Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
For 1) The first die, you have $\frac{6}{6}$. The second die needs to be equal to the first, so you have probability of $\frac{1}{6}$. Then the third die can't be equal to the first and second dice, so it's $\frac{5}{6}$. 
All together you get $1 \cdot \frac{1}{6} \cdot \frac{5}{6}$. And since the next two cases yield the same results, then the probability that the same number apears on exactly two of the three dice is $$ 3 \cdot \left(1 \cdot\frac{1}{6} \cdot \frac{5}{6}\right)=\frac{5}{12}$$
Did I do this correctly?
Thank you. 

As a check: here is a different approach. There are three choices for the odd die out: left, middle, or right. For each choice of odd die out, there are $6$ choices for its value and then $5$ values for the value of the pair. That makes $3\cdot6\cdot5=90$ outcomes for two of three the same. There are a total of $6^3=216$ total possibilities, giving a probability of $\frac{90}{216}=\frac{5}{12}$.
