6. A basketball player is practicing shooting. If he makes the previous shot, the probability of making the next shot is $\frac{2}{3}$; if he misses the previous shot, the probability of making the next shot is $\frac{1}{3}$. It is known that the player has a probability of $\frac{2}{3}$ of making the 1st shot. Then the probability of him making the 4th shot is
6. $\frac{41}{81}$.

Let the probability that the player makes the $(n-1)$-th shot be $a_{n-1}$, and the probability of missing be $1-a_{n-1}$. Then the probability that he makes the $n$-th shot is
$$
\begin{array}{l}
a_{n}=\frac{2}{3} a_{n-1}+\frac{1}{3}\left(1-a_{n-1}\right)=\frac{1}{3}+\frac{1}{3} a_{n-1} \\
\Rightarrow a_{n}-\frac{1}{2}=\frac{1}{3}\left(a_{n-1}-\frac{1}{2}\right) . \\
\text { Also, } a_{n}-\frac{1}{2} \\
=\frac{a_{n}-\frac{1}{2}}{a_{n-1}-\frac{1}{2}} \cdot \frac{a_{n-1}-\frac{1}{2}}{a_{n-2}-\frac{1}{2}} \cdots \cdot \frac{a_{2}-\frac{1}{2}}{a_{1}-\frac{1}{2}}\left(a_{1}-\frac{1}{2}\right) \\
=\left(\frac{1}{3}\right)^{n-1} \times \frac{1}{6}=\frac{1}{2}\left(\frac{1}{3}\right)^{n} \\
\Rightarrow a_{n}=\frac{1}{2}+\frac{1}{2} \times \frac{1}{3^{n}} .
\end{array}
$$

Given $a_{1}=\frac{2}{3}$, we know $a_{n}=\frac{1}{2}\left(1+\frac{1}{3^{n}}\right)$.
Thus, $a_{4}=\frac{1}{2}\left(1+\frac{1}{3^{4}}\right)=\frac{41}{81}$.