The diameter of the Cayley graph of the Rubik's Cube group is the fewest number of turns needed to solve the Cube from any initial configurations. For the 2$\times$2$\times$2 Cube, the diameter is 11 in the half-turn metric, 14 in the quarter-turn metric, 19 in the semi-quarter-turn metric, and 10 in the bi-quarter-turn metric. For the 3$\times$3$\times$3 Cube, the diameter was determined by Rokicki et al. to be 20 in the half-turn metric and 26 in the quarter-turn metric. This study shows that a modified version of the coupon collector's problem in probabilistic theory can predict the diameters correctly for both 2$\times$2$\times$2 and 3$\times$3$\times$3 Cubes insofar as the quarter-turn metric is adopted. In the half-turn metric, the diameters are overestimated by one and two, respectively, for the 2$\times$2$\times$2 and 3$\times$3$\times$3 Cubes, whereas for the 2$\times$2$\times$2 Cube in the semi-quarter-turn and bi-quarter-turn metrics, they are overestimated by two and underestimated by one, respectively. Invoking the same probabilistic logic, the diameters of the 4$\times$4$\times$4 and 5$\times$5$\times$5 Cubes are predicted to be 48 (41) and 68 (58) in the quarter-turn (half-turn) metric, whose precise determinations are far beyond reach of classical supercomputing. It is shown that the probabilistically estimated diameter is approximated by $\ln N / \ln r + \ln N / r$, where $N$ is the number of configurations and $r$ is the branching ratio.

翻译：鲁比克魔方群的Cayley图直径是指从任意初始状态求解魔方所需的最少转动次数。对于2×2×2魔方，在四分之一转度量下直径为11，在quarter-turn度量下为14，在半四分之一转度量下为19，在双四分之一转度量下为10。对于3×3×3魔方，Rokicki等人确定其在half-turn度量下直径为20，在quarter-turn度量下为26。本研究表明，概率论中改进的优惠券收集问题能够正确预测2×2×2和3×3×3魔方在quarter-turn度量下的直径。对于half-turn度量，2×2×2和3×3×3魔方的直径分别被高估了1和2；而对于2×2×2魔方在半四分之一转和双四分之一转度量下，直径分别被高估2和低估1。运用相同的概率逻辑，预测4×4×4和5×5×5魔方在quarter-turn（half-turn）度量下的直径分别为48（41）和68（58），其精确求解远超经典超级计算机的能力范围。研究表明，概率估计直径近似为\(\ln N / \ln r + \ln N / r\)，其中\(N\)为状态数，\(r\)为分支比。