Two traders each arrive at a uniformly random time between 12:00 and 13:00, independently, and each waits 15 minutes before leaving. What is the probability that they meet?

Two traders each arrive at a uniformly random time between 12:00 and 13:00, independently, and each waits 15 minutes before leaving. What is the probability that they meet?

Approach: Plot the two arrival times as a point in the unit square, shade the band where the times differ by less than a quarter of an hour, then compare that area with the whole square.

7/16. This is a geometric probability: scale the hour to 1 and let X and Y be the two uniform arrival times, so the pair is uniform on the unit square. They meet when |X - Y| <= 1/4. The complement is two right triangles with legs 3/4, of total area 2 * (1/2)(3/4)^2 = 9/16, so the meeting probability is 1 - 9/16 = 7/16 = 0.4375. With a wait of w hours the answer is 1 - (1 - w)^2, whose derivative at w = 0 is 2, so the first five minutes of patience buy about twice what an average five minutes buys and more than twenty times what the last five minutes buy.

Follow-up: How long must each trader wait for the meeting probability to reach 90%?

Key concepts: geometric probability, unit square, complement, uniform arrival times.