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X unbiased coins. What is the probability that half of them exactly are heads. Please answer question where X= 20, 30, and 20000.

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marked as duplicate by MJD, Amzoti, Thomas Andrews, Jared, Cameron Buie Aug 2 '13 at 5:32

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I don't think there's "half" of them for $X = 3$. For the $X = 2$ case, we just find the probability of $HT$ and $TH$, making it $\frac{1}{2}$. For 20,000, though, it's ${N \choose N/2}2^{-N}$, where $N$ is 20,000.

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