So I've been thinking about this for a while, but the only answer I can seem to get is 7/8 for corners and 11/12 for edges.

I assume that:

[1] "Correctly placed" means that the corner/edge may be unoriented or oriented in its location.

[2] You're allowing any random configuration of the 3x3x3 cube (including configurations in which only a few pieces are unsolved).

Can someone help me out here, and tell me where I'm going wrong?

You have to take the cycle classes (cycle types) into account as you're doing a calculation like this. Despite all of the possible cycle types, for this type of problem, we just group all cycle types which involve the same number of pieces with each other. So we can use my formula (which I did research on and developed

here...you can see that post to see all of the cycle types, of you're interested):

\( \sum\limits_{k=p-x}^{p}{\left( \begin{matrix}

p \\

k \\

\end{matrix} \right)!k} \)

Where:

*p* = the total number of the piece type. For wing edges,

*p* = 24, for middle edges,

*p* = 12, and for corners,

*p* = 8.

No more than

*x* of each piece type are solved. \( 0\le x\le p \).

and

*k* is just the index of the summation.

**Corners**
As you probably know, the corners have 8! different permutations. The sum of all of the possible cycle types (and all versions of each cycle type) for 8 objects is equal to this amount. That is, if we let

*x* = 8 (meaning that we sum all cycle types involving 0 objects to all 8 objects...all cycle types), we have:

\( \sum\limits_{k=8-8}^{8}{\left( \begin{matrix}

8 \\

k \\

\end{matrix} \right)!k}=\sum\limits_{k=0}^{8}{\left( \begin{matrix}

8 \\

k \\

\end{matrix} \right)!k}=8! \)

[

Link]

But since we are just trying to count how many permutations involve exactly 7 pieces (because involving 7 pieces is equivalent to saying that one piece is left alone/solved), we just use the expression \( \left( \begin{matrix}

8 \\

k \\

\end{matrix} \right)!k \) and not the summation. That is, we substitute 7 for

*k* in the expression, and we get

14, 832.

So we simply do (14832/8!)*100 =

36.8%.

**Edges**
We use the same reasoning as we used for the corners, and we also get

36.8%.

**Corners and Edges**
If we want to calculate the chance that only one corner and one edge are correctly placed, we simply multiply them together, but we must be careful to not forget to multiply by two because the parity of the edges and corners must match (

cube law of permutations).

\( \left( \frac{\left( \begin{matrix}

8 \\

7 \\

\end{matrix} \right)!7}{8!} \right)\left( \frac{\left( \begin{matrix}

12 \\

11 \\

\end{matrix} \right)!11}{\frac{12!}{2}} \right)\times 100= \)

27%.