Let l be the length of the needle, d be the distance between the parallel lines, x be the distance from the center of the needle to the closest line, and $\mathrm{\θ}$ be the acute angle between the needle and a line.

The uniform probability density function of x between 0 and $\frac{d}{2}$ is $\left\{\begin{array}{cc}\frac{2}{d}& 0 \le x \le \frac{d}{2}\\ 0& \mathrm{elsewhere}\end{array}\right.$.

The uniform probability density function of $\mathrm{\θ}$ between 0 and $\frac{\mathrm{\π}}{2}$ is $\left\{\begin{array}{cc}\frac{2}{\mathrm{\π}}& 0 \le \mathrm{\θ} \le \frac{\mathrm{\π}}{2}\\ 0& \mathrm{elsewhere}\end{array}\right.$.

So, the joint probability density function of the independent random variables x and $\mathrm{\θ}$ is the product of their individual probability density functions: $\left\{\begin{array}{cc}\frac{4}{d \mathrm{\π}}& 0 \le x \le \frac{d}{2}\, 0 \le \mathrm{\θ} \le \frac{\mathrm{\π}}{2}\\ 0& \mathrm{elsewhere}\end{array}\right.$.

The needle will cross a line only if  $x \le \frac{l}{2} \sin \left(\mathrm{\θ}\right)$.

This solution can be found simply by using an iterated integral. Assuming that $l \le d$, integrating the joint probability density function gives the probability that the needle will cross a line:

$P \= {\int }_0^{\frac{\mathrm{\π}}{2}} {\int }_0^{\frac{l}{2}\sin \left(\mathrm{\θ}\right)}\frac{4}{d \mathrm{\π}} \ⅆx \ⅆ\mathrm{\θ}$

$P\=\frac{2l}{d\mathrm{\π}}$

We can also calculate the probability, $P$, of the needle crossing a line as the product of $P_1$ and $P_2$, where $P_1$ is the probability that the center of the needle falls close enough to a line to possibly cross it and $P_2$ is the probability that the needle actually crosses the line, given that its center is within reach.

Let $l$ represent the length of the needle and $d$ represent the width of each piece of wood (that is, the distance between two lines).

The needle can possibly cross a line if its center is within $\frac{l}{2} \mathrm{units}$ of either side of the line. So, adding $\frac{l}{2} \+ \frac{l}{2}$ to account for the needle falling on either side of the line, then dividing by the total distance between this line and the next, $d$, we get

$P_1 \= \frac{\left(\frac{l}{2}\+\frac{l}{2}\right)}{d} \= \frac{l}{d}$.

Now, we assume the center is within reach of crossing a line, meaning it lies $\frac{l}{2}\mathrm{units}$ or less from a line.

Recall that the needle will cross a line for a given $x$ when $x \le \frac{l}{2} \sin \left(\mathrm{\θ}\right)$, or $\arcsin \left(\frac{2 x}{l}\right) \le \mathrm{\θ}$. The probability of this happening is thus $\frac{\frac{\mathrm{\π}}{2}-\arcsin \left(\frac{2 x}{l}\right)}{\frac{\mathrm{\π}}{2}-0}$, since we assume $\mathrm{\θ}$ ranges uniformly between 0 and $\frac{\mathrm{\π}}{2}$, independently of $x$. Taking the average overall possible values of $x$ between 0 and $\frac{l}{2}$, we find that:

$P_2 \= \frac{1}{\frac{l}{2}}{\int }_0^{\frac{l}{2}}\mathit{ }\frac{\frac{\mathrm{\pi }}{2}-\arcsin \left(\frac{2 x}{l}\right)}{\frac{\mathrm{\pi }}{2}-0}\ⅆx$

$P_2\=\frac{2}{\mathrm{\π}}$

Putting this all together, we obtain $P \= P_1 P_2 \= \frac{l}{d} \frac{2}{\mathrm{\π}} \= \frac{2 l}{d \mathrm{\π}}$.

The formula from the solution to the "short needle" case above can be rearranged to $\mathrm{\π}\=\frac{2 l}{d P}$.  So, if we conduct an experiment to estimate $P$, we can also find an approximation for $\mathrm{\π}$.

Let's say our experiment involves dropping N needles on the floor and we observe that n of them cross a line, so the observed probability of a needle crossing a line is $P \approx \frac{n}{N}$.

Thus, our approximation of π is: $\mathrm{\π} \approx \frac{2 l N}{d n}$.

This solution can also be found simply by using an iterated integral. Assuming that $l \> d$, integrating the joint probability density function gives us the probability that the needle will cross a line:

$P \&equals; {\int }_0^{\frac{\mathrm{\&pi;}}{2}} {\int }_0^{m\left(\mathrm{\&theta;}\right)}\frac{4}{d \mathrm{\&pi;}} \&DifferentialD;x \&DifferentialD;\mathrm{\&theta;}$, where $m\left(\mathrm{\&theta;}\right)$ is the minimum of $\left\{\left(\frac{l}{2}\right) \sin \left(\mathrm{\&theta;}\right) \&comma; \frac{d}{2}\right\}$. Splitting into the cases $d\&gt;l \sin \left(\mathrm{\&theta;}\right)$ and $d<l \sin \left(\mathrm{\theta }\right)$, we get:

$P \&equals; {\int }_0^{\arcsin \left(\frac{d}{l}\right)} {\int }_0^{\frac{l}{2}\sin \left(\mathrm{\theta }\right)} \frac{4}{d \mathrm{\&pi;}}\&DifferentialD;x \&DifferentialD;\mathrm{\theta }\&plus;{\int }_{\arcsin \left(\frac{d}{l}\right)}^{\frac{\mathrm{\pi }}{2}} {\int }_0^{\frac{d}{2}}\frac{4}{d \mathrm{\pi }} \&DifferentialD;x \&DifferentialD;\mathrm{\&theta;}$
$\&equals;{\int }_0^{\arcsin \left(\frac{d}{l}\right)}\frac{2 l}{d \mathrm{\&pi;}} \sin \left(\mathrm{\theta }\right) \&DifferentialD;\mathrm{\theta }\&plus;{\int }_{\arcsin \left(\frac{d}{l}\right)}^{\frac{\mathrm{\pi }}{2}} \frac{2}{\mathrm{\&pi;}} \&DifferentialD;\mathrm{\&theta;}$
= $\frac{2}{\mathrm{\&pi;}}\left(\frac{l}{d}\left(1-\sqrt{1-\frac{d^2}{l^2}}\right)\&plus;\frac{\mathrm{\&pi;}}{2}-\arcsin \left(\frac{d}{l}\right)\right)$
$\&equals; \frac{2}{\mathrm{\&pi;}}\left(\frac{l-\sqrt{l^2-d^2}}{d}\&plus;\arccos \left(\frac{d}{l}\right)\right)$.

Thus, when $l \&gt;d$, a Monte-Carlo style approximation of π is available via:

$\mathrm{\pi } \approx \frac{2 N}{n}\left(\frac{l-\sqrt{l^2-d^2}}{d}\&plus;\arccos \left(\frac{d}{l}\right)\right)$.