Situate the origin $O$ at the dodecahedron's center, and call the four random points $P_{i}$, where $1 \leq i \leq 4$. To any tetrahedron $P_{1} P_{2} P_{3} P_{4}$ we can associate a quadruple $\left(\epsilon_{(i j k)}\right)$, where $(i j k)$ ranges over all conjugates of the cycle (123) in the alternating group $A_{4}: \epsilon_{i j k}$ is the sign of the directed volume $\left[O P_{i} P_{j} P_{k}\right]$. Assume that, for a given tetrahedron $P_{1} P_{2} P_{3} P_{4}$, all members of its quadruple are nonzero (this happens with probability 1). For $1 \leq i \leq$ 4, if we replace $P_{i}$ with its reflection through the origin, the three members of the tetrahedron's quadruple that involve $P_{i}$ all flip sign, because each $\left[O P_{i} P_{j} P_{k}\right]$ is a linear function of the vector $\overrightarrow{O P}$. Thus, if we consider the 16 sister tetrahedra obtained by choosing independently whether to flip each $P_{i}$ through the origin, the quadruples range through all 16 possibilities (namely, all the quadruples consisting of $\pm 1 \mathrm{~s}$). Two of these 16 tetrahedra, namely those with quadruples $(1,1,1,1)$ and $(-1,-1,-1,-1)$, will contain the origin. So the answer is $2 / 16=1 / 8$.