Buffon's Needle Revisited

The answer is exactly $3/\pi \approx \boxed{0.955}$ .

Let $\theta$ be the angle between the needle and the horizontal in counterclockwise direction. Without loss of generality we can limit ourselves to the case $0 \leq \theta < \pi/2$ . Also, without loss of generality we can choose the origin of a coordinate system, $(0,0)$ , to be the grid crossing to the left and below the bottom-left end of the needle. Then the coordinates of the endpoints of the needle are $(x,y);\ \ (x+\cos\theta, y + \sin\theta).$ Let $C(x,y,\theta) = 1$ if the needle crosses a line of the grid (which will be the line $x = 1$ or $y = 1$ , or both), and $= 0$ if it doesn't. Then $C(x,y,\delta) = \begin{cases} 0 & \text{if}\ x \geq 1-\cos \theta\ \text{and}\ y \geq 1-\sin \theta \\ 1 & \text{otherwise} \end{cases}$ The probability for any given value of $\delta$ is $\mathbb P(\theta) = \int_0^1 \int_0^1 C(x,y,\theta)\:dx\:dy = 1 - (1-\cos \theta)(1 - \sin \theta) \\ = \cos\theta + \sin\theta - \cos\theta\sin\theta = \sqrt 2\sin(\theta + \tfrac\pi 4) - \tfrac12\sin 2\theta.$ We take the average over all angles by integration: $\mathbb P = \frac 1{\pi/2} \int_0^{\pi/2} \mathbb P(\theta)\:d\theta = \frac 2\pi\left(\int_0^{\pi/2} \sin(\theta + \tfrac\pi 4)\:d\theta - \tfrac12\int_0^{\pi/2} \sin 2\theta\:d\theta \right) \\ = \frac 2\pi\left[-\sqrt 2\cos(\theta + \tfrac\pi 4) + \tfrac14\cos 2\theta \right]_0^{\pi/2} = \frac 2\pi\left(-\sqrt 2(-\tfrac12\sqrt 2 - \tfrac12\sqrt2) + \tfrac14(-1 - 1)\right) = \frac 2\pi(2 - \tfrac12) = \boxed{\dfrac 3\pi}.$