<s>[INST] 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?
 [/INST] The probability of getting a non-$6$ the first five times is $(5/6)^5$.
$$
\begin{align}
& \Pr(\text{a non-}6\text{ on the 1st }5\text{ trials}) \\  \\
& = \Pr(\text{a non-}6\text{ on the first trial and a non-}6\text{ on the 2nd trial and a non-}6\text{ on the 3rd trial and }\ldots) \\  \\
& = \Pr(\text{a non-}6\text{ on the 1st trial})\cdot\Pr(\text{a non-}6\text{ on the 2nd trial})\cdot\Pr(\text{a non-}6\text{ on the 3rd trial})\cdots\cdots \\  \\
& = \frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56.
\end{align}
$$
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