A rectangular grid with cylindrical holes of radius r is…

A rectangular grid with cylindrical holes of radius $r$ is given; the distances between the axes of neighbouring holes are $a$ and $b$. Determine the probability that a ball of radius $d$ falls into a hole in one throw without aiming, if the ball falls perpendicular to the plane of the grid.

Step 1: One cell is enough

The grid is periodic: every $a\times b$ rectangle around a hole looks the same. The centre of the ball lands uniformly in such a cell, of area $ab$ (we assume the holes do not overlap, $2r\le\min(a,b)$).

Step 2: When does the ball fall through?

The ball (radius $d$) passes through the hole (radius $r$) if the whole ball's cross-section fits inside the hole: the distance from the ball's centre to the hole's axis must be at most $r-d$. This is only possible if $d\lt r$.

Step 3: The probability

The favourable region is a disc of radius $r-d$, area $\pi(r-d)^2$:

\[ \boxed{P = \frac{\pi(r-d)^2}{ab} \quad (d\lt r), \qquad P=0 \quad (d\ge r).} \]

Step 4: Example

Holes of radius $r=2$ cm with axes $a=b=5$ cm apart, ball of radius $d=1$ cm:

\[ P = \frac{\pi\cdot1^2}{25}\approx0.126. \]

A smaller ball, $d=0.5$ cm, gives $\pi\cdot1.5^2/25\approx0.283$.

Step 5: Why subtract $d$?

A common mistake is to use the full hole area $\pi r^2$: that would be correct for a point-sized ball. A ball of radius $d$ touching the edge of the hole is stopped, so its centre must stay at least $d$ away from the edge.