For $\alpha=\frac{d}{2}$, the desired probabilities ($\frac{29}{64}, \frac{8}{33},\ldots$) are yielded by \begin{equation} \label{Hou1} P(\alpha) =\Sigma_{i=0}^\infty f(\alpha+i), \end{equation} where \begin{equation} \label{Hou2} f(\alpha) = P(\alpha)-P(\alpha +1) = \frac{ q(\alpha) 2^{-4 \alpha -6} \Gamma{(3 \alpha +\frac{5}{2})} \Gamma{(5 \alpha +2})}{6 \Gamma{(\alpha +1)} \Gamma{(2 \alpha +3)} \Gamma{(5 \alpha +\frac{13}{2})}}, \end{equation} and \begin{equation} \label{Hou3} q(\alpha) = 185000 \alpha ^5+779750 \alpha ^4+1289125 \alpha ^3+1042015 \alpha ^2+410694 \alpha +63000 = \end{equation} \begin{equation} \alpha \bigg(5 \alpha \Big(25 \alpha \big(2 \alpha (740 \alpha +3119)+10313\big)+208403\Big)+410694\bigg)+63000. \end{equation} (Qing-Hu Hou helped in the derivation of this formula, using "Zeilberger's algorithm" ["creative telescoping"] ConciseFormula.)
An equivalent formula--also to be presented herenow employing $d$ rather than $\alpha=\frac{d}{2}$--was given by C. F. Dunkl in App. D of MasterLovasAndai .) \begin{equation} \mathcal{P}(d) =3456^{d}\frac{\left( \frac{1}{2}\right) _{d/2}% ^{3}\left( \frac{7}{6}\right) _{d/2}^{2}\left( \frac{5}{6}\right) _{d/2}^{2}\left( 2d\right) !}{\left( \frac{d}{2}\right) !\left( 3\right) _{5d}}\sum_{i\geq0,j\geq0}^{i+j\leq d/2}\frac{\left( -\frac{d}{2}\right) _{i+j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}\left( 2+3d\right) _{i}\left( 1+d\right) _{i}}{\left( 2+\frac{5d}{2}\right) _{i+j}\left( 1+\frac{d}{2}\right) _{j}i!j!\left( -2d\right) _{i}}. \end{equation}
However, for dimensions $n$ or $m$ greater than 2, no analogous formulas are yet available.