Buffon needle problem , scenario $\ell>d$ suppose we have the classic problem of buffon's needle , let $\ell$ be the length of the needle and $d$ the distance between the parallel lines . I have solved the case which $\ell \leq d$ and i understand why $P(\text{needle cross the line})= \frac{2\ell}{\pi d}$. I know this doesn't work for $\ell>d$ because we can have the last probability $> 1$ for $\ell>\frac{\pi d}{2}$. But i cannot understand what doesn't work geometrically.

This is not a full answer, but only an explanation of the integral appearing in the wikipedia article referenced above.

The integral must be slipt in two : $$p= \frac{4}{t\pi}\int_{\theta=0}^{\pi/2} \int_{x=0}^{m(\theta}dx\:d\theta$$

$$p= \frac{4}{t\pi}\int_{\theta=0}^{\pi/2} \int_{x=0}^{ \frac{l}{2} \sin^{-1}(t/l)}\sin(\theta)dx\:d\theta + \frac{4}{t\pi}\int_{\theta=0}^{\pi/2} \int_{x=\frac{l}{2}\sin^{-1}(t/l)}^{t/2}dx\:d\theta $$

$$p=\frac{2l}{t\pi}\left[-\cos(\theta)\right]_{0}^{\sin^{-1}(t/l)} +\frac{2}{\pi}\left(\frac{\pi}{2}-\sin^{-1}(t/l)\right)$$

$$p=\frac{2l}{t\pi}\left(1-\sqrt{1-\frac{t^2}{l^2}}\right) +1-\frac{2}{\pi}\sin^{-1}\left(\frac{t}{l}\right)$$