Chance of n heads in 2n tosses

Richard Hamming discusses an interesting example his excellent book The Art of Probability:

What is the probability of getting exactly \(n\) heads in \(2n\) tosses.

I like this example because I suspect most people would think that the chance of exactly half the tosses turning up as heads would grow with \(n\).¹  Of course, in reality, the opposite is the case. Moreover, even for small values of \(n\), the chance that heads and tails exactly balance is not as high as one might think.

The analytic expression to calculate this probability is straightforward:

$$ \text{P}(n \text{ heads in } 2n \text{ tosses}) = \frac{\binom{2n}{n}}{2^{2n}} = \big(2n!\big)/\big({n!n!} 2^{2n}\big) $$ Using Stirling’s approximation to simplify the expression yields:

$$ \text{P}(n \text{ heads in } 2n \text{ tosses}) \sim 1/\sqrt{\pi n} $$

For \(n = 8 \), the probability is ~ 20%, and tends to drop-off rapidly as the number of tosses grows. In general, it is often the case that exact outcomes tend to be unlikely.