Chance of dissassembling a cube to solve it

[Feryll]

As per the thread title, I saw the thread here and wondered "What about the chance of taking pieces out and randomly placing them in spots? I am in no way sure of if my work is correct, but whatever.

Spoiler:

This is first assuming that you can take the middle pieces out. There are 6! different combinations of how to put on the centers from one angle. But many are the same but just from another angle. This reduces it by a factor of 2(6), so the number drops to 60.
There are 8! different places to put the corners on, and that is multiplied by the flip coefficient (or whatever) of 3 different flips and 8 places the flips occur, so 3^8. That gives us 8! x 3^8. But as before, there are 24 clones per the actual number, so it becomes (8! x 3^8)/24, or 7! x 3^7.
There are 12! different places to place the edges, with 2 flips, so it becomes 12! x 2^12. But once again, there are clones in that calculation. There are 24different places to use as your point of view when looking at a cube (Don't say it's only six, because you can rotate any face 4 times), so it's reduced by a factor of 24. So the number becomes (12! x 2 ^ 12)/24, or 11! x 2^11.
Finally, it's time to multiply 60 (centers) by 11,022,480 (corners) by 81,749,606,400 (edges). If you divide it by the 4.3 quintillion possibilities, this yields a "1 in 9.6 chance" or 10.4%.

But the thing is, I thought it was a 1 in twelve chance. If anybody could find an error in my proof (using my method, not the orientated and permeated simple one), I'd be grateful And you get the opportunity to facepalm me!

[joey]

Well, it's cos you also counted centres, which "don't come off".

[Cyrus C.]

I got:

Spoiler:

1/2 chance of EO
1/3 Chance of CO
1/2 Chance of EP
1/1 Chance sf CP
1/12 Chance of solvable cube.

I don't think there's any error in your proof, You just counted centers, & usually people don't take then off when dissambling the cube.

[Micael]

I got:

Spoiler:

1/2 chance of EO
1/3 Chance of CO
1/2 Chance of EP
1/1 Chance sf CP
1/12 Chance of solvable cube.

I don't think there's any error in your proof, You just counted centers, & usually people don't take then off when dissambling the cube.

I think this is correct. This is 8.3% (1/12). So I do not believe 10.4% is correct because it should reduce the chance if you also remove center caps.

[Sa967St]

Spoiler:

An interesting fact about Rubik's Cube is that if you take it apart and put it together randomly, there will be only a 1 in 12 chance of it actually being solvable by legal moves (that is, without taking the cube apart again).
...
Combining these laws, only 1/12 (1/2 * 1/2 * 1/3) of the conceivable cube states are reachable by legal moves.

^that

[Cyrus C.]

Let me try with centers:

Spoiler:

1/12 EO CO EP
1/64 CP
1/768 Total

I feel like that number is way too low, did I do something wrong?

[Thomas09]

The answer is 42.

[Feryll]

Let me try with centers:

Spoiler:

1/12 EO CO EP
1/64 CP
1/768 Total

I feel like that number is way too low, did I do something wrong?

Actually, its 1/60 for centers. The answer probably is 1/12 x 1/60, or 1/720. I just wonder where I went wrong...
Actually, I counted centers because in a truly mathematical problem, the model would probably have removable centers. And if you're 4x4x4 fell apart completely (like my storeRIPOFFbought), it would be useful info. I don't know anything about 4x4 BLD, so I don't know how to simply go about the 4x4's chance of dissassembling.

[Logan]

The answer is 42.

yep.

[joey]

Chances of a solvable 4x4 = chances that CO is correct.