Probability of a random polygon that does not contain the center of a regular polygon Problem:
Let $m$ be an even number such that $m \geq 6$, and let $n$ be a natural number such that $3 \leq n \leq \cfrac{m}{2}$.
Number each vertex of a regular $m$-sided polygon clockwise as $1,2,\cdots,m$.
Randomly choose $n$ vertices among the $m$ vertices and construct an $n$-sided polygon.
We assume that two $n$-sided polygons that do not share at least one vertex  are different from each other. Congruent polygons anchored on different vertices (rotated or reflected) are considered different.
Let $P_{m,n}$ be the probability that the center of the $m$-sided polygon is not located inside nor on an edge of the $n$-sided polygon.
Then, find the probability $P_{m,n}$.
What I tried:
First, I calculated $P_{m,n}$ for $(m,n)=(8,3),(8,4),(10,3),(10,4)$:
$P_{8,3} = \cfrac{3 \times 8}{_8C_3} \approx 0.4286$
$P_{8,4} = \cfrac{1 \times 8}{_8C_4} \approx 0.1143$
$P_{10,3} = \cfrac{6 \times 10}{_{10}C_3} = 0.5$
$P_{10,4} = \cfrac{4 \times 10}{_{10}C_4} \approx 0.1905$
For example, when $(m,n)=(8,3)$, we can freely choose vertex 1 by symmetry. And we choose two other vertices $v_1, v_2 (v_1 < v_2)$ among the half side of the regular $8$-sided polygon so that we can simply multiply the combinations by 8. For $v_1 = 2$ we must have $v_2 = 3,4$. For $v_2 = 3$ we must have $v_2 = 4$. Therefore, $P_{8,3} = \cfrac{3 \times 8}{_8C_3}$.
For $n=3$, the possible combination of the triangle is:
$m=8 \Rightarrow (1 + 2) \times 8 = 24$
$m=10 \Rightarrow (1 + 2 + 3) \times 10 = 60$
Thus, $$P_{m,3} = \cfrac{m \sum^{(\frac{m}{2}-3)+1}_{k=1}{k}}{_mC_3} = \cfrac{\cfrac{m}{2}(\cfrac{m^2}{4}-\cfrac{3m}{2}+2)}{\cfrac{m(m-1)(m-2)}{3 \cdot 2 \cdot 1}} = \cfrac{3}{4}\cdot\cfrac{m^2-6m+8}{m^2-3m+2}$$
However, I don't know how to generalize it and find $P_{m,m}$.
Could someone help me with this problem?

I think there has to be a diameter with all chosen points on one side of the diameter.  I can't prove that at the moment.  So there is one point that is furthest clockwise.
You are then free to choose any $n-1$ of the $\frac{m}2-1$ points anticlockwise from the first point.  
