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    $\begingroup$ Possibly helpful: cs.umb.edu/~eb/piday/whypi.pdf $\endgroup$ – Ethan Bolker Nov 25 '18 at 23:38
  • $\begingroup$ @EthanBolker I guess so! Still only deals with the situation when $l = d$. was really hoping for some hard-hitting intuition when $l > d$, but maybe it's just not intuitive and that's all there is to it. $\endgroup$ – 0k33 Nov 25 '18 at 23:44
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    $\begingroup$ I think that discussion covers the case you are interested in since it explains (or at least asserts) that the crossing number in fact depends on the ratio of $d$ to $l$. $\endgroup$ – Ethan Bolker Nov 26 '18 at 0:02
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    $\begingroup$ When $l \lt d$, the probability of crossing is equal to the expected number of crossings for precisely the reason you give: the number of crossings can only be $0$ or $1$. When $l\gt d$ then the $\frac{2l}{\pi d}$ gives the number of expected crossings but not the (smaller) probability of of at least one crossing - the expression is obviously not the probability when $\frac{l}{d}l \gt \frac{\pi}{2}$ since the expression will be greater than $1$ $\endgroup$ – Henry Nov 26 '18 at 0:36
  • $\begingroup$ @Henry Yes, of course. $\endgroup$ – Ethan Bolker Nov 26 '18 at 0:42
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It is clear that we can rescale the problem and take, wlog, $d=1$ and $l/d=r$.
Therefore we can take the lines to be the vertical lines at $x \in \mathbb Z$.
Consider the needle placed with one end at $(s,0)$ and forming an angle $\alpha$ wrt the $x$ axis: we can sketch the following scheme

Ago_Buffon_2

Considering the simmetry of the problem, we can limit to the I and II quadrants.
Also, the variable $s$ will be limited to the range $\left[ {0,1} \right)$.
However, there is a symmetry around $s=1/2$, so we will reduce our analysis to $0 \le s < 1/2$, considering $s$ and $1-s$ to be equivalent.

The circle with center in $(s,0)$ and radius $r$ encompasses the abscissas $s-r \le x \le s+r$.
The set of lines that the needle can cross are those given by $$ x = n\quad \left| {\;\left\lceil {s - r} \right\rceil \le n \le \left\lfloor {s + r} \right\rfloor } \right. $$

It is convenient to extend the values of $n$ by two additional elements at the extremes, and define a set of boundary values for $x$ and for the angle $\alpha$ defined as follows $$ \left\{ \matrix{ N = \left\{ {n\quad \left| {\;\left\lceil {s - r} \right\rceil - 1 \le n \le \left\lfloor {s + r} \right\rfloor + 1} \right.} \right\} \hfill \cr X = \left\{ {x(n)} \right\} = \left\{ {\left( {s - r} \right),\;\left\lceil {s - r} \right\rceil ,\;\left\lceil {s - r} \right\rceil + 1,\; \cdots ,\;0, \;1, \cdots ,\left\lfloor {s + r} \right\rfloor ,\left( {s + r} \right)} \right\} \hfill \cr A = \left\{ {\alpha (n) = \arccos \left( {{{x(n) - s} \over r}} \right)} \right\} = \left\{ {\pi ,\;\arccos \left( {{{\left\lceil {s - r} \right\rceil - s} \over r}} \right),\; \cdots ,\; \arccos \left( {{{\left\lfloor {s + r} \right\rfloor - s} \over r}} \right),\;0} \right\} \hfill \cr} \right. $$ where the set $A$ is in non-increasing order, contrary to the others.

In this way, the arc corresponding to $q$ intersections will be individuated by the values of $x$ such that $$ \bbox[lightyellow] { x \in \left( {\left( { - q, - q + 1} \right] \cup \left[ {q,q + 1} \right)} \right) \cap \left[ {s - r,\;s + r} \right] } \tag{1}$$ so that we have in general two arcs, except
- at $q=0$ in which case we have just one range;
- (possibly) at the extremes , where the range could be void or of null measure, depending on the values of $r$ and $s$.

In an another perspective, by the above we are assigning a value $q$ to the intervals delimited by the points in $X$,
and correspondingly to the arcs delimited by the angles in $A$.
Thus we are constructing a measure of the angle $Ang(q,s;r)$ as the sum of one or two angles.

The position $s$ and the angle $\alpha$ are supposed independent and uniformly distributed, thus the
probability of having $N$ intersections
is given by $$ \bbox[lightyellow] { \eqalign{ & dP(q,\,s;\;r) = dP(q,\,1 - s;\;r)\quad \left| \matrix{ \;0 \le s < 1/2 \hfill \cr \;0 < r \hfill \cr \;0 \le q \in Z \hfill \cr} \right. = \cr & = {1 \over \pi }{{ds} \over {1/2}}\left( {\alpha \left( { - q} \right) - \alpha \left( { - q + 1} \right) + \alpha \left( q \right) - \alpha \left( {q + 1} \right)} \right) \cr} } \tag{2}$$

After that, since $$ \eqalign{ & \int {\arccos \left( {{{n - s} \over r}} \right)ds} = - r\int {\arccos \left( {{{n - s} \over r}} \right)d\left( {{{n - s} \over r}} \right)} = \cr & = r\left( {\sqrt {1 - \left( {{{n - s} \over r}} \right)^{\,2} } - \left( {{{n - s} \over r}} \right)\arccos \left( {{{n - s} \over r}} \right)} \right) \cr} $$ we can integrate the above for $0 \le s < 1/2$, with due consideration for the variation in $s$ of the intervals:
the $n$ indicated above may vary $\pm 1$ at varying $s$, which will require to split the integral.

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  • $\begingroup$ This is fantastic. The diagram is incredibly illustrative. Thank you for such a thorough explanation! $\endgroup$ – 0k33 Nov 27 '18 at 3:32
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    $\begingroup$ @Ok33: actually, in my previous version, I missed some important details: sorry. I re-casted the answer to deal more precisely with them. $\endgroup$ – G Cab Nov 28 '18 at 15:20
  • $\begingroup$ @G Cab thank you so much for the follow up! $\endgroup$ – 0k33 Nov 29 '18 at 23:58
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    $\begingroup$ @Ok33 : the argument was interesting for me as well, but it is a pleasure to help people so nice to leave a thank! $\endgroup$ – G Cab Nov 30 '18 at 0:32

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