Let $A_{i_{1}} A_{i_{2}} \cdots A_{i_{k}}$ be a convex $k$-gon that meets the conditions, and let $P=\left\{A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k}}\right\}$. Suppose there are $x_{t}$ points in $X$ between $A_{i_{t}}$ and $A_{i_{t+1}}$ that do not belong to $P \left(t=1,3, \cdots, k, \text{ with } A_{i_{k+1}}=A_{i_{1}}\right)$, then
$$x_{1}+x_{2}+\cdots+x_{k}=n-k,$$

where $x_{1} \geqslant r_{1}, \cdots \cdots x_{k} \geqslant r_{k}$. Thus,
$$\begin{array}{l} \left(x_{1}-r_{1}+1\right)+\left(x_{2}-r_{2}+1\right)+\cdots+\left(x_{k}-r_{k}+1\right) \\ =n-S(r), \end{array}$$

where $S(r)=r_{1}+r_{2}+\cdots+r_{k}$.
This way, each combination $P$ corresponds one-to-one with the positive integer solutions of equation (1)
$$\left(x_{1}-r_{1}+1, x_{2}-r_{2}+1, \cdots, x_{k}-r_{k}+1\right)$$

The number of positive integer solutions of equation (1) is $\mathrm{C}_{n-s(r)-1}^{k-1}$, so there are $\mathrm{C}_{n-s(r)-1}^{k-1}$ combinations $P$.

Placing each combination on a circle, there are $n$ different arrangements (each vertex rotates once), but each convex $k$-gon has $k$ vertices,

and each vertex is considered as $B_{1}$ with the corresponding intervals (solutions of the indeterminate equation) counted once, thus, each convex $k$-gon is counted $k$ times (for example, the triangle $A 2 B 3 C 4$, when points $A, B, C$ are considered as $B_{1}$, it is counted once in the interval patterns $2, 3, 4$, $3, 4, 2$, and $4, 2, 3$). Therefore, the number of all convex $k$-gons that meet the conditions is $\frac{n}{k} \mathrm{C}_{n-S(r)-1}^{k-1}$.

Since there are $\mathrm{C}_{n}^{k}$ ways to choose $k$ points from $n$ points on a circle, the required probability is
$$\frac{n \mathrm{C}_{n-s(r)-1}^{k-1}}{k \mathrm{C}_{n}^{k}}=\frac{n \mathrm{C}_{n-s(r)-1}^{k-1}}{n \mathrm{C}_{n-1}^{k-1}}=\frac{\mathrm{C}_{n-s(r)-1}^{k-1}}{\mathrm{C}_{n-1}^{k-1}},$$

where $S(r)=r_{1}+r_{2}+\cdots+r_{k}$.