How long must they wait for a coin-flip meeting?

Question

Two friends each arrive at a time uniform on the hour (noon to 1pm), independently, and each waits the same amount of time $w$ before leaving. They want the odds of meeting to be a coin flip.

For what waiting time $w$ (in minutes) is the probability they meet exactly $\tfrac{1}{2}$ ?

Accepted Answer

With equal waits w in a window T = 60, the meeting probability is P(	ext{meet}) = 1 - \left(1 - \frac{w}{T}ight)^2. Set it equal to 	frac12 and solve: 1 - \left(1 - \frac{w}{60}ight)^2 = \frac{1}{2} \implies \left(1 - \frac{w}{60}ight)^2 = \frac{1}{2} \implies 1 - \frac{w}{60} = \frac{1}{\sqrt{2}} \approx 0.7071. So \frac{w}{60} = 1 - 0.7071 = 0.2929 \implies w = 60 	imes 0.2929 \approx 17.57 	ext{ minutes}. That is about 17 minutes and 34 seconds. A quick check: at w = 15 the probability is \frac{7}{16} = 0.4375 (below half), and at w = 20 it is \frac{5}{9} \approx 0.556 (above hal…

How long must they wait for a coin-flip meeting?

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