We solve for general $n$ where $n=100$ in the problem. Notice that the points $\left(i, A_{i}\right)$ 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 $$\begin{aligned} & \frac{1}{n} \sum_{i=1}^{n} \frac{1}{(i-1)!!!(n+1-i)!(n-i)!} \\ & =\frac{1}{n \cdot n!^{2}} \sum_{i=1}^{n} \frac{n!^{2}}{(i-1)!i!(n+1-i)!(n-i)!} \\ & =\frac{1}{n \cdot n!^{2}} \sum_{i=1}^{n}\binom{n}{i-1}\binom{n}{n-i}= \\ & =\frac{1}{n \cdot n!^{2}}\binom{2n}{n-1} \end{aligned}$$ Plugging in $n=100$ gives the desired answer.