Two friends and a ten-minute wait

Question

Two friends agree to meet at a cafe, each arriving at a uniformly random time in a 30-minute window, independently. Whoever comes first waits up to 10 minutes for the other.

What is the probability that they meet?

Accepted Answer

Drop the pair of arrival times as a point (x, y) uniformly into a 30-by-30 square; all points are equally likely, so probability is area based. They meet when their arrivals are within 10 minutes: x - y \le 10. The complement is the two corner triangles where the gap exceeds 10 minutes, each with legs 30 - 10 = 20. Their combined area is 2 \cdot \frac{1}{2}(20)(20) = 20^2 = 400. The whole square has area 30^2 = 900, so P(	ext{meet}) = 1 - \frac{400}{900} = 1 - \left(\frac{20}{30}ight)^2 = 1 - \frac{4}{9} = \frac{5}{9} \approx 0.556. Compared with the traders' 15-minute wait in an hour, her…

Two friends and a ten-minute wait

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