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.

The probability that at least two are the same is one minus the probability that all four are different.  This is
\begin{align}
& \Pr(\text{2nd differs from 1st}) \\[8pt]
\times {} & \Pr(\text{3rd differs from 1st and 2nd} \mid \text{2nd differs from 1st}) \\[8pt]
\times {} & \Pr(\text{4th differs from 1st, 2nd, and 3rd}\mid \text{first three differ}) \\[10pt]
= {} & \frac 5 6 \cdot \frac 4 6 \cdot \frac 3 6 = \frac 5 {18}.
\end{align}
