, 2018

Solution:
Answer: $\frac{\pi+8}{16}$
Consider the two intersection points of the circle and the square, which are either on the same side of the square or adjacent sides of the square. In order for the circle to intersect a side of the square twice, it must be at distance at most $1$ from that side and at least $1$ from all other sides. The region of points where the center could be forms a $2 \times 1$ rectangle.

In the other case, a square intersects a pair of adjacent sides once each if it is at distance at most one from the corner, so that the circle contains the corner. The region of points where the center could be is a quarter-circle of radius $1$.

The total area of the regions where the center could be is $\pi+8$, so the probability is $\frac{\pi+8}{16}$.