Odds of a near-birthday in a group of ten

Question

In a group of 10 people (birthdays independent and uniform over 365 days), call two people a near-match if their birthdays are within 1 day of each other. What is the probability that at least one near-match exists in the group?

Accepted Answer

Start with the per-pair chance. Fixing one person's day, the other near-matches on that day or the two adjacent days, 3 of 365: p = \frac{3}{365} \approx 0.00822. There are \binom{10}{2} = 45 pairs. Treating the pairs as roughly independent (a good approximation), the chance of no near-match anywhere is P(	ext{no near-match}) \approx e^{-45\,p} = e^{-45 	imes 0.00822} = e^{-0.370} \approx 0.691. Therefore P(	ext{at least one near-match}) \approx 1 - 0.691 \approx 0.309, about 31\%. An equivalent quick view: the expected number of near-match pairs is 45 	imes 3/365 \approx 0.37, and "at le…

Odds of a near-birthday in a group of ten

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