The probability of getting a double six in one throw of two dice
$\begin{aligned}
&=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36} \\
\therefore \quad & \mathrm{p}=\frac{1}{36} \\
& \mathrm{q}=1-\mathrm{p} \\
&=1-\frac{1}{36}=\frac{35}{36} \\
& \quad \begin{array}{l}
\text { Now, }(\mathrm{p}+\mathrm{q})^{\mathrm{m}} \\
\mathrm{q}^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_{1} \mathrm{q}^{\mathrm{n}-1} \mathrm{p}+{ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{q}^{\mathrm{n}-2} \mathrm{p}^{2}+\ldots \\
+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \mathrm{p}^{\mathrm{r}}+\ldots+\mathrm{p}^{\mathrm{n}}
\end{array}
\end{aligned}$
The probability of getting atleast one double six in $\mathrm{n}$ throws with two dice.
$\begin{aligned}
&=(\mathrm{q}+\mathrm{p})^{\mathrm{n}}-\mathrm{q}^{\mathrm{n}} \\
&=1-\mathrm{q}^{\mathrm{n}}=1-\left(\frac{35}{36}\right)^{\mathrm{n}} \\
\therefore \quad & 1-\left(\frac{35}{36}\right)^{\mathrm{n}}>0.99 \\
\Rightarrow \quad &\left(\frac{35}{36}\right)^{\mathrm{n}} < 0.01
\end{aligned}$
$\begin{array}{l}
\Rightarrow \quad \mathrm{n}(\log 35-\log 36) < \log 0.01 \\
\Rightarrow \quad \mathrm{n}[15441-15563] < -2 \\
\Rightarrow \quad-0.0122 \mathrm{n} < -2 \\
\Rightarrow \quad 0.0122 \mathrm{n}>2 \Rightarrow \mathrm{n}>163.9
\end{array}$
So, the least value of $\mathrm{n}$ is 164.