Trying to be certain

The probability of not getting a 6 after $n$ throws is $q = \left(\dfrac 56\right)^n$ . Therefore, the probability of getting at least a 6 after $n$ throws is $p = 1 - q = 1 - \left(\dfrac 56\right)^n$ . Then we have:

$\begin{aligned} p & \ge 0.99 \\ 1 - \left(\dfrac 56\right)^n & \ge 0.99 \\ \implies \left(\dfrac 56\right)^n & \le 0.01 & \small \color{#3D99F6} \text{Taking }\log_{10} \text{ both sides} \\ n \left(\log_{10} 5 - \log_{10} 6\right) & \le - 2 \\ \implies n & \ge \frac 2{\log_{10} 6 - \log_{10} 5} \approx 25.259 \\ n & = \boxed{26} \end{aligned}$

The probability of rolling a 6 in $n$ rolls is $1-(\text{probability of rolling a }6)^n = 1-\left(\frac{5}{6}\right)^n$ $\begin{aligned} 1-\left(\frac{5}{6}\right)^n &\geq 0.99 \\ \left(\frac{5}{6}\right)^n &\leq 0.01 \\ n &\geq \frac{\log\left(\frac{5}{6}\right)}{\log(.01)}\approx 25.26 \end{aligned}$ Thus you'd have to roll the die at least $\boxed{26}$ to guarantee a $99\%$ chance of a 6.