What is the probablity that a Rubik's cube is solvable if you randomly switch two squares?

Question

I am trying to learn how to solve a Rubik's cube and the 4x4 rubic cube at school is missing 4 squares, that got me thinking: if you randomly switched two squares, would it still be solvable?

I know that the flat picture (example of what I mean) where you slide a square around on the xy plane are solvable omitting if you switch the last two squares when it is in its identity location. For example if cut a picture into 16 squares and then removed the 16th (bottom righthand) and then you switched the 14th and the 15th square then it would be unsolvable.

I really don't know how to say that better mathematicaly

Answer 1

For the standard Rubik cube, the probability is zero if you switch two squares that are actually of different colour:

The Rubik cube has squares at face centers, which don't move at all; edge cubelets, which may rotate and move around, but always stay edge cubelets; vertex cublets, which may rotate and move around, but always stay vertex cubelets.

• If you switch two face centers, the cube becomes unsolvable because no suitable vertex cubelets will be available for at least one vertex.
• If you switch the two faces of the same edge cubelet, the cube becomes unsolvable because flipping such a cubelet is not en element of the group.
• If you switch two faces of the same vertex cubelet, the cube becomes unsolveable because that cubelet no longer fits (wrong orientation of the colours on it)
• If you switch squares between different edge cubelets, this would require to not alter the set of edge cubelets. E.g., you must not produce a second green-yellow edge. This means that you need to switch between two cubelets sharing a colour (say switch yellow and red between the gren-yellow and the green-red edge cubelet). If I remember correctly$^1$, it is not possible to switch two edge cubelets belonging to the same face while keeping their orientation correct (with respect to the common face)
• If you switch squares between differnt vertx cubelets, this makes the cube unsolvable. Indeed, knowing two of the squares of a vertex cubelet, its correct placement is already completely determined.

$^1$ This is the only point I'm not 100% sure of and would have to check.

EDIT: After checking at Wikipedia, I got rid of the doubts mentioned in the footnote: Even considering only position, not orientation of cubelets, the group operates only as $A_{12}$, not $S_{12}$ on the edges.