27.10. (Belgium, 81). Two players $A$ and $B$ observe a boy who continuously flips a coin. The results of the flips are recorded sequentially using letters: on the $k$-th position in the sequence, the letter $O$ or the letter $P$ is placed depending on whether "heads" or "tails" appears on the $k$-th flip, respectively. Player $A$ claims that the triplet 000 will appear in the record before the triplet $O P O$. Player $B$ bets that the opposite will happen. Which player has a better chance of winning this bet?
27.10. After the first letter $O$ (from the moment the boy is observed, with probability $I$ the letter $O$ will appear at least once), with equal probability of $1 / 4$, one of 364 combinations can follow:

$$
P O, \quad O O, \quad P P, \quad O P
$$

In the first case, player $B$ wins, in the second case, player $A$ wins. If the third case occurs, then after this, the players will have the same chances as at the beginning of the game. In the fourth case, with probability $1 / 2$ the letter $O$ will follow and player $B$ will win, and with probability $1 / 2$ the letter $P$ will follow, after which the players will have the same chances as at the beginning of the game. Thus, with probability $1 / 4$ player $A$ wins, with probability $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{2}=\frac{3}{8}$ player $B$ wins, and with probability $3 / 8$ a situation will arise where the players will have the same chances as at the beginning of the game. Therefore, player $B$ has a higher chance of winning than player $A$.