Probability of rolling three dice without getting a 6 I am having trouble understanding how you get $91/216$ as the answer to this question. 

say a die is rolled three times
what is the probability that at least one roll is 6?


There are two answers already that express the probability as $$1-\left(\frac56\right)^3 = \frac{91}{216},$$
I'd like to point out that a more complicated, but more direct calculation gets to the same place.  Let's let 6 represent a die that comes up a 6, and X a die that comes up with something else.  Then we might distinguish eight cases for how the dice can come up:
666

66X
6X6
X66

6XX
X6X
XX6

XXX

We can easily calculate the probabilities for each of these eight cases.  Each die has a $\frac16$ probability of showing a 6, and a $\frac56$ probability of showing something else, which we represented with X.  To get the probability for a combination like 6X6 we multiply the three probabilities for the three dice; in this case $\frac16\cdot\frac56\cdot\frac16 = \frac5{216}$.  This yields the following probabilities:
$$\begin{array}{|r|ll|}
\hline
\mathtt{666} & \frac16\cdot\frac16\cdot\frac16 & = \frac{1}{216} \\
\hline
\mathtt{66X} & \frac16\cdot\frac16\cdot\frac56 & = \frac{5}{216} \\
\mathtt{6X6} & \frac16\cdot\frac56\cdot\frac16 & = \frac{5}{216} \\
\mathtt{X66} & \frac56\cdot\frac16\cdot\frac16 & = \frac{5}{216} \\
\hline
\mathtt{6XX} & \frac16\cdot\frac56\cdot\frac56 & = \frac{25}{216} \\
\mathtt{X6X} & \frac56\cdot\frac16\cdot\frac56 & = \frac{25}{216} \\
\mathtt{XX6} & \frac56\cdot\frac56\cdot\frac16 & = \frac{25}{216} \\
\hline
\mathtt{XXX} & \frac56\cdot\frac56\cdot\frac56 & = \frac{125}{216} \\
\hline
\end{array}
$$
The cases that we want are those that have at least one 6, which are the first seven lines of the table, and the sum of the probabilities for these lines is $$\frac{1}{216}+\frac{5}{216}+\frac{5}{216}+\frac{5}{216}+
\frac{25}{216}+\frac{25}{216}+\frac{25}{216} = \color{red}{\frac{91}{216}}$$
just as everyone else said.  
Since the first 7 lines together  with the 8th line account for all possible throws of the dice, together they add up to a probability of $\frac{216}{216} = 1$, and that leads to the easier way to get to the correct answer: instead of calculating and adding the first 7 lines, just calculate the 8th line, $\frac{125}{216}$ and subtract it from 1.
