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


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*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.
A: I'm posting a new answer, because existing answer deals with 3×3 cube, which is very different in this regard from a 4×4 cube.
To actually answer the question, YES a 4×4 cube will be solvable if you randomly exchange any two centers. 
On a 4×4 cube, the parity of centerpieces permutation must always be the same as the parity of the cornerpieces permutation, however, odd and even parity of centers permutation is not distinguishable, as there are 4 identical centers of each color.
This can be demonstrated by a simple commutator-algorithm such as this one. This algorithm exchanges three centerpieces from the mechanical perspective, however, due to the fact that two of those are of the same color, the resulting effect is that only two centerpieces have been exchanged.
