In the center of a rectangular billiard table that is 3 m long and 1 m wide, there is a billiard ball. It is struck by a cue in a random direction. After the strike, the ball stops after traveling exactly 2 m. Find the expected number of bounces off the table edges.
Let's represent a grid with a cell of 3 m $\times$ 1 m. The central cell of the grid is a table, highlighted in light green. We need to extend the grid in all directions so that it can accommodate a circle with a radius of 2 m, centered at the center of the table $O$.

We will allow the ball to "pass through the table edges," rolling straight along the grid. Reflection from the table edge corresponds to crossing one grid line. Clearly, we can consider only one quarter of the circle, i.e., all directions leading the ball to one of the points on the arc $A K$.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-30.jpg?height=1132&width=1354&top_left_y=0&top_left_x=352)

From the figure, it is clear that there are two intersections if the ball hits the arc $A B$ or the arc $C L$. If the ball hits the arc $B C$ or $K L$, there is one intersection.

$\smile A B=\arccos \frac{O M}{O B}=\arccos \frac{3}{4}, \smile C L=\arccos \frac{O P}{O L}-\arcsin \frac{O N}{O C}=\arccos \frac{1}{4}-\arcsin \frac{3}{4}$.

Thus, the probability $p_{2}$ of two intersections is $\frac{2}{\pi}\left(\arccos \frac{3}{4}+\arccos \frac{1}{4}-\arcsin \frac{3}{4}\right) \approx 0.7594$.

The probability $p_{1}$ of one intersection is $1-p_{2}$.

The expected number of intersections is $p_{1}+2 p_{2}=1+p_{2}=1+\frac{2}{\pi}\left(\arccos \frac{3}{4}+\arccos \frac{1}{4}-\arcsin \frac{3}{4}\right) \approx 1.7594$.

## Answer

$1+\frac{2}{\pi}\left(\arccos \frac{3}{4}+\arccos \frac{1}{4}-\arcsin \frac{3}{4}\right) \approx 1.7594$