Strangers until one shares your birth-hour of the week

Question

Label each person by the hour of the week they were born, one of $168$ equally likely slots ( $7$ days times $24$ hours). Yours is fixed.

How many random strangers do you need before it is more likely than not that at least one shares your hour-of-the-week slot?

Accepted Answer

You are matching each stranger against your one fixed slot, so use the complement, nobody matches you. With n strangers, each missing your slot with probability \frac{167}{168}, P(	ext{no match}) = \left(\frac{167}{168}ight)^{n}. We want this to drop below 	frac12: \left(\frac{167}{168}ight)^{n} < 	frac12. Take logarithms: n \frac{\ln 0.5}{\ln(167/168)} = \frac{-0.6931}{-0.005970} \approx 116.1, so you need n = 117 strangers. A useful rule: matching a fixed 1-in-d target reaches even odds at about n \approx d \ln 2 \approx 0.693\,d. Here 0.693 	imes 168 \approx 116.4, rounding up to 1…

Strangers until one shares your birth-hour of the week

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