6. A basketball player is practicing shooting. If the $n$th shot is made, the probability of making the $(n+1)$th shot is $\frac{2}{3}$; if the $n$th shot is missed, the probability of making the $(n+1)$th shot is $\frac{1}{3}$. If the player has a $\frac{2}{3}$ probability of making the 1st shot, then the probability of making the 4th shot is . $\qquad$
The probability of the player making the $n$-th shot is $p_{n}$, then $p_{n+1}=\frac{2}{3} p_{n}+\frac{1}{3}\left(1-p_{n}\right)=\frac{1}{3} p_{n}+\frac{1}{3}$ $\Rightarrow p_{n+1}-\frac{1}{2}=\frac{1}{3}\left(p_{n}-\frac{1}{2}\right) \Rightarrow p_{n}-\frac{1}{2}=\frac{1}{6}\left(\frac{1}{3}\right)^{n-1} \Rightarrow p_{n}=\frac{1}{2}\left[1+\left(\frac{1}{3}\right)^{n}\right]$.

Therefore, the probability of making the 4th shot is $p_{4}=\frac{41}{81}$.