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

Let the number of throws by Xiao Ming and Xiao Hong be $\xi$ and $\eta$, respectively. Then the required probability is
$$
\begin{aligned}
\sum_{i=1}^{+\infty}[P(\xi=\eta=i)+P(\xi=i, \eta=i+1)+ \\
P(\xi=i+1, \eta=i)] .
\end{aligned}
$$

By independence, the required probability is
$$
\begin{array}{l}
\sum_{i=1}^{+\infty}[P(\xi=i) P(\eta=i)+P(\xi=i) P(\eta=i+1)+ \\
P(\xi=i+1) P(\eta=i)] \\
=\sum_{i=1}^{+\infty}\left[\left(\frac{5}{6}\right)^{i-1} \times \frac{1}{6} \times\left(\frac{5}{6}\right)^{i-1} \times \frac{1}{6}+\right. \\
\left.2 \times\left(\frac{5}{6}\right)^{i-1} \times \frac{1}{6} \times\left(\frac{5}{6}\right)^{i} \times \frac{1}{6}\right] \\
=\frac{8}{33} .
\end{array}
$$