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Birthday problem - Wikipedia, the free encyclopedia

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Saved by 10 people (-4 private), first by anonymouse user on 2006-11-24


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For 57 or more people, the probability is more than 99%, tending toward 100% as the pool of people grows

Highlighted by johnwprior

a group of 23 people there are 23*22/2=253 pairs

Highlighted by johnwprior

with one person they have 365 opportunities to have a different birthday. The second person only has 364 possibilities to have a different birthday than the first person. The third person has 363 days, and so on.

Highlighted by johnwprior

\bar p(n) = 1 \cdot \left(1-\frac{1}{365}\right) \cdot \left(1-\frac{2}{365}\right) \cdots \left(1-\frac{n-1}{365}\right) = { 365 \cdot 364 \cdots (365-n+1) \over 365^n } = { 365! \over 365^n (365-n)!}

Highlighted by johnwprior