Three-Way Duel With Rational Players
Alice, Bob, and Chad are in a three-way duel. They take turns shooting in the fixed order Alice, Bob, Chad, repeating until only one person remains. Their guns have the following hit probabilities:
- Alice: $1/3$
- Bob: $2/3$
- Chad: $1$ (never misses)
If a shot misses the target, it hits no one. Each player plays optimally to maximize their own probability of survival. On each turn, a player may choose to shoot at either opponent -- or choose not to shoot at all.
Under optimal play from all three players, what is the probability that Alice is the last one standing?
Hints
- Each player prefers facing the weakest remaining opponent. Who is the biggest threat, and who should each player target?
- Alice's best first move is counterintuitive: she should deliberately not shoot. If she kills either opponent, she's left facing the stronger survivor.
- After Alice passes, the first round resolves into Alice vs Bob (probability $2/3$) or Alice vs Chad (probability $1/3$). Each two-player duel can be solved with a geometric series.
Worked Solution
How to Think About It: In a three-way duel, the strongest player is the biggest threat to everyone. Chad never misses, so both Alice and Bob want Chad dead first. But here's the subtle part: Alice, being the weakest shooter, actually benefits from doing nothing in the first round. Why? If she shoots at Chad and hits (probability $1/3$), she's now in a two-player duel against Bob, who shoots with $2/3$ accuracy -- a tough spot. If she shoots at Bob and hits, she faces Chad, who never misses -- even worse. Her best move is to deliberately miss (or not shoot) and let Bob and Chad fight it out.
Quick Estimate: Bob will shoot at Chad (the bigger threat). He hits with probability $2/3$. If Bob hits, Chad dies and Alice faces Bob with Alice shooting first -- decent odds. If Bob misses (probability $1/3$), Chad shoots Bob dead (he never misses). Then Alice faces Chad. Rough estimate: $P(\text{Alice wins}) \approx (2/3) \times (\text{Alice vs Bob}) + (1/3) \times (\text{Alice vs Chad})$. Alice vs Chad: she must hit first shot ($1/3$). Alice vs Bob: maybe around $3/7$. So roughly $(2/3)(3/7) + (1/3)(1/3) = 6/21 + 7/63 \approx 0.40$.
Approach: Analyze Alice's optimal first move, then reduce to two-player duels.
Formal Solution:
Step 1: Alice's optimal first-round strategy.
Every player prefers facing a weaker opponent. Chad ($p = 1$) is the strongest, so: - Bob will shoot at Chad (eliminating the bigger threat). - Chad (if alive) will shoot at Bob (the bigger threat to Chad, since Bob is more accurate than Alice). - Alice should deliberately not shoot. Shooting at Chad risks killing him and leaving her against Bob. Shooting at Bob risks leaving her against Chad. Doing nothing lets Bob and Chad eliminate each other.
Step 2: What happens in the first round.
Alice passes. Bob shoots at Chad: - With probability $2/3$: Bob hits, Chad dies. Survivors: Alice and Bob, with Alice's turn next. - With probability $1/3$: Bob misses. Chad then shoots Bob dead (probability 1). Survivors: Alice and Chad, with Alice's turn next.
Step 3: Alice vs Bob (Alice shoots first).
This is a turn-based duel where Alice hits with probability $p = 1/3$ and Bob hits with $q = 2/3$, with Alice going first.
Let $a$ be Alice's win probability. On Alice's turn: - She hits Bob with probability $1/3$ -- she wins. - She misses with probability $2/3$. Then Bob shoots: - He hits Alice with probability $2/3$ -- Bob wins. - He misses with probability $1/3$ -- back to the start.
$$a = \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3} \cdot a$$
Solving: $a = \frac{1}{3} + \frac{2}{9}a$, so $\frac{7}{9}a = \frac{1}{3}$, giving $a = \frac{3}{7}$.
Step 4: Alice vs Chad (Alice shoots first).
Chad never misses, so if Alice doesn't kill Chad on her first shot, Chad kills her on his turn. Alice's win probability is simply $1/3$.
Step 5: Combine by total probability.
$$P(\text{Alice wins}) = \frac{2}{3} \cdot \frac{3}{7} + \frac{1}{3} \cdot \frac{1}{3} = \frac{6}{21} + \frac{1}{9} = \frac{18}{63} + \frac{7}{63} = \frac{25}{63}$$
Answer: Under optimal play, Alice's probability of being the last one standing is $\dfrac{25}{63} \approx 0.397$.
Intuition
This is the classic "truel" problem, and the most surprising result is that the weakest player has the best strategy -- and a surprisingly high win probability (nearly 40%). Alice exploits the fact that the two stronger players are bigger threats to each other than she is to either of them. By doing nothing, she guarantees that one of them is eliminated in the first round, and then she gets to shoot first in the ensuing two-player duel.
The broader lesson is about strategic patience. In game theory, having the option to "pass" or do nothing is genuinely valuable when other players will act in ways that benefit you. This shows up in trading contexts too: sometimes the optimal strategy is to wait and let other market participants trade against each other, then pick off the survivor. The weakest player's edge comes not from firepower but from being non-threatening enough that no one targets them first.