Imagine that an $\epsilon$-radius hole is punched in the plane centered on every integer-coordinate point. Now a point "ball" is dropped on the plane at a random spot $p$. If $p$ has not already dropped through a hole, the plane is tilted in a random direction, and the ball rolls along the gradient vector $u$.

What is the expected length $L(\epsilon)$ of its roll before it falls through a hole?

If $p$ is in a hole upon generation, then $L=0$.
Otherwise $L$ is determined by the first encounter with a hole,
i.e., the ray from $p$ along direction $u$ passes within distance
$\epsilon$ of some hole center $c$.
For example, here $\epsilon=\frac{1}{4}$ and $c=(2,1)$, with $L
\approx 1.6$:

I am particularly interested in the growth rate of $L(\epsilon)$.

Intuition from another viewpoint suggests $\sim \frac{1}{\epsilon}$. Shrink the holes to points, and grow the ball to an $\epsilon$-radius disk. This sweeps out a rectangle of area $2 \epsilon L$ for a roll of length $L$. Very crudely, when this area reaches $1$ (or $1-\pi \epsilon^2$), I would expect it to include a lattice point. So perhaps $L \approx \frac{1}{2\epsilon}$. But this reasoning is surely not sound.

Perhaps there is an approach that employs techniques from rational approximations? All ideas welcomed—Thanks!

**Addendum**. Here are two histograms for $L(\epsilon)$ with $\epsilon=\frac{1}{4}$.
The first is of 200 random points, the second of 300 random points.

They begin to illustrate Doug Zare's point that very long paths occur not infrequently.
In the 300-point example, one path has $L \approx 63$, while more than half the paths
have $L \le 1$. The probability that $L \le 10$ is $>98$%, even though the expected
length is $\infty$!