, 2018

Solution:

We assume that $x$, $y$, $z$ are all nonzero, since the other case contributes zero to the total probability.

If $x$, $y$, $z$ are all positive or all negative then the equation is obviously true. Otherwise, since flipping the signs of all three variables or permuting them does not change the equality, we assume WLOG that $x, y > 0$ and $z < 0$.

If $x+y+z > 0$, then the LHS of the original equation becomes $x + y - z + x + y = z = 2x + 2y$, and the RHS becomes $x + y + |x+z| + |y+z|$, so we need $|x+z| + |y+z| = x + y$. But this is impossible when $-x-y < z < 0$, since the equality is achieved only at the endpoints and all the values in between make the LHS smaller than the RHS. (This can be verified via simple casework.)

If $x+y+z < 0$, then $x+z, y+z < 0$ as well, so the LHS of the original equation becomes $x + y - z - x - y - z = -2z$ and the RHS becomes $x + y - x - z - y - z = -2z$. In this case, the equality holds true.

Thus, we seek the volume of all points $(x, y, z) \in [0,1]^3$ that satisfy $x + y - z < 0$ (we flip the sign of $z$ here for convenience). The equation $x + y - z = 0$ represents a plane through the vertices $(1,0,1)$, $(0,0,0)$, $(0,1,1)$, and the desired region is the triangular pyramid, above the plane inside the unit cube, which has vertices $(1,0,1)$, $(0,0,0)$, $(0,1,1)$, $(0,0,1)$. This pyramid has volume $\frac{1}{6}$.

So the total volume of all points in $[-1,1]^3$ that satisfy the equation is $2 \cdot 1 + 6 \cdot \frac{1}{6} = 3$, out of $2^3 = 8$, so the probability is $\frac{3}{8}$.