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Easiest way to approach this problem without dealing with nasty factorials

Did
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adam
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    http://en.wikipedia.org/wiki/Birthday_problem – lab bhattacharjee Nov 25 '13 at 07:55
  • Already asked on the site, perhaps even as many times as the answer (23). – Did Nov 25 '13 at 08:09
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    Deal with nice factorials instead. – copper.hat Nov 25 '13 at 08:15
  • I don't think it was fair to mark this question as a duplicate. adam asked for a solution without factorials, which the "duplicate" question does not. This was clearly not a useless question, as I presented a solution which uses no factorials. Unfortunately I do not have the reputation to cast a re-open vote... – dinoboy Nov 26 '13 at 21:26

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Suppose a year has $n$ days. Now if we have $k$ people in a room the probability no two people have the same birthday is: $$ p = \prod_{i=1}^{k}\left ( 1 - \frac{i-1}{n} \right ) \ge \prod_{i=1}^{k} e^{\frac{-i+1}{n}} = \text{exp} \left ( \frac{k(k-1)}{2n} \right )$$

So we need roughly $\sqrt{2n \ln 2}$ people. This is pretty accurate for large $n$, as then our approximations get closer and closer to correct.

dinoboy
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