In his 1988 book Group Representations in Probability and Statistics , Diaconis considers mixing times of the 15-puzzle. He states:
Here is a simplified version: Consider the blank as a $16$th block, and consider the puzzle on a "torus." ... It is not hard to show that it takes order $n^3$ steps to randomize a single square (on an $n\times n$ grid). I presume that order $n^3\log n$ steps suffice to randomize everything. For a $4\times 4$, this gives about $90$ "moves" to randomize. ... Similar questions can be asked for other puzzles such as Rubic's cube. (page 90, § G(4). Emphasis added).
Quanta magazine describes how recently Chu and Hough have solved Diaconis' version of the mixing time of the 15-puzzle by studying the eigenvalues of the corresponding transition (adjacency) matrix. Chu and Hough showed that, contrary to Diaconis's conjecture, an $(n^2-1)$-puzzle takes $n^4\log n$ random steps to have a uniform probability of being in one of the $\frac{n!}{2}$ positions. I believe instead of only $90$ moves, it may take up to $4^4\log 4 \approx 355$ moves to sufficiently scramble the 15-puzzle.
Of course, both the the 140-year old 15-puzzle and the 40-year old Rubik's cube have had storied histories in mathematics and culture. It's natural to ask:
How much of the results of Chu and Hough carry over to address the mixing time of the $3\times 3$ Rubik's cube? Under, say, the half-turn metric, or if easier under the quarter-turn metric?
Given that Chu and Hough's solution of the 15-puzzle mixing time is longer than the intuition of Diaconis, perhaps the mixing time of the Rubik's cube under the half-turn metric (quarter-turn metric) is much longer than the diameter of $20$ ($26$).
