Roll six dice. Probability of at least one pair. I roll 6 fair dice. What is the probability that at least one pair shows up?
 A: If I understand you correctly, you have six dice which you are rolling, and you want to know the probability of at least two dice of the same number appearing in that roll. One way to solve this is to subtract the probability of there not being a pair in your six dice from unity. There are $6!$ permutations of the six dice where each die is unique out of a total of $6^{6}$ possible permutations. This means that there are $6^{6} - 6!$ possible rolls of your dice that contain at least a pair. Therefore, $$1 - 6!/6^{6} \approx 0.98.$$
A: The number of possible output$=6^6$ (each dice can have 6 possible outputs)
If  we don't allow pair, then the 1st can have $6$ possible outputs, the 2nd can have $(6-1)$ possible outputs(excluding the previous result), the 3rd $(6-2)$ possible outputs(excluding the previous two results) and so on.
So, the number of possible output with no pair is $6!$ 
The probability of no pair is $\dfrac{6!}{6^6}=\frac{5}{324}$
The  probability of at least one pair will be $1-\dfrac{6!}{6^6}=1-\frac{5}{324}=\frac{319}{324}$.
(as the  probability of at least one $= 1- \text{the  probability of none)}$.
