The Meeting Problem

Two friends agree to meet at a cafe sometime between noon and one o'clock. Each arrives at a uniformly random minute in that hour, independently, and waits five minutes for the other before leaving. What is the probability that they meet?

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#Map it to a square

Let $X$ and $Y$ be the two arrival times in minutes after noon, each uniform on $[0,60]$ and independent. The pair $(X,Y)$ is then uniform over the square $[0,60]^2$ , so every probability is an area divided by $60^2$ . The friends meet exactly when their arrivals fall within five minutes of each other,

\lvert X - Y \rvert \le 5. \tag{1}

#Measure the band by its complement

The meeting set is the diagonal band $\lvert X - Y \rvert \le 5$ . The miss set is cleaner to measure, two right triangles where one friend arrives more than five minutes before the other. Each triangle has legs $60 - 5 = 55$ , so the misses cover

2 \cdot \tfrac{1}{2} \cdot 55^2 = 55^2 \quad\Longrightarrow\quad \PP(\text{miss}) = \frac{55^2}{60^2} = \left(\frac{11}{12}\right)^2 = \frac{121}{144}. \tag{2}

The pair meets only inside the diagonal band where the arrivals differ by at most five minutes. The two miss triangles each have legs 55, so the misses total (55/60)^2 and the meeting chance is 23/144.

#Read it off

\PP(\text{meet}) = 1 - \frac{121}{144} = \frac{23}{144} \approx 0.16. \tag{3}

Just under one chance in six. The five-minute grace is a twelfth of the hour, and the band straddles the diagonal on both sides, which is why the answer lands near twice that twelfth rather than equal to it.