N points on a Circle

Since the problem is trivial for $n \leq 2$, suppose that $n \geq 3$. Without loss of generality, assume that the circle has circumference one and that the first of $n$ points is located on its top (the next $n-1$ points are drawn independently and uniformly on the circumference).
Then, it is readily seen that the problem is equivalent to the following one. Let $U_1,\ldots,U_{n-1}$ be i.i.d. uniform rv's on $[0,1]$. What is the probability that $U_{\max} - U_{\min} \leq 1/2$?

As is well known (cf. this, for example) and easy to show (simple exercise in order statistics, upon conditioning on $U_{\max}$), $$ {\rm P}(U_{\rm max} - U_{\rm min} \leq x) = (n-1)x^{n - 2} - (n - 2)x^{n-1}, \;\; x \in [0,1]. $$ Substituting $x=1/2$ gives the desired result of $\frac{n}{{2^{n - 1} }}$.

Suppose the points are $x_1,\ldots,x_n$. If the points are indeed contained in a semicircle, then we can define a "leftmost" point, i.e. a point $x_i$ such that $x_j \in [x_i, x_i+\pi)$ (addition is modulo $2\pi$). There is only one leftmost point, and the probability that $x_i$ is leftmost is $1/2^{n-1}$, since the probability that $x_j \in [x_i, x_i+\pi)$ is $1/2$. Since there are $n$ possible leftmost points, we get $n/2^{n-1}$.

Now generalize the argument to $d$ dimensions.

Edit: The argument generalizes to show that the probability that all points lie in an $\alpha$-section of the circle is $n\alpha^{n-1}$ for $\alpha \leq 1/2$. You can challenge yourself by thinking about the case $\alpha > 1/2$.