<s>[INST] What is the probability we need k rolls of a fair sided die to get at least one five and one six? I HAVE checked the questions and none of them seem to specify any method to compensate for at least one 5 AND one 6. 
Here is how I tried it. 
$$ P(k\ Rolls \ and \ last \ is \ a \ six)= P(\ k -1 \ rolls \ have \ at \ least\ one \ 5 )*P(\ k \ is \ a \ six)$$
$$ = \sum_{j=1}^{k-1} {k-1\choose j}(\frac{1}{6})^j (\frac{4}{6})^j  $$
$$ = (\frac{5}{6})^{k-1}- {k-1\choose0}(\frac{1}{6})^0(\frac{4}{6})^{k-1}  $$
$$ =(\frac{5}{6})^4 - (\frac{4}{6})^4$$
 [/INST] It's not clear to me if the question is whether $k$ rolls suffice or whether exactly $k$ rolls are required.  I try to cover both possibilities below.
Let's say the first roll on which we have at least one 5 and at least one 6 is roll $K$.
Let's also say $A$ is the event of rolling no 5 in $k$ rolls and $B$ is the event of rolling no 6 in $k$ rolls.  Then the probability that we have at least one 5 and at least one 6 by roll $k$ is
$$\begin{align}
\Pr(K \le k) &=\Pr(A^c \cap B^c) \\
&= 1 - \Pr(A \cup B) \\
 &= 1 - \Pr(A) - \Pr(B) + \Pr(A \cap B) \\
 &= 1 - 2 \left( \frac{5}{6} \right)^k + \left( \frac{4}{6} \right)^k \\
\end{align}$$
So the probability that the first roll on which we have at least one 5 and at least one 6 is roll $k$ is
$$\Pr(K = k) = \Pr(K \le k) - \Pr(K \le k-1) = \frac{1}{3} \left( \frac{5}{6} \right)^{k-1} - \frac{1}{3} \left( \frac{4}{6} \right)^{k-1}$$
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