What is the probability of getting a random scramble of a 7x7 with one center fully solved? 4x4?
I think all you need to do to calculate this for the nxnxn is just take out the positions that one composite center contributes to the total number of positions on the nxnxn. To see how I got the following piece (in the denominator of the fraction below) of the formula for the number of positions on the regular nxnxn, see
this post.
\( \frac{20!^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }\left( \frac{1}{4!^{5}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }}{24!^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }\left( \frac{1}{4!^{6}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \)
There are \( \left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor \) orbits (sets) of 24 non-fixed center pieces. 4 are in each face. Therefore we change 24! to 20! (24-4=20). For the other factor, we change 4!^6 to 4!^5 because we are pretending to have 5 composite centers instead of 6.
\( =\left( \frac{\frac{20!}{4!^{5}}}{\frac{24!}{4!^{6}}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\left( \frac{20!}{4!^{5}}\frac{4!^{6}}{24!} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor } \)\( =\left( \frac{20!}{24\left( 23 \right)\left( 22 \right)\left( 21 \right)20!}\frac{4!^{6}}{4!^{5}} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\left( \frac{1}{\left( 23 \right)\left( 22 \right)\left( 21 \right)} \right)^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }=\frac{1}{10626^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \)
All you have to do is substitute a value for
n, corresponding to the cube size you want. See
this link for the probability of a random composite center for cube sizes 2 through 11 (you can change the value of
n for larger cube sizes).
