Buffon's experiment with squares Say, we'd like to make the Buffon's experiment but with squares instead of needles.
Notation:


*

*$d$ is the distance between lines

*$b$ is the square side length

*$y$ is the distance from the center of the square to the nearest line

*$\alpha$ is the acute angle between one of the diagonals and the vertical line


My attempt to illustrate it:

Now we note that $y$ is uniformly distributed in $[0,\frac{d}{2}]$  and $\alpha$ in $[0,\frac{\pi}{4}]$. Furthermore, the square crosses the closest line if $$y\leq \frac{\sqrt{2}}{2}b \cos\ \alpha$$
Now if $E$ is the the event when the square crosses a line then: $$P(E)=\frac{\int_0^{\pi/4} \frac{\sqrt{2}}{2}b \cos\alpha \ \text{d}\alpha}{\frac{d}{2}\frac{\pi}{4}}=\frac{4b}{\pi d}$$
Therefore $$\pi= \frac{4b}{P(E)d}$$
Does this make sense?
Unfortunately, my textbook gives a different answer: $$\pi= \frac{4b(\sqrt{2}-1)}{P(E)d}$$
Any clarifications or alternative solutions are highly appreciated.
 A: You definitely need to know something about the relative sizes of $b$ and $d$. The discrepancy between your answer and the book's answer may have something to do with this.
If $b \leq \frac{d}{\sqrt{2}}$, the boundary of the square intersects the line in either $0$ or $2$ places with probability $1$, and $P(E)$ is the probability that it intersects in $2$ places. The Buffon's noodle argument tells us that a curve of length $L$ expects to intersect lines at distance $d$ in $\frac{2L}{\pi d}$ places. As the perimeter of the square is of length $4b$, the expected number of intersections is $\frac{8b}{\pi d}$; since those intersections always come in pairs, the probability of intersection is $\frac{4b}{\pi d}$ as you've calculated.
On the other hand, if $b > \frac{d}{\sqrt{2}}$, the boundary of the square could intersect the lines in $0$, $2$, $4$, or maybe even more places. So knowing the expected number of intersections is no longer good enough to tell us anything about the probability (though it does give a bound: certainly the probability of intersection is still at most $\frac{4b}{\pi d}$).
