Chance of a double in three dice What is the chance of rolling doubles in three six sided dice?

The answer I have is:
$$
\frac{1}{6}•\frac{1}{6}•\frac{3}{2} = \frac{1}{24}
$$
The number of elements in the sample space = 216 or (6*6*6) 

Of the three numbers two of them to be same is (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) = 6 doublets. 
 Every one of this can happen in 3 different ways, that is (1,1,x) (1,x,1) (x,1,1) = 3 ways. 
 Now x can be any of the remaining 5 numbers.
 So total number of ways two of them can be same = 6 x 3 x 5 = 90 

So probability of the event = 90/216 = 5/12
There are 36 possible permutations when rolling the first and second die.  6 of these produce doubles.  Therefore the probability of rolling doubles with the first and second die is 6/36, or 1/6.  There is also a 1/6 chance of rolling doubles with the first and third die.  And a 1/6 chance of rolling doubles with the second and third die.  Therefore the probability of rolling doubles with all three dice is 1/6 + 1/6 + 1/6 = 3/6 or 1/2 (including triples).  The probability of rolling triples is 1/36 so the probability of rolling doubles with no triples is 18/36 - 1/36 = 17/36.