How long does it take for rain drops cover a square? Consider a square divided into $n \times n$ grid. Each rain drop falls from the sky and covers an $r \times r$ grid chosen uniformly at random within the square. What is the expected number of rain drops needed to cover the whole square? 
To avoid boundary conditions, we can assume that the square is actually a torus. In other words, the top and the bottom of the square are glued together, while the left side and the right side of the squares are guled together.
If $r=1$, then this is the classical coupon collector problem. It also gives an upper bound of $n^2 \log(n^2)$ for this rain drop model. I wonder if this model has already been studied? It seems to be natural extension of coupon collector.
 A: I ran $N = 10^5$ simulations for the case $r = 2$, $n = 8$, and the result fit extremely well to an inverse gamma distribution with parameters $(a, b) = (14.6313, 987.144)$, where I have used the parametrization $$f_X(x) =  \frac{e^{-b/x} \left(b/x\right)^a}{x \Gamma (a)}, \quad x > 0.$$  This would give a mean and variance of $\mu = 72.4176$, $\sigma^2 = 415.185$.  
Of course, we should not expect this to be the same as $r = 1$, $n = 4$, which is the classic coupon collector's problem. 
A simulation with $r = 3$, $n = 8$ gives another good inverse gamma fit, with $(a,b) = (12.1359, 329.206)$.
Perhaps someone else can find some relationship between $r, n$ and $a, b$, but I believe that the distribution of the number of drops needed is well-modeled as inverse gamma, even though I have no theoretical basis for it.
I am curious what would happen in the continuous case, with $\rho = r/n = 1/4$ but $n \to \infty$.  My intuition says that in the asymptotic case, the expectation and variance increase.
