Megaminx parity I have an old 12-colored Megaminx that I put all new stickers on because the old ones were falling off. This Megaminx was in more of a state of disrepair than I originally thought, though, and when I was solving it 2 of the pieces (1 edge and 1 corner) popped out and fell on the floor. I wasn't paying attention to those particular pieces, so I had no idea which way they were facing when they popped out.
I plugged them back into the puzzle. I had no idea if they had the correct orientation or not though. Surprisingly, I was still able to complete the puzzle without disassembling it.
I know that a Rubik's Cube has parity; only $\frac{1}{12}$ of the ways to assemble its cubelets results in solvable cubes. My intuition tells me that the Megaminx obeys the same parity rules (when I first got my Megaminx back in high school I was able to solve it using only the algorithms that come with a Rubik's Cube, with a few minor tweaks); however, I lack the mathematics background to verify this.
My question: If I were to completely disassemble a Megaminx and reassemble it at random, what are the odds that the resulting state would be solvable?
 A: A simple reasoning shows that a 12-colors megaminx must have parity...
Consider for a moment each piece to be of a solid color (i.e. the two faces of each edge and the three faces of each corner must be of the same color) and consider a single 72 degree face turn. The move will rotate 5 edge pieces and 5 corners, generating two 5-cycles.
Obviously the product of two independent 5-cycles is an even permutation and this means that any legal scramble of the megaminx (being a sequence of face turns) will produce an even permutation of the pieces.
Thus even if you paint all faces of each piece with a solid color (i.e. even if you just consider the simplified version of 12-colors megaminx where no orientation is taken into account) you have at most 50% probability of getting a solvable puzzle by randomly assembling the pieces.
For example if you take out just two edges and swap them leaving all other pieces in solved position, (this is an odd permutation) no matter the orientation you put them back in, you will always end up with an unsolvable megaminx. The same if you just swap two corners.
There's actually more; while for example on the 3x3x3 rubik's cube you can swap a pair of corners and a pair of edges (R' U L' U2 R U' R' U2 R L' U': "j" permutation) this is not possible on the 12-colors megaminx; a 5-cycle is an even permutation, thus even if all edges were black there is no way on the 12-colors megaminx to swap just two corners using face turns. The same is true if you paint all corners black: there is no way to swap just two edges.
The fact that edge position parity and corner position parity are independent (differently from the 3x3x3 cube) thus clearly implies that the probability of getting a valid 12-colors megaminx permutation by disassembling and randomly reassembling the parts cannot be greater than 25%.
For a 6-color megaminx this simple reasoning is not enough as pieces are not unique and distinct permutations of the pieces are "the" solution.
