The birthday problem

Question

There are 23 people in a room. Assume birthdays are independent and uniform over 365 days (ignore leap years). What is the probability that at least two people share a birthday? Why is 23 the famous threshold?

Accepted Answer

Use the complement: all birthdays distinct. Seat people one at a time; each must avoid all previous birthdays: P(	ext{all distinct}) = \frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}\cdots\frac{343}{365} = \prod{k=0}^{22}\frac{365-k}{365} \approx 0.4927. Therefore P(	ext{shared birthday}) = 1 - 0.4927 \approx 0.507. Just over one half, so 23 is the smallest group size at which a match is more likely than not (with 22 people the match probability is \approx 0.476). Why so few? The right unit is pairs, not people. With n = 23 people there are \binom{23}{2} = 253 pairs, each matching w…

The birthday problem

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