This problem comes from an old math contest which has the answer, but no solution.
Two friends get home at 2:30 P.M. They both go to bed 10 P.M. If both friends will go to a park twice for 15 minutes, what is the probability that they will be able to see each other at any random time between the two time intervals? Assume that the travel to and from the park takes no time.
The standard method to determine the probability of meeting if they go to the park once yields $\frac{30^2-29^2}{30^2} = \frac{59}{900}$ (by considering the area of $|y-x| \leq 1$ within the square $0 \leq x \leq 30, 0 \leq y \leq 30$). How would you generalize this geometric argument to handle going to the park twice?
The listed answer is $\frac{29}{225}$. I noticed that this is the same as $\frac{30^2-28^2}{30^2} = \frac{15^2-14^2}{15^2}$, which is the probability that they see each other assuming they go to the park just once, but staying for 30 minutes (ie like the first situation, but going back-to-back). It seems strange to me that the probability would work out to this; it seems like this is a lower bound on the probability, since they could also meet within a 15-minute interval, and go another time completely removed from this time interval. Is the listed answer correct, and why? Or is there a problem with the answer?
