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.
 A: You answer for "exactly two the same" counts some cases twice - when you get two pairs ($4545$, for example.)
The case of $2$ the same the others different counts to $6\cdot 5\cdot 4\cdot\binom{4}{2}=720$.
The case of two pair is $\binom{6}{2}\cdot\binom{4}{2}=90$.
Those two values add up to $810$, and you over-counted by $90$ - that is, you counted each "two pair" result twice. 
This gives a total of $720+90+120+6=936=1296-360$.
This sort of problem is much easier to do by calculating the probability of the opposite (that they are all different) and subtract that from $1$. The probability that they are all different is $\frac{6\cdot5\cdot 4\cdot 3}{6^4} = \frac{360}{1296}$.
A: 
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.
A: 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}
A: 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%
