Two friends, equal short waits

Question

Two friends each arrive at a bench at a uniformly random time during a $45$ -minute window, independently, and each waits $15$ minutes for the other (never past the end of the window).

What is the probability they meet?

Accepted Answer

Let a, b be arrival minutes, each uniform on [0, 45]. They meet iff a - b \le 15. They miss iff a - b 15 \quad	ext{or}\quad b - a 15 , two congruent triangles with legs 45 - 15 = 30 inside the 45 	imes 45 square: P(	ext{miss}) = \frac{2 \cdot 	frac12 (30)^2}{45^2} = \frac{900}{2025} = \frac{4}{9} . So P(	ext{meet}) = 1 - 	frac49 = 	frac59 \approx 0.556. Takeaway. Equal waits make the two miss triangles congruent, recovering the classic 1 - \left(1 - 	frac{w}{T}ight)^2. Here 1 - \left(	frac{30}{45}ight)^2 = 	frac59.

Two friends, equal short waits

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