1. Paint a $1\times 1$ square in blue.
2. Take $k$ points randomly and uniformly from the square.
3. Paint $k$ disks of area $1/k$ centered at each point in red.
What is the expected remaining blue area?
Remark: the sum of the areas of the disks is $k\cdot 1/k=1$, the same as the area of the square, but we have to account for:
On
Here is some C code using OpenGL to estimate numerically. And a plot of the results (I plotted different image sizes to get a sense of how well it approximates the true value). The limit seems to agree with Ross's analytic answer.
A naive approach that will be close for large $k$ is to say a point has $\frac {k-1}k$ chance of not being covered by a given disk, so $\left(\frac{k-1}k\right)^k$ chance of not being covered at all. The limit in large $k$ is that $\frac 1e$ is not covered. This is not quite right as points near the edge have lower chance of being covered. The corner points can only be covered if the center of the disk is in an area $\frac 1{4k}$ so the chance of coverage by one disk is $\frac {4k-1}{4k}$. Raising that to the $k$ power has limit $e^{-1/4}\approx 0.7788$ of not being covered. The side points have a chance of $e^{-1/2}$, but the area of reduced probability of coverage gets small as $k$ gets large.