probability

There are altogether 4x3x2= 24 possible combination of putting the cubes into the holes, or you can simply list out all combinations as follows :
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 4 1
3 2 1 4
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1

Among those, you can find only the following combinations fulfill the requirement that every cube is in the wrong hole :

2 1 4 3
2 3 4 1
2 4 1 3
3 1 4 2
3 4 1 2
3 4 2 1
4 1 2 3
4 3 1 2
4 3 2 1

So there are totally 9 combination fulfilling the requirement.

P (every cube is in the wrong hole)
=9/24
=3/8
=0.375

http://pstock.idv.tw/files-pstock/9062003.pdf

First consider 3 holes and 3 cubes. There are 3! ways to put the cubes.
Way to have all 3 correct = 1
Way to have 2 correct : not possible, 2 correct => all 3 correct
Ways to have 1 correct: 3C1 (select 1 from 3) = 3
Ways to have all 3 wrong = 3!-1-3=2 ...(A)

Now consider 4 holes and 4 cubes. There are 4! ways to put the cubes.
Way to have all 4 correct=1
Way to have 3 correct: not possible
Ways to have 2 correct = 4C2 (select 2 from 4) x 1
Ways to have 1 correct = 4C1 (select 1 from 4) x (the rest 3 are wrong) = 4C1 x 2 from (A) above
No of ways to have all 4 wrong = 4!-1-6-8=9

Probability = 9/24=3/8