, 2023

Solution:

We solve for general $n$ where $n = 100$ in the problem. Notice that the points $(i, A_{i})$ must form a convex hull, so there is some unique maximal element $A_{i}$. Consider the $i-1$ points $A_{1}, \ldots, A_{i-1}$ left of $i$, and the $i$ slopes formed between these points of segments $\overline{A_{0}A_{1}}, \ldots, \overline{A_{i-1}A_{i}}$. Notice that we must choose the $i-1$ points to be decreasing. Ignoring cases where they have some shared $y$-coordinates since this happens with probability $0$, we have a $\frac{1}{(i-1)!}$ chance of picking them in ascending order. Now, we order the differences

$$\left\{A_{1} - A_{0}, A_{2} - A_{1}, \ldots, A_{i} - A_{i-1}\right\}$$

in descending order, obtaining some new list

$$\left\{d_{1}, d_{2}, \ldots, d_{i}\right\}$$

and redefining $A_{k} = \sum_{j=1}^{k} d_{j}$. Notice that this procedure almost surely maps $(i-1)!i!$ possible sequences of points $A_{1}, A_{2}, \ldots, A_{i-1}$ to a valid convex hull, so the chance that the points left of $A_{i}$ are valid is $\frac{1}{(i-1)!i!}$. Similarly, the chance that the points on the right work is given by $\frac{1}{(n+1-i)!(n-i)!}$. So, for a maximum value at $A_{i}$ the chance that we get a valid convex hull is $\frac{1}{(i-1)!i!(n+1-i)!(n-i)!}$.

To finish, note that each point is equally likely to be the peak. Our answer is

$$\begin{aligned} & \frac{1}{n} \sum_{i=1}^{n} \frac{1}{(i-1)!i!(n+1-i)!(n-i)!} \\ & = \frac{1}{n \cdot n!^{2}} \sum_{i=1}^{n} \frac{n!^{2}}{(i-1)!i!(n+1-i)!(n-i)!} \\ & = \frac{1}{n \cdot n!^{2}} \sum_{i=1}^{n} \binom{n}{i-1} \binom{n}{n-i} \\ & = \frac{1}{n \cdot n!^{2}} \binom{2n}{n-1} \end{aligned}$$

Plugging in $n = 100$ gives the desired answer.