<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] Let's count all desired possibilities out the $6^3$ total potential results:


*

*${3 \choose 3} \times 3!=6$ ways of getting all three original values in some order

*${3 \choose 2} \times {6-3 \choose 1} \times 3!= 54$ ways of getting two different original values and one non-original value  in some order

*${3 \choose 1} \times {2 \choose 1} \times \frac{3!}{2!} = 18$ ways of getting a double original value and a single original value  in some order


making the probability $\dfrac{6+54+18}{6^3} = \dfrac{78}{216}=\dfrac{13}{36}\approx 0.361$, the same as drhab found another way 
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