Solution
A Truel with Better Aim
Show the problem again
Three marksmen A, B and C have accuracies $1/2$, $2/3$ and $1$ respectively: each shot by a marksman hits its chosen target independently with that probability. They fire in the fixed order A, B, C, A, B, C, ... (skipping anyone who has died) until only one is left alive. On his turn a marksman may aim at any living opponent or deliberately miss. Each marksman maximises his own probability of being the sole survivor, and this is common knowledge. Use the equilibrium in which the strategy depends only on who is alive and whose turn it is (it is unique here). What is A's probability of surviving? Give a reduced fraction (a decimal accurate to four places is also accepted).
Worked solution
The answer is 17/30. Start with the two-person endgames. In a duel where the mover hits with probability $a$ and the other with $b$, the mover survives with probability $a/(1-(1-a)(1-b))$, and shooting always beats passing. This gives: A moving first against B, A survives with $3/5$ (and B with $2/5$); A moving first against C, A survives with $1/2$; B moving first against A, A survives with only $1/5$; C moving first against anyone wins for certain. Now the opening. If A shoots B and hits, C kills A: 0. If A shoots C and hits, B moves first against A and A survives with only $1/5$. So shooting is poor. If A misses on purpose, B must act. Killing A hands C the move, and C kills B, so B shoots C. If B hits C (probability $2/3$), A moves first against B, so A survives with $3/5$. If B misses (probability $1/3$), C kills B, the most dangerous opponent, and A then moves first against C, surviving with $1/2$. Hence A survives with $\frac23\cdot\frac35+\frac13\cdot\frac12=\frac25+\frac16=\frac{17}{30}$. Aiming at C instead would give $\frac12\cdot\frac15+\frac12\cdot\frac{17}{30}=\frac{23}{60}<\frac{17}{30}$, so A is right to fire into the air. An exhaustive check of all 27 stationary strategy profiles in exact arithmetic finds exactly this equilibrium, with survival probabilities $(17/30,\,4/15,\,1/6)$.
Source: Truel analysis in the style of Mosteller and of Kilgour and Brams, with new accuracies. Statement written for AxiomIQ.