27.11. (Belgium, 77). Three shooters $A, B, C$ decided to duel simultaneously. They positioned themselves at the vertices of an equilateral triangle and agreed on the following: the first shot is taken by $A$, the second by $B$, the third by $C$, and so on in a circle; if one of the shooters is eliminated, the duel continues between the two remaining. It is known that shooter $A$ hits the target with a probability of 0.3, shooter $C$ with a probability of 0.5, and shooter $B$ never misses. Each shooter aims at one of the two others or at the air in such a way as to maximize their probability of winning the duel. Where should shooter $A$ direct his first shot: 1) at shooter $C$; 2) at shooter $B$; 3) at the air?
27.11. Let's consider three events that may occur after the first shot of shooter $A$.

1) $C$ is hit. Then with probability 1, shooter $A$ will be hit by the first shot of $B$.
2) $B$ is hit. Then: either with probability 0.5, shooter $C$ will hit $A$ with his first shot, or with probability $0.5 \cdot 0.3$, shooter $A$ will hit $C$ with his second shot, or with probability $0.5 \cdot 0.7 \cdot 0.5$, shooter $C$ will hit $A$ with his second shot, or with probability $0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3$, shooter $A$ will hit $C$ with his third shot, and so on. Therefore, the probability for $A$ to win the duel in this case is

$$
\begin{aligned}
& 0.5 \cdot 0.3 + 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3 + 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.7 \cdot 0.5 \cdot 0.3 + \ldots = \\
& \quad = 0.15 \left(1 + 0.35 + 0.35^2 + \ldots \right) = 0.15 \frac{1}{1 - 0.35} = \frac{15}{100} \cdot \frac{100}{65} = \frac{3}{13}
\end{aligned}
$$

3) No one is hit. After this, $B$ will shoot at $C$ (as the more accurate of his opponents) and hit him. Then $A$ with probability 0.3 will hit $B$, winning the duel.

Thus, the most advantageous situation for shooter $A$ is when no one is hit after his shot. Therefore, he should shoot into the air the first time.