Find the relative measure of the space defined by
$$ Z\cdot a \geq 0, \quad Z \cdot b \geq 0, \quad Z \cdot 1=0 $$ to the unconstrained problem $$ \quad Z \cdot 1 = 0 $$ where $Z, a, b, 1 \in R^d$ and $d$ is 6 (dice rolling) in the example but I think the minimum viable dimension is probably lower for this kind of problem.
I think this might be equivalent to the relative surface area of the sphere. I was looking for something in the solid angle space but am wondering if there are any simple geometric approaches to solving this. Is there something other than "solid-angle" to look at here.
Write the orthogonal decomposition $\mathbb R^d=W\oplus\langle1\rangle$, where $W=\langle1\rangle^\perp$ is the set of all $Z$ such that $Z\cdot 1=0$. Assume, without loss of generality, that $a,b\in W$ are unit vectors. You may as well ask for the fraction of $W$ cut out by your inequalities $Z\cdot a\ge 0, Z\cdot b \ge0$. Here "fraction" is measured with respect to a rotationally symmetric probability measure on $W$, such as uniform measure on the unit ball, or unit sphere, or rotationally symmetric gaussian measure.
To answer this question, look at the further decomposition $W=V\oplus V^\perp$, where $V=\langle a,b\rangle$ is the plane determined by $a$ abd $b$, and ask for the fraction of $Z\in V$ satisfying $Z\cdot a\ge 0, Z\cdot b \ge0$. The intersection of these two half planes is a wedge, whose angle is $\pi-|\arccos(a\cdot b)|$, from which the fraction of the plane $W$ cut out is $f=\frac 1 2-|\arccos(a\cdot b)|/2\pi$.
Use Cavalieri's principle to lift from $V$ to $W$: the desired fraction of $W$ is an integral over $V^\perp$ of the fraction of a translate of $V$; the Cavalieri integrand is the constant $f$.
The initial choice of rotationally symmetric probability measure on $W$ does not affect the final answer, as everything factors through the rotational invariance of the circle in the plane, for which there is a unique invariant probability measure.