7. Person A and Person B independently and repeatedly roll a fair die until the first 6 appears. The probability that the number of rolls by A and B differs by no more than 1 is $\qquad$
7. $\frac{8}{33}$.

Let the number of throws made by A and B be $\xi$ and $\eta$, respectively. Then the required probability is
$$
\begin{array}{l}
\sum_{i=1}^{\infty}(P(\xi=\eta=i)+P(\xi=i, \eta=i+1)+ \\
P(\xi=i+1, \eta=i)) .
\end{array}
$$

By independence, the required probability is
$$
\begin{array}{l}
\sum_{i=1}^{\infty}(P(\xi=i) P(\eta=i)+P(\xi=i) . \\
P(\eta=i+1)+P(\xi=i+1) P(\eta=i)) \\
=\sum_{i=1}^{\infty}\left(\left(\frac{5}{6}\right)^{i-1} \frac{1}{6}\left(\frac{5}{6}\right)^{i-1} \frac{1}{6}+\right. \\
\left(\frac{5}{6}\right)^{i-1} \frac{1}{6}\left(\frac{5}{6}\right)^{i} \frac{1}{6}+ \\
\left.\quad \frac{1}{6}\left(\frac{5}{6}\right)^{i} \frac{1}{6}\left(\frac{5}{6}\right)^{i-1} \frac{1}{6}\right) \\
=\frac{8}{33} .
\end{array}
$$