I have four sets--let us denote them $A_i, i=1,\ldots,4$--of quantum-information-theoretic interest (https://arxiv.org/abs/2004.06745, https://arxiv.org/abs/1905.09228). The set $A_1$ contains the other three subsets properly, and we set its measure $m(A_1)$ to equal 1. ($A_1$ and $A_2$ are known to be convex.) Now, we know from exact integrations, that \begin{equation} m(A_2)=\frac{8 \pi }{27 \sqrt{3}} \approx 0.537422, \end{equation} \begin{equation} m(A_3)=\frac{1}{81} \left(27+\sqrt{3} \log \left(97+56 \sqrt{3}\right)\right) \approx 0.445977, \end{equation} and \begin{equation} m(A_2 \land A_3)=\frac{2}{81} \left(4 \sqrt{3} \pi -21\right) \approx 0.0189305. \end{equation} (We note that $1-m(A_2)-m(A_2 \land A_3)=m(A_1)-m(A_2)-m(A_2 \land A_3)=\frac{13}{27}$.)
Further, numerical (quasirandom estimation http://extremelearning.com.au) procedures lead us to conjectures that \begin{equation} m(A_2 \land A_3 \land \neg A_4)=\frac{25}{69984}=\frac{5^2}{2^5 \cdot 3^7} \approx 0.000357225, \end{equation} \begin{equation} m(A_2 \land A_3 \land A_4)=\frac{2}{121}= \frac{2}{11^2} \approx 0.0165289, \end{equation}\begin{equation} m(A_3 \land A_4)=\frac{680}{1573} =\frac{2^3 \cdot 5 \cdot 17}{11^2 \cdot 13} \approx 0.432295. \end{equation} and \begin{equation} m(\neg A_3 \land \neg A_4) =\frac{21}{44} =\frac{3 \cdot 7}{2^2 \cdot 11} \approx 0.477273. \end{equation}
So, can these relations be employed to deduce others (or might they possibly be inconsistent)? For example, from the last relation, $m(\neg A_3 \land \neg A_4) =\frac{21}{44}$, Boolean logic tells us that $m( A_3 \lor A_4) =\frac{21}{44}$.
In particular, I am interested in an exact ("entanglement probability") formula for $m(A_2 \land (A_3 \lor A_4)$, which I estimate as 0.0815031.
Also, of particular interest, in such regards, are $m(A_3 \lor A_4)$, $m(A_2 \land A_4)$ and $m(A_4)$ which I estimate as 0.5227319, 0.0791279 and 0.50900311, respectively.
I also have estimates of several other union-intersection measures--$m(\neg A_3 \lor \neg A_4)$, $m(A_2 \land \neg A_3 \land A_4)$, and $m(A_2 \land (\neg A_3 \lor A_4))$ of 0.5677507, 0.455919 and 0.520893, respectively. (My estimates should continue to improve as I continue with the quasirandom process.)