Question 190: Person A and Person B each roll a six-sided die, stopping when they roll a "6". Let the probability that the number of times they roll the dice differs by at most one be $\mathrm{P}$, then $\mathrm{P}=$ $\qquad$ -
Question 190, Answer: According to the conditions, the probability of rolling the dice exactly $\mathrm{k}$ times is $\mathrm{p}_{\mathrm{k}}=\left(\frac{5}{6}\right)^{\mathrm{k}-1} \cdot \frac{1}{6}$.
If A stops after one roll, then B rolls once or twice; if A stops after $k(k \geq 2)$ rolls, then B rolls $k-1$, $k$, or $k+1$ times. Thus we know:
$$
P=p_{1}\left(p_{1}+p_{2}\right)+\sum_{k=2}^{+\infty} p_{k}\left(p_{k-1}+p_{k}+p_{k+1}\right)=\frac{8}{33}
$$

Therefore, the required probability $P=\frac{8}{33}$.