What is the probability that when they leave that none of them is wearing either of their own shoes?

Question

Three ladies get together for a bit of fun. When they arrive they all kick off their shoes. Unfortunately, when they leave they are not quite in the same frame of mind so they each grab a left shoe and a right shoe at random, and put them on.
What is the probability that when they leave that none of them is wearing either of their own shoes?

Answer 1

If the shoes are named
L1 R1 ==> Women 1 (W1)
L2 R2 ==> Women 2 (W2)
L3 R3 ==> Women 3 (W3)

Now all the possible favourable events for the given probability is
[L2 R2 (W1)]×[L3 R3 (W2)]×[L1 R1 (W3)] = 1/9 × 1 × 1/4 = 1/36
[L3 R3 (W1)]×[L1 R1 (W2)]×[L2 R2 (W3)] = 1/9 × 1/4 × 1 = 1/36
[L2 R3 (W1)]×[L3 R1 (W2)]×[L1 R2 (W3)] = 1/9 × 1/2 × 1/2 = 1/36
[L3 R2 (W1)]×[L1 R3 (W2)]×[L2 R1 (W3)] = 1/9 × 1/2 × 1/2 = 1/36

Required Probability = 1/36 + 1/36 + 1/36 + 1/36 = 1/9.

Answer 2

1/9
.
Let's look at the left foot first.
The chance that none has their own shoe is 2/6.
1/2/3 - at least 1 correct
1/3/2 - at least 1 correct
2/1/3 - at least 1 correct
2/3/1 - all wrong
3/1/2 - all wrong
3/2/1 - at least 1 correct
.
The same holds for the right shoe.
Thus the change that none has either shoe correct is 2/6 * 2/6 = 4/36 = 1/9