Just two tennis balls!

Suppose that both balls have radius $d$ . If their centres are a distance $R$ apart, then the range of angle through which one ball can be kicked in order to hit the target ball is $4\alpha$ , where $\alpha$ is the angle between the line of centres and the tangent from the centre of the first ball to the second. Then $\alpha = \sin^{-1}\tfrac{d}{R}$ .

Suppose now that the room has side length $2X$ . The target ball $A$ is placed at the centre of the room. If the other ball $B$ is placed randomly (and uniformly) along one wall, and its (signed) distance from the centre of that wall is $x$ , then the distance between the centres of these two balls is $\sqrt{(X-d)^2 + x^2}$ , and $x$ is uniformly distributed over $[-(X-d),(X-d)]$ . Note that the centre of the ball $B$ cannot be at the wall's edge, due to its radius, and so we have to use $X-d$ instead of $X$ . Since ball $B$ is kicked away from the wall, the probability that the two balls collide from this position is $\frac{4}{\pi}\sin^{-1}\left(\frac{d}{\sqrt{(X-d)^2 + x^2}}\right)$ and so, conditioning on the starting position of ball $B$ , the probability that they collide is $p \; = \; \frac{1}{2(X-d)}\int_{-(X-d)}^{X-d} \frac{4}{\pi}\sin^{-1}\left(\frac{d}{\sqrt{(X-d)^2 + x^2}}\right)\,dx \; = \; \frac{4}{\pi(X-d)}\int_0^{X-d}\sin^{-1}\left(\frac{d}{\sqrt{(X-d)^2 + x^2}}\right)\,dx$ The substitution $x = (X-d)\tan\theta$ and $\beta = \tfrac{d}{X-d}$ gives us $p \; = \; \frac{4}{\pi}\int_0^{\frac14\pi} \sin^{-1}\left(\frac{d}{X-d}\cos\theta\right)\sec^2\theta\,d\theta \; = \; \frac{4}{\pi}\int_0^{\frac14\pi} \sin^{-1}(\beta\cos\theta)\sec^2\theta\,d\theta$ In this problem, $X = 3.75$ and $d = 0.04$ , and so $\beta = 0.0107817$ .

This integral for $p$ can be evaluated in terms of $F_1$ , one of the Appell two-variable hypergeometric functions, but this is not very informative. The integral for $p$ can be performed numerically, obtaining the answer $\boxed{0.0120994}$ . Alternatively, the fact that $\beta$ is small means that we can use the Maclaurin series expansion of $\sin^{-1}$ to obtain a series expansion for $p$ . All of the terms of this series are integrals which can be performed exactly, so we obtain the approximation $p \approx \tfrac{4}{\pi}\beta \ln(\sqrt{2}+1) + \tfrac{\sqrt{2}}{3\pi}\beta^3 + \tfrac{5\sqrt{2}}{8\pi}\beta^5$ which gives us $7$ DP accuracy.