Probability of exactly 50 heads in 100 flips

Question

You flip a fair coin 100 times. Estimate the probability of getting exactly 50 heads. Derive the approximation, no calculator.

Accepted Answer

The exact answer is P = \binom{100}{50}\left(\frac{1}{2}\right)^{100}. Apply Stirling, n! \approx \sqrt{2\pi n}\,(n/e)^n, to the central binomial coefficient with n = 50: \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \approx \frac{\sqrt{4\pi n}\,(2n/e)^{2n}}{2\pi n\,(n/e)^{2n}} = \frac{4^n}{\sqrt{\pi n}}. Therefore P = \frac{\binom{100}{50}}{2^{100}} \approx \frac{1}{\sqrt{\pi \cdot 50}} = \frac{1}{\sqrt{157.1}} \approx 0.0798. The exact value is 0.0796, the estimate is good to three decimal places. So "exactly half heads" happens only about 8% of the time: the most likely single count, yet still rare…

Probability of exactly 50 heads in 100 flips

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