If 4 fair dice are thrown simulatenously, what is the probability of getting at least one pair?

(I was able to compute the solution using Permutations, but I wanted to try calculating one pair and two pair probabilities)

Method1 -

One pair => there are 12/2 = 6 arrangements for a pair in XXYZ


(6*1*5*4) 6
----------- [First two dice form a pair, the other two don't]
(6*6*6*6)

=20/36 = 10/18

Similarly, Two different pair => there are 3 arrangements for a pair in XXYY

(6*1*5*1)*3
-----------
6*6*6*6

=5/72

Two same pair (i.e. four dice with the same number) => 6/(6*6*6*6) = 1/6*6*6

Total = 10/18 + 5/72 + 1/6*6*6 != 13/18, which is the correct answer. CRASHED

Please help


Method 2 (using perm) => 1-(6*5*4*3/6*6*6*6) = 13/18

Thanks