As the estimate answer is a bit long winded, I've posted this mathematical answer separately.
Given a random triangle in a circle, the chance of a random point landing in the triangle is the area of a triangle divided by the area of a circle.
So what we need is the average area of a triangle in a circle. The average area of a triangle for a unit disk (r=1) can be determined with the following formula as per Worlfram Disk Triangle Picking
$$A_t = \frac{35}{48 \pi} \approx 0.232100... $$
As we all know the area of a circle is
$$A_c = \pi r^2 $$
Because r=1 for a unit disk it is simplified to
$$A_c = \pi $$
So the chance of the fourth random dot landing in a random triangle is
$$p_4 = \frac{A_t}{A_c} = \frac{35}{48 \pi^2} \approx 0.07388003$$
However with four points we are forming four triangles, each point will have the same chance to land in the triangle formed by the three other points, so the probability is.
$$p = 4 \frac{A_t}{A_c} = 4 \frac{35}{48 \pi^2} = \frac{35}{12π^2} \approx 0.295520119$$
The above formula is only valid for
A unit disk (r=1). I've not yet come across any proofs for the average area of a triangle in a circle with radius r. Does it increase at the same rate as the area of the circle at $r^2$?*
Four random points only. The extended question asking for more points is not yet covered.
EDIT: Probability remains the same for all r > 0
Now I've looked at more than 4 points as well and generated sets of random points (100000 times) and tallied up how many points landed in a triangles. Some interesting results which I've not worked out the math for yet.
╔═════════════╦═════════╦═════════╦═════════╦═════════╦═════════╗
║ Points ║ 4 ║ 5 ║ 6 ║ 7 ║ 8 ║
╠═════════════╬═════════╬═════════╬═════════╬═════════╬═════════╣
║ Triangles ║ 4 ║ 10 ║ 20 ║ 35 ║ 56 ║
║ Any ║ 0.29845 ║ 0.64469 ║ 0.86461 ║ 0.96053 ║ 0.99096 ║
║ 0 ║ 0.70155 ║ 0.35531 ║ 0.13539 ║ 0.03947 ║ 0.00904 ║
║ 1 ║ 0.29845 ║ 0 ║ 0 ║ 0 ║ 0 ║
║ 2 ║ 0 ║ 0.54789 ║ 0 ║ 0 ║ 0 ║
║ 3 ║ 0 ║ 0 ║ 0.24551 ║ 0 ║ 0 ║
║ 4 ║ 0 ║ 0.0968 ║ 0.20193 ║ 0.08295 ║ 0 ║
║ 5 ║ 0 ║ 0 ║ 0.04365 ║ 0 ║ 0.02192 ║
║ 6 ║ 0 ║ 0 ║ 0.12658 ║ 0.12148 ║ 0 ║
║ 7 ║ 0 ║ 0 ║ 0.14877 ║ 0 ║ 0 ║
║ 8 ║ 0 ║ 0 ║ 0.06697 ║ 0.12974 ║ 0.02969 ║
║ 9 ║ 0 ║ 0 ║ 0.00447 ║ 0 ║ 0.01231 ║
║ 10 ║ 0 ║ 0 ║ 0.01476 ║ 0.16427 ║ 0.01963 ║
║ 11 ║ 0 ║ 0 ║ 0.0096 ║ 0 ║ 0.01344 ║
║ 12 ║ 0 ║ 0 ║ 0.00237 ║ 0.19293 ║ 0.01909 ║
║ 13 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.06144 ║
║ 14 ║ 0 ║ 0 ║ 0 ║ 0.11626 ║ 0.02674 ║
║ 15 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00612 ║
║ 16 ║ 0 ║ 0 ║ 0 ║ 0.08477 ║ 0.05515 ║
║ 17 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.05771 ║
║ 18 ║ 0 ║ 0 ║ 0 ║ 0.04646 ║ 0.05092 ║
║ 19 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.04256 ║
║ 20 ║ 0 ║ 0 ║ 0 ║ 0.01522 ║ 0.04368 ║
║ 21 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.07326 ║
║ 22 ║ 0 ║ 0 ║ 0 ║ 0.00392 ║ 0.05944 ║
║ 23 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.03105 ║
║ 24 ║ 0 ║ 0 ║ 0 ║ 0.00192 ║ 0.0459 ║
║ 25 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.06104 ║
║ 26 ║ 0 ║ 0 ║ 0 ║ 0.00061 ║ 0.04756 ║
║ 27 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.02972 ║
║ 28 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.03464 ║
║ 29 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.03928 ║
║ 30 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.02749 ║
║ 31 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.01559 ║
║ 32 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.01359 ║
║ 33 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.01606 ║
║ 34 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.01174 ║
║ 35 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00531 ║
║ 36 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00451 ║
║ 37 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00515 ║
║ 38 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00333 ║
║ 39 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00132 ║
║ 40 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00119 ║
║ 41 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00121 ║
║ 42 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.0009 ║
║ 43 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00036 ║
║ 44 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00014 ║
║ 45 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00025 ║
║ 46 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00017 ║
║ 47 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00009 ║
║ 48 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.0001 ║
║ 49 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.0001 ║
║ 50 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00006 ║
║ 51 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0.00001 ║
║ 52 ║ 0 ║ 0 ║ 0 ║ 0 ║ 0 ║
╚═════════════╩═════════╩═════════╩═════════╩═════════╩═════════╝
We arbitrarily assume $S$ is the unit sphere, $O$ is the origin, $A=(1,0,0)$ and $B$ lies in the upper $xy$-plane; this can always be done by a suitable rotation and $x=\angle AOB$ has pdf $\frac12\sin x$ for $x\in[0,\pi]$. Looking at the sphere on the $z$-axis then gives two cases, depending on the comparison between $x$ and $\pi/2$:
If $x<\pi/2$ then $A$ and $B$ are connected. $C$ then only needs to lie in either of the hemispheres defined by $A$ and $B$; the probability of this happening is the proportion of the disc covered by the two hemispheres in the top-down view. The angle covered by either hemisphere is $\pi+x$.
If $x>\pi/2$ then $A$ and $B$ are not connected, so $C$ must lie in both their hemispheres. In the top-down view, the angle covered by both hemispheres is $\pi-x$.
Thus we have the final probability that $A,B,C$ are connected as
$$\int_0^{\pi/2}\frac12\left(\frac{\pi+x}{2\pi}\right)\sin x\,dx+\int_{\pi/2}^\pi\frac12\left(\frac{\pi-x}{2\pi}\right)\sin x\,dx$$
$$=\frac1{4\pi}\left(\int_0^{\pi/2}(\pi+x)\sin x\,dx+\int_{\pi/2}^\pi(\pi-x)\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\int_0^{\pi/2}(\pi+x)\sin x\,dx+\int_0^{\pi/2}x\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\pi\int_0^{\pi/2}\sin x\,dx+2\int_0^{\pi/2}x\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\pi+2\left([-x\cos x]_0^{\pi/2}-\int_0^{\pi/2}-\cos x\,dx\right)\right)$$
$$=\frac1{4\pi}\left(\pi+2[\sin x]_0^{\pi/2}\right)=\frac{\pi+2}{4\pi}$$
Best Answer
To simplify the problem, let's assume that $w = 1$ (since the probability is preserved by scaling the figure) and that the figure is centered at the origin. To make it more precise:
Let $X_1, X_2$ be uniformly distributed on $[-2, 2]$, and let $Y_1, Y_2$ be uniformly distributed on $[-1, 1]$, all independent from one another. Then the points $(X_1, Y_1)$ and $(X_2, Y_2)$ are uniformly distributed on the rectangle $[-2, 2] \times [-1, 1]$. Their midpoint is given by $$M = (M_x, M_y) = \left(\frac{X_1 + X_2}{2}, \frac{Y_1 + Y_2}{2}\right).$$ Our eventual goal is to determine if $M$ lies in the disk centered at the origin with radius $1/2$.
Since $M_x$ and $M_y$ are the average of two i.i.d. uniform variables, they have a triangular distribution; that is, the density of $M_x$ is $\frac 1 2 - \left| \frac x 4 \right|$ on $[-2, 2]$, and the density of $M_y$ is $1 - |y|$ on $[-1, 1]$. (The motivation for these density functions is that they make symmetric triangles centered at $0$ with total area of $1$ on the regions forming their support.) To check that we're on the right track, we note that $$\int_{-2}^2 \int_{-1}^1 (1 - |y|) \left( \frac 1 2 - \left| \frac x 4 \right| \right) \, \textrm d y \, \textrm d x = 1$$ which confirms that we have at least written down a valid joint density function for the two coordinates.
Our task now is to recompute that same integral, but this time integrating over the aforementioned disk instead of on the entire rectangle. To get rid of the absolute values, let's exploit the symmetry of the problem and restrict ourselves to quadrant 1 by computing $$\color{blue}{4} \cdot \iint_D (1 - y) \left( \frac 1 2 - \frac x 4 \right) \, \textrm dy \, \textrm dx$$ where $D$ is the portion of the disk that lies in Quadrant 1. Converting to polar coordinates gives \begin{align*} 4 \cdot \int_0^{1/2} \int_0^{\pi/2} (1 - r \sin \theta) \left( \frac 1 2 - \frac 1 4 r \cos \theta \right) \cdot r \, \textrm d \theta \, \textrm d r \end{align*} which is equal to $\fbox{$\frac{\pi}{8} - \frac{15}{128} \approx 0.275512$}$, confirming @Stephen Donovan's numerical calculations. (Note: I used Wolfram Alpha for laziness, but that integral is perfectly doable by hand.)