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.


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)?

Yes.  Your cases are


*

*1 quadruplet: $\binom{4}{4}\times \binom{6}{1}$ arrangements.

*1 triplet, 1 singleton: $\binom{4}{3,1}\times \binom{6}{1}\times \binom{5}{2}$ arrangements.

*1 doublet, 2 singletons: $\binom{4}{2,1,1}\times \binom{6}{1}\times \binom{5}{2}$ arrangements.

*2 doublets: $\binom{4}{2,2}\times \binom{6}{2}$ arrangements.


And the complement is the remaining case of 4 singletons: $\binom{6}{4}$ arrangements.
Finally, the total space is, of course, of $6^4$ arrangements.

NB: $\dbinom{4}{2,1,1} = \dfrac{4!}{2!\,1!\,1!}$ is a multinomial coefficient.
