Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?
I'm talking about orientation parity. Perhaps I am using the wrong term, since true parity should only be either odd or even, whereas corner orientation is tristate, but I don't know of a better term to use, so I'm sticking with that one for now. If you turn just one corner clockwise 120 degrees, you reach an illegal cube state. If you then turn any other corner counter-clockwise 120 degrees, the cube state is made legal again. This is what I'm talking about.
I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.
Am I not following your argument? Am I making a mistake in my logic somewhere?
Okay, let me give some examples. I have here a solved 2x2 with yellow on top, blue in front, and red on the right. Grab your 2x2 and you can follow along at home. For the purposes of this discussion, X will mean the left side sticker of the DBL corner, and Y will mean the right side sticker of the DBR corner.
As I look at the solved cube, X is orange and Y is red, and since I know the DB corners contain only one red sticker and one orange sticker, I know the full state of the cube without looking at the D or B faces.
Now apply this
swap algorithm: U F U' F D R D2 R F D2 F. It swaps the DB corners while preserving which stickers are exposed at X and Y.
The result is that now X is red and Y is orange. This is still a uniquely identifiable state.
Now apply this
rotation algorithm: R F' D R' D R D2 F2 R' D'. It rotates the DBR corner clockwise 120 degrees and the DBL corner counterclockwise 120 degrees. (Between these two algorithms, we can bring about any of the six legal states of those two corners without affecting the rest of the cube.)
As we look at the cube, X and Y are both green. If we were to execute the above swap algorithm again, they would still both be green, and so this state is not uniquely identifiable.
Now execute the rotation algorithm again. X and Y are now both white, and the same argument applies. Because we can use the swap algorithm to reach a new state that looks the same, this position is also confusable. So far, we have operated entirely inside one orientation parity state of the DB corner pair.
Now we will execute a new algorithm which I will call the
parity algorithm: R U2 R' U R2 U' R2 U R' U2. This one changes the parity state of the DB corners by twisting the UFR corner. This change in parity is readily observable without viewing the B or D faces.
If you've followed along properly, you should see that X is now white and Y is now green. White exists on both of the DB corners, and green exists on both of the DB corners. But I assert that this state is still uniquely identifiable without viewing the D or B faces. To demonstrate why, we will cycle through all six possible DB corner pair states for the current orientation parity state:
State | X | Y | Notes |
#1 | White | Green | This is the starting state. Now perform the rotation algorithm. |
#2 | Red | White | Now perform the rotation algorithm again. |
#3 | Green | Orange | Now perform the swap algorithm. |
#4 | Orange | Green | Now perform the rotation algorithm again. |
#5 | White | Red | Now perform the rotation algorithm one last time. |
#6 | Green | White | This is the sixth and final state. |
Notice that each of these six states has a distinct set of values for X and Y. This means that if you see X is white and Y is green, you know that the DB corners are in state #1, whereas if you see that X is green and Y is white, you know that they are in state #6. There is no ambiguity here because of parity.
In order to reach the other White/Green states, you would have to change the orientation of one of the other six corners, which would be visible without viewing the D or B faces, and so the observer, understanding parity perfectly, would be able to figure out what you had done and would still not be fooled. This is why the answer to question A is 7/6 (~85.714%) and not 16/21 (~76.19%).
Does this sufficiently answer your question?