<s>[INST] What is the probability of getting at least two different numbers from first roll when three six-sided dice are rolled a second time? I roll three six sided fair dice. Three different numbers show. What is the probability that at least two different numbers show in the second roll that occurred in the first roll? For example, first roll (1,2,3), second roll (3,1,4)?
Attempt
\begin{align*}
\Pr(\text{of at least two the same}) & = 1 - \Pr(\text{exactly $0$ the same}) - \Pr(\text{exactly $1$ the same})\\
& = 1 - \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} - \frac{1}{3} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot 3\\ 
& = 0.625
\end{align*}
In the second term, I am multiplying by $3$ to account for the number of placements the number can be in. Is this approach correct? Thanks!
 [/INST] Your answer is wrong.
The probability of "exactly $0$ the same" is indeed $\frac36\frac36\frac36=\frac{27}{216}$ as you suggest. 
But the probability of "exactly $1$ the same" equals:$$\left[\frac16\frac16\frac16+3\frac16\frac16\frac36+3\frac16\frac36\frac36\right]3=\frac{111}{216}$$
If e.g. the first roll was $(1,2,3)$ then between the brackets $[$ and $]$ you find the probability that number $1$ appears in the second roll and the numbers $2$ and $3$ do not. The first term of the three gives the probability on roll $(1,1,1)$, the second term gives the probability of a roll with exactly twice a $1$ in it and the third term gives the probability of a roll with once a $1$ in it.
The final result is: $$1-\frac{27}{216}-\frac{111}{216}=\frac{78}{216}\approx0.3611$$
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