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This is the coupon collector's problem.

http://en.wikipedia.org/wiki/Coupon_collector%27s_problem

The formula in the article (1/2 + n*gamma + n log n) gives 2365 for the expected number, but this is different than "at least 50% chance".

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I don't think this is exactly the same as the proposed problem. What it's giving you is the expected value. This includes in its calculation the lower probability cases and the higher probability cases. For instance, there's a slim change of this happening when the number of employees is 365. It's been a while since I took probability, but I think the expected value is going to be larger than the minimum value needed to achieve a .5 probability if the probability graph is skewed, which in this case it is.

Well... you made my day complete. My brute force approach got me 2364.646, glad to know my math intuition isn't entirely broken.

Thanks for the link!

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