and they are always accurate. However, the way for A to maximise the chances of winning this Truel is to not aim at B or C in the first turn, but rather to intentionally miss both of them, by shooting the gun straight into the air! To show why this is the case, we will have to analyze several possible outcomes. Firstly, If A aims at B : If A hits B (a 33% chance) then A will definitely lose, because C will kill A right away. If A misses B, then B will aim at C (because if B doesn’t kill C, C will kill B next turn as B is more dangerous
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than A). If B hits C, then A and B will shoot at each other in turns until one of them is killed. If B misses C, then C will kill B, and then A will have 1 chance of shooting C, if A misses C will shoot A and the truel will be over. As a result the probability of A losing if he aims at B
= P(A hits B)+ P(A misses C)[P(B misses C)xP(A misses C)]+[P(B hits C)xP(B will win duel when A goes first)].
P(B will win duel when A goes first)]
= 1 − [(1/3) + (2/3)(1/3)2 + (2/3)2(1/3)2. . . . ] = 1 − (1/3)/(1 − 2/9) = 4/7
Therefore the probability of A losing if he aims at B = 1/3 + 2/3[(1/3 x 2/3) + (2/3 x 4/7)] = 139/189 . So if A aims at B, he will only have a 40/189 (26%) chance of winning. If A aims at C: If A hits C, then B and A will have a duel until one dies (with B going first), if A misses C, then B will aim at C, and if B hits C then A and B will have a duel until one dies (with A going first). If B misses C then C will kill B and A can aim at C, if A misses C then C will also kill A. Therefore, the probability of A losing if he aims at C first is: 8 =[P(A hits C)xP(B wins duel when B goes first)]+P(A misses C)[P(B hits C)xP(B wins duel when A goes first)+P(B misses C) x P(A misses C)]
P(B wins duel when B goes first)= (2/3)+(1/3)(2/3) 2 +(1/3) 2 (2/3) 3 ....=(2/3)/(1- 2/9)=6/7.
8 http://www.readthehook.com/79248/strange-true-aim-high-avoid-first-shot-truel
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