Which exact-half count is likeliest?

Question

Consider three fair-coin experiments: 10 flips, 100 flips, and 1000 flips. In which is "exactly half heads" the most likely, and by roughly how much? Rank the three, no calculator.

Accepted Answer

For 2n flips, P(	ext{exactly } n 	ext{ heads}) \approx 	frac{1}{\sqrt{\pi n}}. The three experiments have n = 5, 50, and 500: P{10} \approx \frac{1}{\sqrt{5\pi}} = \frac{1}{\sqrt{15.7}} \approx 0.252, P{100} \approx \frac{1}{\sqrt{50\pi}} = \frac{1}{\sqrt{157}} \approx 0.0798, P{1000} \approx \frac{1}{\sqrt{500\pi}} = \frac{1}{\sqrt{1571}} \approx 0.0252. So fewer flips make exact-half more likely: 25\% at 10 flips, 8\% at 100, 2.5\% at 1000. Each tenfold increase in flips divides the probability by about \sqrt{10} \approx 3.16, which is exactly the gap you see (0.252 	o 0.0798 	o 0.0252…

Which exact-half count is likeliest?

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