Suppose we are in an arena of height $H$ and width $L$, with paddles of width $w$ parallel to the height on either side. I'll try to attach a picture in just a bit. Suppose that we shoot a ball (that's infinitely small, a point) at a uniformly random angle $\theta\in[0,\pi]$ from the left paddle to the right paddle. Specifically, suppose that we shoot the ball from a height $h$ off the left side, and we are trying to hit a paddle of width $w$ on the right side, whose center is at height $p$ off the bottom wall.
First, let's consider directly hitting the paddle without bounces. To find the range $\Theta$ of hitting angles, notice that we get a triangle formed with height $L$ and width $w$. Using the two formulas we know for the area of a triangle, we get:
$\frac{Lw}{2}=\frac{r_1r_2\sin(\Theta)}{2}$
where $r_1=\sqrt{L^2+(h-(p-w/2))^2}$ and $r_2=\sqrt{L^2+(h+(p-w/2))^2}$, where we can now solve for $\Theta$ giving:
$\Theta=\arcsin\left(\frac{Lw}{r_1r_2}\right)$.
Now we need to figure out all other possible ways to hit the paddle, with bounces. This becomes easy if we stack arena's one atop the other in a periodic fashion, where we see that hitting the paddle is equivalent to crossing arena's. In other words, we need to find $\theta$ angles which hit paddles centered at $p+kH$ where $k\in\mathbb{Z}$. The idea is exactly the same as before, and we get:
$\Theta_k=\arcsin\left(\frac{Lw}{r_1(k)r_2(k)}\right)$
where now every instance of $p$ in the definition of $r_1,r_2$ is replaced by $p(k)=p+kH$. Thus the probability of hitting the paddle will be the sum over all these chunks (convince yourself that they are disjoint geometrically) normalized by $\pi$:
$\frac{\sum_{k=-\infty}^\infty\Theta_k}{\pi}$
which I am not sure how to further simplify...