Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
Since $x,y,z$ are all reals located in $[0, n]$, the number of choices for each one is continuous so we use geometric probability.
WLOG([Without loss of generality](https://artofproblemsolving.com/wiki/index.php/Without_loss_of_generality)), assume that $n\geq x \geq y \geq z \geq 0$. Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$. The volume is $V_1=\dfrac{n^3}{6}$, as shown in the first figure in red.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2,-1,2/3);  // three - currentprojection, orthographic draw((1,1,0)--(0,1,0)--(0,0,0),dashed); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1),dashed); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);  draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red); draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((1,1,1)--(1,0,0), red); [/asy]

Now we will find the region with points satisfying  $|x-y|\geq1$, $|y-z|\geq1$, $|z-x|\geq1$.
Since $n\geq x \geq y \geq z \geq 0$, we have $x-y\geq1$, $y-z\geq1$.
The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3);  // three - currentprojection,  orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green);  draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green);    draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red);  draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed);  [/asy]
The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$.
So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$.
Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\frac{512}{1000}>\frac{1}{2}$, when $n=9$, $p=\frac{343}{729}<\frac{1}{2}$. So $n\geq 10$.
So the answer is $\boxed{\textbf{(D)}\ 10}$.