Dice Question Probability A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times? 
So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)^{2} \frac{1}{6} + \cdots + \left(\frac{5}{6} \right)^{4} \frac{1}{6}$$
So just add $-1$ to this?
 A: The probability of getting a non-$6$ the first five times is $(5/6)^5$.
$$
\begin{align}
& \Pr(\text{a non-}6\text{ on the 1st }5\text{ trials}) \\  \\
& = \Pr(\text{a non-}6\text{ on the first trial and a non-}6\text{ on the 2nd trial and a non-}6\text{ on the 3rd trial and }\ldots) \\  \\
& = \Pr(\text{a non-}6\text{ on the 1st trial})\cdot\Pr(\text{a non-}6\text{ on the 2nd trial})\cdot\Pr(\text{a non-}6\text{ on the 3rd trial})\cdots\cdots \\  \\
& = \frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56.
\end{align}
$$
A: For your solution, you'd subtract what you have from 1. It is fine.
It can be done a bit more simply though:
The probability that it has to be cast more than 5 times to obtain the first 6 is exactly the probability that each of the first 5 rolls do not result in a 6. This would be
$ 
 \bigl({5\over 6}\bigr)^5
$, 
assuming independence of the rolls.
A: The easiest way to do this is simply to raise the probability that a six is not rolled to the fifth power, as these probabilities are independent, which gives $$P(\text{cast more than five times}) = P(\text{individual roll is not 6})^5 = \left(\frac{5}{6}\right)^5$$
This is equivalent to $1 - P$ where $P$ is what you wrote, as
$$1 - P = 1 - \frac{1}{6}\left(\frac{5}{6} + \cdots + \left(\frac{5}{6}\right)^4\right) = 1 - \frac{1}{6}\frac{1-(5/6)^5}{1 - 5/6} = \frac{1}{6}\frac{(5/6)^5}{1/6} = \left(\frac{5}{6}\right)^5$$
