6. Given $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$

Then the number of ordered pairs $(a, r)$ that satisfy the condition is
6. 62.

From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$

Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.

By the universal formula for two angles.
The probability of returning to the starting point after $n$ steps is
$$
\begin{array}{l}
\left.\frac{5}{3}=\frac{\cos A \cdot \mathrm{c}}{\sin A}-p_{n}\right) \\
\Rightarrow p_{n}=\frac{2}{3}\left(-\frac{1}{2}\right)^{n}+\frac{1}{3} .
\end{array}
$$

Substituting $n=10$ into the above formula yields.