Let $R$ be a convex region of the plane with unit area, and choose $n\ge 3$ points at random from the region. We will derive a general expression for the probability $P_{n}$ that the convex hull of these points contains a particular point $\tau \in R$ (which we will take as the origin of our coordinate system). We can then evaluate that expression for the particular question posed (in which $n=3$, $R$ is an equilateral triangle, and $\tau$ is its center).
Suppose the hull does not contain $\tau$, and assume that $\tau$ is not collinear with any two of the points (this is true with probability 1). Then we can draw a ray from the center point through one of the $n$ points (the "leftmost" point) such that the remaining $n-1$ points are to the right of the ray. (Note that this condition is both necessary and sufficient. Note also that if the convex hull does contain $\tau$, then there is no "leftmost" point.) We can calculate the desired probability by integrating over all such configurations. In particular, take $f(\varphi)$ to be the area of the subregion that lies to the right of the radial ray at angle $\varphi$, and take $r(\varphi)$ to be the distance from $\tau$ to the boundary of $R$ along that ray. Then the probability that a particular point will lie in the angular interval $[\varphi, \varphi+d\varphi]$ is $da = \frac{1}{2}r(\varphi)^2 d\varphi$, and the probability that each of the other $n-1$ points will lie to its right is $f(\varphi)$. Since there are $n$ points to choose as the leftmost, the overall probability of not capturing $\tau$ in the convex hull is
$$
1-P_{n} = n \int f(\varphi)^{n-1} da = n \int_{-\pi}^{\pi} \frac{1}{2} f(\varphi)^{n-1} r(\varphi)^2 d\varphi,
$$
We can make a few observations at this point. First, if the region $R$ is symmetric under reflection through $\tau$, then $f(\varphi)$ is identically $1/2$: each radial line splits the region in half. The result then is
$$
P_{n} = 1 - \frac{n}{2^{n-1}},
$$
which gives the known result $P_{3} = 1/4$ for polygons with an even number of sides. Second, if the region $R$ instead has bilateral symmetry across the $x$-axis, then $r(\varphi)$ is an even function, and $f(\varphi) = 1/2 + f^{-}(\varphi)$, where $f^{-}$ is an odd function. Then terms in the integral with odd powers of $f^{-}$ must vanish: in particular, we have
$$
P_{3} = \frac{1}{4} - 3 \int{{f^{-}(\varphi)}^2 da} = \frac{1}{4} - 3\int_{-\pi}^{\pi}\frac{1}{2}{f^{-}(\varphi)}^2 r(\varphi)^2 d\phi \le \frac{1}{4},
$$
and interestingly the relation $P_{4} = 2P_{3}$ continues to hold.
It remains to evaluate $f^{-}(\varphi)$, $r(\varphi)$, and the resulting integral in the case of an equilateral triangle. The figure above shows an equilateral triangle with vertices at $(-x,0)$ and $(x/2, \pm x\sqrt{3}/2)$. For $\varphi \in [0,\pi/3]$, the function $f^{-}(\varphi)$ is the area of the blue triangle minus the area of the red triangle,
$$
f^{-}(\varphi) = \frac{x^2}{8}\tan\varphi - \frac{x^2}{2}\frac{1}{\cot\varphi + \cot\frac{\pi}{6}} = \frac{x^2}{8}\tan\varphi \left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right);
$$
and the radius $r(\varphi) = \frac{1}{2}x\sec\varphi$ over the same domain. By symmetry, the full integral is just six times its value over $[0,\pi/3]$:
$$
\begin{eqnarray}
P_3 &=& \frac{1}{4} - 18\int_{0}^{\pi/3}\frac{x^6}{512}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{36\sqrt{3}}\int_{0}^{\pi/3}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{12}\int_{0}^{1} u^2\left(1 - \frac{4}{1+3u}\right)^2 du,
\end{eqnarray}
$$
where we have used $x = 2/\sqrt{3\sqrt{3}}$ (in order for the triangle to have unit area) and introduced the transformed variable $u = \tan\varphi / \sqrt{3}$ (so $du = (\sec^2\varphi / \sqrt{3}) d\varphi$). The final integral is a straightforward exercise for the reader; the result is
$$
P_3 = \frac{1}{4} - \frac{1}{324}\left(57 - 80\ln{2}\right) = \frac{2}{81}\left(3 + 10\ln{2}\right) = 0.2452215...,
$$
in agreement with the brute-force and numerical results already given. This result can be generalized easily to the regular $m$-gon for any odd $m$ (changing only the values of some constants), and to the case where $n>3$ (complicating the final integral).
The question is about points on a circle, so that's the one I'm answering here. The triangle is right angled precisely if two of the three points lie on a diameter of the circle. This is Thales' theorem. As already noted, the probability that this occurs is $0$. The triangle is obtuse precisely if all three points lie on the same side of some diagonal of the circle. The probability for that is $\frac{3}{4}$ (assuming a uniform distribution on the circle).
One way of seeing this is to introduce for each vertex $v_k$ a random variable $X_k$ as follows:
$$
X_k = \begin{cases}
1 & \textrm{if the other vertices lie on the half circle starting at } v_k \textrm{ in clockwise direction}\\
0 & \textrm{otherwise}
\end{cases}
$$
Then at most one of $X_1, X_2, X_3$ can be equal to one. Therefore
$$
\mathbb{P}(\textrm{triangle is obtuse}) = \mathbb{P}(X_1+X_2+X_3 = 1) = \mathbb{E}(X_1+X_2+X_3) = 3 \mathbb{E}(X_1).
$$
The last equality follows from the fact that $X_1, X_2, X_3$ all have the same probability distribution. Now $\mathbb{E}(X_1) = \frac{1}{4}$ since both $v_2$ and $v_3$ have a probability of $\frac{1}{2}$ to lie on the half circle starting at $v_1$.
Best Answer
Here is my non-intuitive answer.
Assume the circle is $x^2+y^2=1$, and the points are:
$A\space(\cos(-\alpha), \sin(-\alpha))$ where $0<\alpha<2\pi$
$B\space(\cos \beta, \sin \beta)$ where $0<\beta<2\pi$
$C\space(1,0)$
The tangent at $A$ has equation $y=(\cot\alpha)(x-\cos\alpha)-\sin\alpha$.
The tangent at $B$ has equation $y=-(\cot\beta)(x-\cos\beta)+\sin\beta$.
The tangent at $C$ has equation $x=1$.
Assume the chosen side is the one through $C$. This side has length $|(\cot\alpha)(1-\cos\alpha)-\sin\alpha+(\cot\beta)(1-\cos\beta)-\sin\beta|$, so the probability $P$ is
$$P(|(\cot\alpha)(1-\cos\alpha)-\sin\alpha+(\cot\beta)(1-\cos\beta)-\sin\beta|<2).$$
Let $\alpha=Y-X$ and $\beta=Y+X$. Using basic trigonometric identities, and symmetry, $P$ simplifies to
$$P\left(Y<\arccos\left(\frac{-\cos X+\sqrt{1+\sin^2 X}}{2}\right)\right)$$
where $0<X<\pi$, and $0<Y<\frac{\pi}{2}$.
So we have
$$P=\frac{2}{\pi^2}\int_0^\pi f(X)dX$$
where $f(X)=\arccos\left(\frac{-\cos X+\sqrt{1+\sin^2X}}{2}\right)dX$.
Now let $g(X)=f\left(X+\frac{\pi}{2}\right)-\frac{\pi}{4}$.
$$P=\frac{2}{\pi^2}\left(\int_{-\pi/2}^{\pi/2}g(X)dX+\frac{\pi^2}{4}\right)$$
$\sin (g(X)+g(-X))$ simplifies to $0$, so $-g(-X)=g(X)$, so $g(X)$ is an odd function.
$$\therefore P=\frac12$$