From a Goldman Sachs Quant. position test.
Two points are chosen randomly in a unit square. What is the probability that the circle formed using the diameter of the 2 points contains the square's center?
[Math] Probability that a circle formed in a unit square contains the square’s center.
probability
Related Solutions
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 $r_i$: $$ \iiint_{0\le r_i \le 1} dr_1^2 dr_2^2 dr_3^2\;r_1 r_2 \sin \theta = \left(\frac23\right)\left(\frac23\right)\left(1\right) \sin\theta = \left(\frac23\right)^2 \sin\theta $$
Preliminary remark: @Jack D'Aurizio 's answer left me unconvinced. Therefore I ran a few simulations. They revealed that the expected value is about $0.23$, which is definitely larger than Aurizio's value ${2\over3\pi}=0.212$. On the other hand I found out that the problem had been treated in a 1996 paper by B. L. Burrows and R. F. Talbot. Since their paper resides behind a paywall I decided to solve the problem anew from scratch.
We start with the simpler situation that the vertex $C$ of $T$ lies on the rim of the disc $D$. We momentarily put the center of $D$ at $(0,1)$ and fix the vertex $C$ at the origin. The random vertex $A\in D$ is then given by $A=(r\cos\phi,r\sin\phi)$, whereby $$0\leq r\leq 2\sin\phi,\quad 0\leq \phi\leq\pi,\qquad{\rm d}(x,y)=r\>{\rm d}(r,\phi)\ .$$ Let $B=(s\cos\psi,s\sin\psi)$ be the second random vertex, whereby we restrict to $\phi\leq\psi\leq\pi$ and multiply by $2$ at the end. We then get the following expression for the expected area of $T$: $$E_1={2\over\pi^2}\int_0^\pi d\phi\>\int_\phi^\pi d\psi\>\int_0^{2\sin\phi}dr\>\int_0^{2\sin\psi} ds\left({1\over2}rs\sin(\psi-\phi)\>r\>s\right)\ .$$ Integrating first with respect to $r$ and $s$ we obtain $$E_1={1\over\pi^2}\int_0^\pi \int_\phi^\pi {64\over9} \sin(\psi-\phi)\sin^3\phi\sin^3\psi\>d\psi d\phi ={35\over36\pi}\ .$$ We now move $D$ back to the origin and put $Q:=\max\bigl\{|A|, |B|, |C|\bigr\}$. The expected area of $T$, conditioned on the value $q$ of $Q$, is then given by $$E_q={35\over36\pi}q^2\ .$$ The probability density of each of the three radii $|A|$, $|B|$, and $|C|$ is given by $f_R(x)=2x$. It is then an easy exercice to prove that the probability density of $Q$ is $f_Q(q)=3q^2\cdot2q=6q^5$. Therefore the expectation of $Q^2$ comes to $$E(Q^2)=\int_0^1 q^2\cdot 6q^5\>dq={3\over4}\ ,$$ and the final result is $$E={35\over36\pi}E(Q^2)={35\over48\pi}=0.232101\ .$$
Best Answer
Let $X$ and $Y$ denote the randomly drawn points. If $O$ denotes the center of the square, the event of interest occurs iff $\angle XOY$ is not acute. Now draw the line orthogonal to $OX$ at $O$. We want the probability that $X$ and $Y$ don't fall on the same side of this line. Given $X$, the conditional probability is $\frac{1}{2}$ because the line of interest cuts the area of the square in half and $Y$ is drawn uniformly at random from the square and independent of $X$. Taking average (i.e., expectation) with respect to $X$ we deduce that the probability is $\frac{1}{2}$.