Buffon's needle: expected number of intersections & pmf when $l > d$ 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!
 A: 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

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.
