What are the odds of all $4$ of them making a quadrilateral that has no sides of icosagon $(\text{I})$?

Question

Given just an icosagon $(\text{I})$, whose $4$ of the vertices randomly chosen. What are the odds of all $4$ of them making a quadrilateral that has no sides of icosagon $(\text{I})$?

Suppose that the icosagon $(\text{I})$ be with form $\text{A}_{1}\text{A}_{2}\cdots\text{A}_{19}\text{A}_{20}$, $4$ of the vertices randomly chosen, the number of quadrilaterals is $\textrm{C}_{20}^{4}$. Consider $3$ cases:

1. The number of quadrilaterals has $3$ sides that is icosagon $(\text{I})$'s sides: $20$

2. The number of quadrilaterals has $2$ sides that is icosagon $(\text{I})$'s sides: $\cdots$

   Consider $2$ cases:

   a. A quadrilateral that has $2$ of icosagon $(\text{I})$'s adjacent sides: $\cdots$

   b. A quadrilateral that has $2$ of icosagon $(\text{I})$'s non-adjacent sides: $150$

3. The number of quadrilaterals has $2$ sides that is icosagon $(\text{I})$'s sides: $\cdots$

I used these to prep for my University Entrance Examination, this question has $4$ choices:

1. $443$ per $969$ ($2215$ per $4815$; $2215$ per $\textrm{C}_{20}^{4}$)
2. $473$ per $969$ ($2365$ per $4815$; $2365$ per $\textrm{C}_{20}^{4}$)
3. $365$ per $969$ ($1825$ per $4815$; $1825$ per $\textrm{C}_{20}^{4}$)
4. $395$ per $969$ ($1975$ per $4815$; $1975$ per $\textrm{C}_{20}^{4}$)

So I guess the fourth choice, maybe it added the number of quadrilaterals that have $2$ of icosagon $(\text{I})$'s non-adjacent sides: $150$.