probabilty of random points on perimeter containing center related question: probablity of random pick up three points inside a regular triangle which form a triangle and contain the center
What is the probability that a (possibly degenerate) triangle made by three randonly chosen points on the perimeter of an n-gon contains the centre of an n-gon?
For a square, there is a $\frac{1}{16}$ chance that the points are in configuration a, $\frac{3}{16} $ for configuraion b, and $\frac{3}8$ for c and d. The probility that the points contain the center is $0$ for a and c, $\frac{1}3$ for b (since the center is contained iff one point is on each side of the line TF1 and arbitrarily taking the square to have unit sides yields $2\int_0^1 a-a^2 \mathrm{d} a=\frac{1}3$) and $\frac{1}2$ for d (center contained iff B1 is the opposite side of the line through D1H1 to C1, $\int_0^1 a \mathrm{d}a=\frac{1}2$).Therefore, if I have somehow not made an error, the probability is $\frac{1}4$. 

[edited] The limiting case of a circle is $\frac{2}{\tau}\int_0^{\frac{\tau}2}\frac{a}{\tau}\mathrm{d}a=\frac{1}4$ (using $\tau=2\pi$ just to be controversial)
 A: The answer is in fact always $\frac14$, independent of $n$; you forgot to multiply by $2$ for symmetry and divide by $\tau$ for normalization in the circle result.
Wherever the first point is chosen, the "diameter" on which it lies divides the $n$-gon into two symmetric halves, and the second and third points must be in opposite halves. The line connecting them must also lie above the centre (as seen from the first point), and if the second point is at a distance $x$ from the first point along the perimeter (in units of the length of the perimeter), there's an admissible range of length $\frac12-x$ for the third point. Thus the probability is
$$2\int_0^{1/2}\left(\frac12-x\right)\mathrm dx=2\int_0^{1/2}x\,\mathrm dx=\frac14\;,$$
where the factor of $2$ is for symmetry because the second and third point can be interchanged.
A: The answer is 1/4. I did a slight variation in the above answer to make things look easier, if x is length at which the second point lies. The third point can only lie in the equal sector on opposite side with the same angle of the sector and length of arc being x ( Two sector are made by two diameters. Thus 2∫x dx limits from 0 to 1/2 gives 1/4.
A: As @joriki said "Wherever the first point is chosen, the "diameter" on which it lies divides the -gon into two symmetric halves, and the second and third points must be in opposite halves. The line connecting them must also lie above the centre (as seen from the first point), and if the second point is at a distance $x$ from the first point along the perimeter"
We can have the following definitions:
$\nu_1$, $\nu_2$, and $\nu_3$ are the first, the second, and the third point respectively.
event $B$: is to place $\nu_2$ and $\nu_3$ in the opposite halves of n-gon.
event $\bar{B}$: is to place $\nu_2$ and $\nu_3$ in the same half of n-gon.
event $A$: is to form the triangle having the centroid of n-gon within.
$L$ is the perimeter of n-gon.
$x$ is the distance between $\nu_1$ and $\nu_2$ along the shortest perimeter.
Note: Circle is a special case of n-gon, where n is approaching to $\infty$.

With above definition, we can use the following formula to describe what @joriki said:
$$
P(A) = \int_x P(A|x,B)f(x)P(B)dx + \int_x P(A|x,\bar{B})f(x)P(\bar{B})dx
$$
Because we know $P(A|x,\bar{B})\equiv0$ when event $\bar{B}$ happened. Therefore we could simply ignore the part with $\bar{B}$ involved, and rewrite above formula to become:
$$
P(A) = \int_x P(A|x,B)f(x)P(B)dx
$$

Then as we discussed before:
$P(A|x,B)$ is probability of forming the triangle having the centroid of n-gon within, when $\nu_2$ is placed at the distance $x$ away from the $\nu_1$ along the shorter perimeter, and $\nu_2$ and $\nu_3$ have been placed in the opposite halves of a n-gon.
$$
P(A|x,B) = \frac{x}{L/2} = \frac{2x}{L}\\
$$
$f(x)$ is the distribution of possible $x$. Because we could put the second point anywhere on the outline of the n-gon, and $x \in [0,L/2]$, so $f(x)$ is a uniform distribution $U[0, L/2]$
$$
f(x) = \frac{1}{L/2-0} = \frac{2}{L}
$$
$P(B)$ is the probability of placed $\nu_2$ and $\nu_3$ in the opposite halves of a n-gon
$$
P(B) = \frac{{{2}\choose{2}}}{{{2}\choose{2}}+{{2}\choose{1}}} = \frac{\frac{2!}{2!0!}}{\frac{2!}{2!0!}+\frac{2!}{1!1!}} = \frac{2}{2+2} = \frac{1}{2}
$$
The numerator is ${{2}\choose{2}}$, a binomial coefficient, meaning choose 2 halves from 2 halves to hold $\nu_2$ and $\nu_3$ respectively. And denominator is sum of ${{2}\choose{2}}$ and ${{2}\choose{1}}$, where ${{2}\choose{1}}$ means choose 1 half from 2 halves to hold both $\nu_2$ and $\nu_3$

As we discussed before $x \in [0, L/2]$, Thus
$$
\begin{align*} 
P(A) & = \int_0^{L/2} \frac{2x}{L}\cdot\frac{2}{L}\cdot\frac{1}{2}dx\\
P(A) & = \frac{2}{L^2}\int_0^{L/2} x dx\\
P(A) & = \Big[\frac{2}{L^2}\cdot\frac{x^2}{2}\Big]\Big|_0^{L/2}\\
P(A) & = \frac{x^2}{L^2}\Big|_0^{L/2} = \frac{1}{4}
\end{align*} 
$$
