The expected area of a triangle made up three points picked randomly inside a circle of radius, and contains the center Consider the triangle formed by randomly distributing three points Inside a circle. What is the expected area of the triangle that contains the center of the circle. 
And what is the expected area of the intersection of a set of dependent trinagles that contain the center?
Thank you in  advance for your help.
 A: This base part of the problem is not that hard because for suitable chosen coordinates, the condition that the center falls inside the triangle is relatively simple.
WOLOG, assume the circle is the unit circle. 
Let $(r_i, \alpha_i), i = 1,2,3$ be the positions of the 3 random points in polar coordinates.
Because of the symmetry of the problem. It just suffices to look the case where $\alpha_1 = 0$ and $0 \le \alpha_2 \le \pi$. 
Let $\theta = \alpha_2$ and $\phi = 2\pi - \alpha_3$. It is easy to check for the triangle to contain the origin, the condition is given by $0 \le \phi \le \pi$ and $\theta + \phi \ge \pi$. (see image at end for an illustration)
From this, we see the probability for the triangle to contain the origin is given by:
$$\int_0^\pi \frac{d\theta}{\pi}\int_{\pi-\theta}^\pi \frac{d\phi}{2\pi} \iiint_{0\le r_i\le1}  dr_1^2 dr_2^2 dr_3^2 = \frac{1}{2\pi^2}\int_0^\pi\theta d\theta = \frac14$$
When the triangle contains the origin, its area is given by:
$$\frac12 r_1 r_2 \sin\theta + \frac12 r_1 r_3 \sin\phi + \frac12 r_2 r_3 \sin(2\pi-(\theta+\phi))\tag{*}$$
Using symmetry of the problem again and notice $\int_0^1 r_i dr_i^2 = \frac23$, 
the contribution to expected area when the triangle contains the origin is given by:
$$\int_0^\pi \frac{d\theta}{\pi}\int_{\pi-\theta}^\pi \frac{d\phi}{2\pi} \left[\frac32 \left(\frac23\right)^2 \sin\theta \right] = \frac{1}{3\pi^2}\int_0^\pi \theta\sin\theta d\theta = \frac{1}{3\pi}\tag{**}$$
As a result, the conditional expected area of the triangle when it contains the origin is $\frac{4}{3\pi}$.


UPDATE
There are higher dimension generalization on the probability of picking a triangle containing the origin. Quoting from an
article 
by R.Howard and Paul Sisson, we have:

Let $\mathbb{R}^n$ be endowed with a probability measure $\mu$ which is symmetric with
  respect to the origin and such that when $n+1$ points are chosen
  independently with respect to $\mu$, with probability one their convex
  hull is a simplex. Then the probability that the origin is contained
  in the simplex generated by $n+1$ such random points is $\frac{1}{2^n}$.


UPDATE2 What does the contribution to expected area means.
In general, when you uniformly pick 3 points from a circle, it need not enclose the
center. Let $T$ be a variable running through all possible configuration of the 3 points.
Let $\mathscr{A}(T)$ be the area of corresponding triangle and $d\mu(T)$ be the probability density of occurrence of $T$. Let $\mathscr{C}$ be the set of configuration where the triangle contains the origin.
The expected area of the unconstrained triangle is given by 
$$\int \mathscr{A}(T) d\mu(T)$$
The expected area of the triangle conditional to it contains the center is given by:
$$\frac{\int_{\mathscr{C}} \mathscr{A}(T) d\mu(T)}{\int_{\mathscr{C}} d\mu(T)}$$
The numerator here is what I mean "contribution to expected area"
and the denominator $\int_{\mathscr{C}} d\mu(T) = \frac14$ is
the probability for the triangle to contain the origin. 
For $T \in \mathscr{C}$, one can setup coordinates so that $\mathscr{A}(T)$ has
the simple form in $(*)$. By symmetry, the contribution for those 3 pieces in $(*)$
equal to each other. This explains the factor $\frac32$ appear in the integrand
of L.H.S of $(**)$. The remaining part of the integrand in $(**)$ comes from the partial integral over $r_i$:
$$ \iiint_{0\le r_i \le 1} dr_1^2 dr_2^2 dr_3^2\;r_1 r_2 \sin \theta =
   \left(\frac23\right)\left(\frac23\right)\left(1\right) \sin\theta = \left(\frac23\right)^2 \sin\theta $$
