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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!

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
$$
x = nquad left| {;leftlceil {s - r} rightrceil le n le leftlfloor {s + r} rightrfloor } right.
$$

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
$$
left{ matrix{
N = left{ {nquad left| {;leftlceil {s - r} rightrceil - 1 le n le leftlfloor {s + r} rightrfloor + 1} right.} right} hfill cr
X = left{ {x(n)} right} = left{ {left( {s - r} right),;leftlceil {s - r} rightrceil ,;leftlceil {s - r} rightrceil + 1,; cdots ,;0,
;1, cdots ,leftlfloor {s + r} rightrfloor ,left( {s + r} right)} right} hfill cr
A = left{ {alpha (n) = arccos left( {{{x(n) - s} over r}} right)} right}
= left{ {pi ,;arccos left( {{{leftlceil {s - r} rightrceil - s} over r}} right),; cdots ,;
arccos left( {{{leftlfloor {s + r} rightrfloor - s} over r}} right),;0} right} hfill cr} right.
$$
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
$$
eqalign{
& int {arccos left( {{{n - s} over r}} right)ds} = - rint {arccos left( {{{n - s} over r}} right)dleft( {{{n - s} over r}} right)} = cr
& = rleft( {sqrt {1 - left( {{{n - s} over r}} right)^{,2} } - left( {{{n - s} over r}} right)arccos left( {{{n - s} over r}} right)} right) cr}
$$
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.