Variant of the Truel Question: What is the Probability that A wins if A starts by shooting C?

Question

Shooters A, B, and C are having a truel. The probability that shooter A hits his target is 1/3, the probability that B hits his target is 2/3, and the probability that C hits his target is 1. They take turns, A first, then B, then C, and repeating until only one shooter is left. If A decides to shoot at C on his first turn, what is the probability that A wins?

Answer 1

Let us assume that each shooter tries to maximize their chances of survival. In this case, $A$ and $B$ will always aim for $C$ first; if they do not, they are certain to die if they should hit their opponent and $C$ gets their turn. $C$ will aim for $B$ if both $A$ and $B$ should be alive. We can distinguish the following cases:

1. $A$ kills $C$ and wins the duel with $B$ shooting first
2. $A$ misses $C$, $B$ kills $C$ and $A$ wins the duel with $A$ shooting first
3. $A$ misses $C$, $B$ misses $C$, $C$ kills $B$ and $A$ wins the duel with $A$ shooting first

In all other cases, $A$ dies. We find the following probabilities:

• $P(A~\text{kills}~C~\text{in first turn}) = \frac{1}{3}$
• $P(B~\text{wins}, B~\text{starts} | C~\text{dead}) = \frac{2}{3} + \frac{1}{3} \frac{2}{3} P(B~\text{wins}, B~\text{starts} | C~\text{dead}) \iff P(B~\text{wins}, B~\text{starts} | C~\text{dead}) = \frac{6}{7}$
• $P(A~\text{wins}, B~\text{starts} | C~\text{dead}) = 1 - P(B~\text{wins}, B~\text{starts} | C~\text{dead}) = \frac{1}{7}$
• $P(A~\text{misses}~C) = \frac{2}{3}$
• $P(B~\text{kills}~C | A~\text{misses}~C) = \frac{2}{3}$
• $P(A~\text{wins}, A~\text{starts} | C~\text{dead}) = \frac{1}{3} + \frac{2}{3} \frac{1}{3} P(A~\text{wins}, A~\text{starts} | C~\text{dead}) \iff P(A~\text{wins}, A~\text{starts} | C~\text{dead}) = \frac{3}{7}$
• $P(B~\text{misses}~C | A~\text{misses}~C) = \frac{1}{3}$
• $P(C~\text{kills}~B | A~\text{misses}~C, B~\text{misses}~C) = 1$
• $P(A~\text{wins} | B~\text{dead}) = \frac{1}{3}$

Putting this all together, we find:

$$P(A~\text{wins}) = \frac{1}{3} \cdot \frac{1}{7} + \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{3}{7} + \frac{2}{3} \cdot \frac{1}{3} \cdot 1 \cdot \frac{1}{3} = \frac{1}{21} + \frac{4}{21} + \frac{2}{27} = \frac{59}{189} \approx 0.312$$