>Let us consider $n$ points, where $n \geq 2$, placed independently and uniformly at random on the circumference of a circle. > >A set of points is said to lie on a single semicircle if there exists a diameter of the circle such that all $n$ points are contained within the same closed $180\textdegree$ arc. > >Determine the probability $P(n)$ that all points lie on the same semicircle. Let us start with the first point. Since it is alone, the probability of it being on the same semicircle as itself is $\frac{1}{1}$. When we place the second point, the further it can be from the first point is opposite of the first. However, this still forms a $180 \textdegree$arc, so the probability of this second point being on the same semicircle as the first one is also $\frac{1}{1}$. It's only with the third point that the arc containing all the points can grow bigger than $180\textdegree$, making the problem non-trivial. It's tempting to think that if the arc formed by the two first points is $100 \textdegree$, then the point is within a $180\textdegree$ arc if it falls between the first two points, or if it forms an arc up to $80 \textdegree$ with the leftmost or rightmost point. This would lead us into a geometrical solution. There is, however, a much simpler way to look at this problem. The condition that all $n$ points lie on some semicircle is difficult to calculate directly. Rephrasing it will make our work easier. This condition is met if and only if we can find one point, which we will call an *anchor*, such that all other $n-1$ points lie in the $180 \textdegree$ arc starting from the anchor and proceeding clockwise[^1]. Let the points be $P_1,P_2,\ldots,P_n$. Let $E_i$ be the event that $P_i$ is the anchor. For any given arrangement of points, only one point can be the unique starting point of the clockwise $180 \textdegree$ arc containing all others. We can discard the zero-probability case of ties[^2]. Therefore, $E_1,E_2,\ldots,E_n$ are said to be mutually exclusive. Since the positions of the points are independent, the probability that all $n-1$ points $P_2,...,P_n$ fall into the semicircle is: $ \begin{align} P(E_1)=(\frac{1}{2}) \times (\frac{1}{2}) \times \ldots \times (\frac{1}{2}) = (\frac{1}{2})^{n-1} \end{align} $ By symmetry, the probability that any point $P_i$ is the anchor is the same: $P(E_1) = P(E_2) = \ldots = P(E_n) = \left(\frac{1}{2}\right)^{n-1}$ Since the events $E_i$ are mutually exclusive, the total probability $P(n)$ of any of them occurring is the sum of their individual probabilities: $ \begin{align} P(n)=P(E_1)+P(E_2)+\ldots+P(E_n) \\ P(n)=\sum_{i=1}^{n}P(E_i) \\ P(n)=n \cdot (\frac{1}{2})^{n-1} \end{align} $ [^1]: The choice of a clockwise direction is an arbitrary convention; the probability holds identically for a counter-clockwise direction. [^2]: The probability of two points placed independently and uniformly at random on the circumference of a circle is negligible. We assign it a value of $0$. Adding $0$ to the overall probability will not influence our result.