What is the probability that nobody receives the same ranking twice? 
Four players compete in a tournament and are ranked $1$ to $4$. They then compete in another tournament and are again ranked from $1$ to $4$. Suppose that their performances in the second tournament, so that the two sets of rankings are independent.
  What is the probability that nobody receives the same ranking twice?

One way to do it:
Let the notation $(2,3,1,4)$ represent the result that the player who finished 1st
in tournament 1 finished 2nd in tournament 2, the player who finished 2nd
in tournament 1 finished 3rd in tournament 2, the player who finished 3rd in
tournament 1 finished 1st in tournament 2, and the player who finished 4th in
tournament 1 finished 4th in tournament 2.
Total outcomes= $4! = 24$
The results where no player receives the same ranking in the two tournaments
are:
$$(2,1,4,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,1,2) (3,4,2,1), (4,1,2,3), (4,3,1,2),(4,3,2,1)$$
There are nine of these results and so the required probability is $9/24 = 3/8$
If anyone could give me another way of doing this question to give me an intuition into whats going on behind the numbers, it would help a lot.
Thank you!

Inclusion-exclusion formula can be useful. For example, in the first round the allocation is (1,2,3,4). Then in how many ways can we get the legitimate allocation? It is $4! - \binom{4}{1} 3! + \binom{4}{2} 2! - \binom{4}{3} 1! + 1$. Divide this by $4!$ to get the probability. Now keep in mind you have also $4!$ possible allocations in the first round. 
