Do Not Invite Sumo Wrestlers To Your Playground

We can also derive this formula by integrating $n$ variables, too. Check the $n$ Sumos counter-clockwise, if they stand on the same semi-circle, then there are $n!$ ways for them to stand. In each way, for the most right person, the angle corresponding to his position (denoted as $\theta_{1}$ ) ranges from 0 to 2 $\pi$ . the angle of the second person, $\theta_{2}$ ranges from $\theta_{1}$ to $\theta_{1}+\pi$ . With the same logic, the $i^{th}$ angle ranges from $\theta_{i-1}$ to $\theta_{1}+\pi$ . Since each angle $\theta_{i}$ can vary in a interval with width $2\pi$ . The "total area" is $(2\pi)^{n}$ and the probability that the $n$ people stand on the same semi-circle is $n!\times \frac{\int_{0}^{2\pi}d\theta_{1}\int_{\theta_{1}\le \theta_{2}\le \cdots \theta_{n}\le \theta_{1}+\pi}d\theta_{1}d\theta_{2}\cdots d\theta_{n}}{(2\pi)^{n}}$ .

Let's calculate $\int_{\theta_{n-1}}^{\theta_{1}+\pi}d\theta_{n}$ first. It equals $\theta_{1}+\pi-\theta_{n-1}$ . Then we integrate $\int_{\theta_{n-2}}^{\theta_{1}+\pi}(\theta_{1}+\pi-\theta_{n-1})d\theta_{n-1}$ , which equals $\frac{1}{2}(\theta_{1}+\pi-\theta_{n-2})^{2}$ . We can find that the integral is not difficult and has an obvious pattern. The final result of the numerator is $\frac{1}{(n-1)!}\pi^{n-1}\times 2\pi$ . So the total probability equals $\frac{n!\times \frac{1}{(n-1)!}\pi^{n-1}\times 2\pi}{(2\pi)^{n}} = \frac{n}{2^{n-1}}$