Four 6-sided dice are rolled. What is the probability that at least two dice show the same number? Am I doing this right? I split the problem up into the cases of 2 same, 3 same, 4 same, but I feel like something special has to be done for 2 of the same, because what if there are 2 pairs (like two 3's and two 4's)?
This is what I have:
For 2 of the same: $5\times 5\times 6\times {4\choose 2}=900$
For 3 of the same: $5\times 6\times {4\choose 3}=120$
For 4 of the same: $6\times {4\choose 4}=6$
Combined: $900+120+6=1026$
Total possibilities: $6^4=1296$
Probability of at least 2 die the same: $\frac {1026}{1296}\approx 79.17$%
Confirmation that I'm right, or pointing out where I went wrong would be appreciated. Thanks!
Sorry if the formatting could use work, still getting the hang of it.

Indeed, when you count the combinations with pairs, you counted twice each combination that has two pairs.
There are  3 * 5 * 6 = 90 such combinations of 2 pairs. 
So, for 90 combinations, you counted the combination twice in your total of 900.
The number of combinations having one or two pairs is then 900 - 90 = 810
The total number of combination having at least two identical dice is then 
(810 + 120 + 6) / 1296 = 72.2222%
Numerical check: the easy method gives: 
1 - (5/6 * 4/6 * 3/6) also equals to 72.2222%
