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  "body": "\n## Example\n\nAssume that each person has a birthday on one of $365$ days, and that each day is equally likely.\nAssume also that the birthdays of different people are independent.\nLet $p(n)$ be the probability that among $n$ people, at least two have the same birthday.\nBy the [[rule for the complement|0PQEBTYDZM]], for $1 \\le n \\le 365$,\n$$p(n) = 1 - \\frac{365 \\cdot 364 \\cdots (365 - n + 1)}{365^n}.$$\nThen $p(22) \\approx 0.476$ and $p(23) \\approx 0.507$. So $23$ people are enough for a probability above $1/2$.\n\n## Remarks\n\nThe answer is surprising because there are many pairs: $23$ people form $\\binom{23}{2} = 253$ pairs.\nFor $n$ much smaller than $365$, $p(n) \\approx 1 - e^{-n(n-1)/730}$.\nFor $n = 70$, $p(n) \u003e 0.999$. For $n \\ge 366$, $p(n) = 1$ by the pigeonhole principle.\n\n## Explanations\n\n### L1 human\n\nIt is easier to compute the chance that all birthdays are different.\nThe second person must avoid $1$ day, the third person must avoid $2$ days, and so on. These small losses multiply quickly.\n",
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