Dice probability over multiple rolls. What is the probability of rolling one or more 6's using 3 six sided dice (1...6) that are rolled three times?
How does multiple rolls influence the probability, is it simply 3 times the probability of a single roll? 
Please simplify ; )
 A: I find it easier to consider the problem of not rolling a 6 with any of the three dice (the same problem with a different definition).
The chance of not rolling a 6 is 5/6. The chance of not rolling a six on any of your dice is therefor 5/6 * 5/6 * 5/6.
The chance of rolling at least one six is therefore the opposite of this:
1 - (5/6)^3 = 0.421
A: using bernoilles trials , if an event is repeated $n$ times and if probability  of success is $p$ and $q$ the probability of failure . then the event happening $r$ times in $n$ trials is
$$P(r) = ^{n}C_{r}p^{r}q^{n-r}$$
here $r =1,2,or3$.
as you are requiring consecutive terms $p= \frac {1}{6} $ and $q =\frac{5}{6}$.  $n =3$
then
$$P = P(1) + P(2) + P(3)$$
A: Hint:  We have a total of $9$ independent dice rolls. First find the probability of no $6$.
Remark: In direct answer to your question, no, that is not correct, there is no reason to multiply by $3$, and it will not give the right answer.
But you might find the probability of no $6$ on one roll of the three dice, and then find the probability this happens three times in a row, However, the approach suggested in the hint is a little shorter.
