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.
As the probability is the same in all cubes, we can calculate it in the cube $[-1,1]^3$.
We can limit ourselves to the sixth of the cube where $z$ is positive and has the greatest absolute value of the three coordinates. Then the radius of the sphere is $1-z$, and the centre of the cube is in the sphere if $x^2+y^2+z^2\le(1-z)^2$. Thus the admissible area of $(x,y)$ is the intersection of the square $[-z,z]^2$ with the circle $x^2+y^2=1-2z$. A corner of the square lies on the circle if $3z^2=(1-z)^2$, that is, $z=\frac{\sqrt3-1}2$, and a midpoint of the square lies on the circle if $2z^2=(1-z)^2$, that is, $z=\sqrt2-1$.
Thus, for $0\le z\le\frac{\sqrt3-1}2$, the entire square lies within the circle, so the area is $4z^2$.
For $\frac{\sqrt3-1}2\le z\le\sqrt2-1$ the circle and square intersect. The four segments of the circle that extend beyond the square each have area $(1-2z)\arccos\frac z{\sqrt{1-2z}}-z\sqrt{1-2z-z^2}$, so the area is $\pi(1-2z)-4\left((1-2z)\arccos\frac z{\sqrt{1-2z}}-z\sqrt{1-2z-z^2}\right)$.
For $\sqrt2-1\le z\le\frac12$ the entire circle lies within the square, so the area is $\pi(1-2z)$; and for $z\gt\frac12$ the area is $0$.
Thus the desired probability is
$$ \frac68\left(\int_0^\frac{\sqrt3-1}24z^2\mathrm dz+\int_\frac{\sqrt3-1}2^{\sqrt2-1}\left((1-2z)\left(\pi-4\arccos\frac z{\sqrt{1-2z}}\right)+4z\sqrt{1-2z-z^2}\right)\mathrm dz+\int_{\sqrt2-1}^\frac12\pi(1-2z)\mathrm dz\right)\;. $$
The first and last integral evaluate to $\frac43\left(\frac{\sqrt3-1}2\right)^3=\sqrt3-\frac53$ and $\frac\pi4\left(1-2\left(\sqrt2-1\right)\right)^2=\pi\left(\frac{17}4-3\sqrt2\right)$, respectively. Wolfram|Alpha evaluates the indefinite form of the second integral to
$$ -\pi z^2+\pi z+4\sqrt{1-2z-z^2}\left(\frac{z^2}3+\frac z6-\frac56\right)+(6-z)\sqrt{1-2z-z^2}+\frac{15}2\arctan{\frac{1+z}{\sqrt{1-2z-z^2}}}+\frac12\arctan\frac{1-3z}{\sqrt{1-2z-z^2}}-4\arcsin\frac{1+z}{\sqrt2}+4(z-1)z\arccos\frac z{\sqrt{1-2z}} $$
but refuses to evaluate it with limits. Substituting the limits by hand yields
$$ -\pi\left(3-2\sqrt2\right)+\pi\left(\sqrt2-1\right)+\frac{15}2\cdot\frac\pi2-\frac12\cdot\frac\pi2-4\cdot\frac\pi2=\left(3\sqrt2-\frac52\right)\pi $$
at the upper limit and
$$ -\pi\left(1-\frac{\sqrt3}2\right)+\pi\cdot\frac{\sqrt3-1}2+\frac23-\sqrt3+\frac72\sqrt3-4+\frac{15}2\cdot\frac{5\pi}{12}+\frac12\left(-\frac\pi{12}\right)-4\cdot\frac{5\pi}{12}+4\cdot\frac{\sqrt3-3}2\cdot\frac{\sqrt3-1}2\cdot\frac\pi4=-\frac{10}3+\frac52\sqrt3+\frac{17}{12}\pi $$
at the lower limit, so the second integral evaluates to
$$ \frac{10}3-\frac52\sqrt3+\left(3\sqrt2-\frac{47}{12}\right)\pi\;. $$
Thus, the desired probability is
$$ \frac34\left(\sqrt3-\frac53+\frac{10}3-\frac52\sqrt3+\left(3\sqrt2-\frac{47}{12}\right)\pi+\pi\left(\frac{17}4-3\sqrt2\right)\right)\\=\boxed{\frac\pi4+\frac54-\frac98\sqrt3\approx0.086841}\;, $$
in agreement with Aaron’s calculation and simulation.
I'll try an approach using order statistics. Suppose the coordinates of the chosen random point are $(X_1, X_2, X_3)$. We can assume that the coordinates of the chosen point are all positive. (If they are not, we can reflect the point into the first octant.) Hence, $X_1, X_2, X_3$ are independent uniform random variables on $(0, 1)$.
Now, we define the order statistics $Y_1, Y_2, Y_3$ so that $Y_1$ is the smallest of the $X_i$ values, $Y_2$ is the middle value, and $Y_3$ is the largest value. Note that the $Y_i$ variables are neither uniform on $(0, 1)$ nor independent of one another.
There are two variables of interest: $R$, the radius of the sphere, and $D$, the distance from the chosen point to the origin. Note that $R = \min\{1 - X_1, 1 - X_2, 1 - X_3\} = 1 - Y_3 $ and that $D = \sqrt{X_1^2 + X_2^2 + X_3^2} = \sqrt{Y_1^2 + Y_2^2 + Y_3^2}$. The operative question is: what is $\mathbb P(D < R)$? That is, what is $$\mathbb P \left(\sqrt{Y_1^2 + Y_2^2 + Y_3^2} < 1 - Y_3 \right)?$$
First, a bit of algebra to clean this up: \begin{align*} \mathbb P \left(\sqrt{Y_1^2 + Y_2^2 + Y_3^2} < 1 - Y_3 \right) &= \mathbb P \left( Y_1^2 + Y_2^2 + Y_3^2 < (1 - Y_3)^2 \right) \\ &= \mathbb P \left(Y_1^2 + Y_2^2<1-2 Y_3 \right) \end{align*} We know the joint pdf of these order statistics to be $$f(y_1, y_2, y_3) = \begin{cases} 3!, & 0 < y_1 < y_2 < y_3 < 1 \\ 0, & \text{otherwise} \end{cases}$$ so we just need to integrate that density over the set $\{y_1^2 + y_2^2 < 1 - 2 y_3\}$ in the cube $[0, 1]^3$. Note that this requires in particular that $y_3 \leq 1/2$. I claim that this triple integral can be expressed as \begin{align*} \int_0^{1/2} \int_0^{\min\{y_3, \sqrt{1 - 2 y_3}\}} \int_0^{\min\{y_2, \sqrt{1 - 2y_3 - y_2^2}\}} 6 \, \textrm d y_1 \, \textrm d y_2 \, \textrm d y_3. \end{align*}
I don't know the analytic value of that integral (I haven't tried very hard yet), but Wolfram Alpha estimates it to be $\fbox{0.0868}$.
When I do a really long probability computation like this, I always assume I screwed up somewhere and verify my work with a Monte Carlo simulation. Here's that work in R:
spherecube <- function(){
center <- runif(3, min=-1, max=1)
radius <- min(abs(1 - center), abs(1 + center))
sum(center * center) < radius^2
}
mean(replicate(100000, spherecube()))
# 0.08674
Aside from being an interesting problem, it's a great advertisement for the power of Monte Carlo simulations!
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*}
