A three-way duel

Idea. Work backwards from the duels, find who the stronger players aim at, and then compare A's options. A's best move is to miss on purpose.

Step 1: two-player duels, A shooting first.

• A against B: A wins at once with probability $0.3$; otherwise B hits with probability $0.5$; otherwise the duel restarts. So $p=0.3+0.7\cdot 0.5p$, giving $p=\frac{0.3}{0.65}=\frac{6}{13}$.
• A against C: A must hit with the first shot, since C never misses: $\frac{3}{10}$.

Step 2: B and C with three alive. Each prefers to shoot the more dangerous opponent. If B hits A, C then kills B for sure, so B aims at C. C, on its turn with all three alive, kills B, the stronger remaining threat, and then faces A.

Step 3: A's options on the first shot.

• Fire into the air. Then B shoots at C. With probability $\frac{1}{2}$ B hits, and A starts a duel against B: survival $\frac{6}{13}$. With probability $\frac{1}{2}$ B misses, C kills B, and A starts a duel against C: survival $\frac{3}{10}$. Total: $$\frac{1}{2}\cdot \frac{6}{13}+\frac{1}{2}\cdot \frac{3}{10}=\frac{99}{260}\approx \mathrm{0.381.}$$
• Shoot at C. If A hits ($0.3$), B shoots first in a duel against A and A survives with probability only $0.5\cdot \frac{6}{13}=\frac{3}{13}$. If A misses, we are back to the line above. Total $0.3\cdot \frac{3}{13}+0.7\cdot \frac{99}{260}\approx 0.336$.
• Shoot at B. Hitting B leaves A alone with C, who kills A. Worse still.

Step 4. A fires into the air and survives with probability $\frac{99}{260}\approx 0.381$. That is the best chance of the three: B survives with probability $\frac{1}{2}\cdot \frac{7}{13}=\frac{7}{26}\approx 0.27$ and C with $\frac{1}{2}\cdot 0.7=0.35$. The weakest shooter does best because the two strong ones aim at each other.