A random sphere containing the center of the unit cube Inspired by a Putnam problem, I came up with the following question:
A point in randomly chosen in the unit cube, a sphere is then created using the random point as the center such that the sphere must be contained inside the cube (In other words, the largest sphere that fits). What is the probability that the center of the cube is contained inside the sphere created?
No real idea how to approach this one but thought some of you might find this interesting.

Here's the solution, confirming the previous answers, using symmetry as I suggested. I'll reorder my variables to conform with previous discussions. So we're going to consider the cube $[-1,1]^3$ and restrict to centers of the sphere in the pyramid $0\le x\le y\le z\le 1$. This means that the face $z=1$ will be the closest. The origin will be (on or) inside such a sphere if and only if $x^2+y^2+z^2\le (1-z)^2$, i.e., $2z\le 1-(x^2+y^2)$. 
Over what region in the $xy$-plane does our region project? Since $x\le y\le z$, we must have $2y\le 2z\le 1-(x^2+y^2)$, which means $x^2+y^2+2y\le 1$, or $x^2+(y+1)^2\le 2$. This results in the portion of $0\le x\le y$ lying inside the circle $x^2+(y+1)^2\le 2$. Note that $0\le x\le \dfrac{\sqrt3-1}2$.
Setting up the triple integral, the volume we desire is
$$\int_0^{\frac{\sqrt3-1}2}\int_x^{\sqrt{2-x^2}-1}\int_y^{\frac12(1-x^2-y^2)} dz\,dy\,dx,$$
and since we're comparing to the volume of the full pyramid, which is $1/6$, we take $6$ times this answer. 
(We can also set this up nicely in polar coordinates:
$$\int_0^{\pi/4}\int_0^{\sqrt{\cos^2\theta+1}-\cos\theta}\int_{r\sin\theta}^{\frac12(1-r^2)}\,r\,dz\,dr\,d\theta.)$$
The integral, multiplied by $6$, turns into
\begin{align*}
\int_0&^{\frac{\sqrt3-1}2} \big({-}5+4\sqrt{2-x^2}+3x^2-2x^2\sqrt{2-x^2}-(3x-3x^2-4x^3)\big)dx\\ &= \int_0^{\frac{\sqrt3-1}2} \big({-}5-3x+4x^3+4\sqrt{2-x^2}+6x^2-2x^2\sqrt{2-x^2}\big)dx \\ &= \frac54-\frac98\sqrt3+\frac{\pi}4.
\end{align*}
