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Find the smallest number of people you need to choose at random so that the probability that at least one has a birthday today exceeds 1/2.

About all I have so far is that (assuming a 365 day calendar year), the probability of a person having a birthday today is 1/365 and the probability of the opposite is 364/365. Where do I go from here?

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HINT: Suppose you choose $n$ people. If each person has probability $1-p$ of failing to have a birthday today, what is the probability that all $n$ people fail to have a birthday today?

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  • $\begingroup$ by the product rule, wouldn't it be (1-p)^n? (I'm sorry, our professor didn't do a stellar job covering this topic, and our book is a little confusing to read, for me at least) $\endgroup$
    – user266529
    Commented Oct 26, 2015 at 4:42
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    $\begingroup$ Ah, I see. So it would be N people when 1 - (364/365)^N > 1/2. $\endgroup$
    – user266529
    Commented Oct 26, 2015 at 4:55
  • $\begingroup$ That would require $N \gt \frac{\log(1-1/2)}{\log(1 - 1/365)}$ so $N=253$ or more. It is not a coincidence that $253=\frac{23(23-1)}{2}$ $\endgroup$
    – Henry
    Commented Aug 13, 2022 at 11:31

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