The "truel"
Smith, Jones & Brown are fighting a three-person duel. Smith is a sure shot, killing 100% of the time. Jones kills with an 80% accuracy rate, and Brown a mere 50%. Each of them in turn gets one shot at a time, and they roll a 6-sided die to see who goes first, with each of them having a 1/3 chance of winning (and then roll again to see who goes second, with the two remaining contestants each having half a chance.)
Assuming they all use the best strategy, who has the best chance of winning the truel?
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1 solution
Since Smith & Jones will target each other as long as they're both alive, Brown's best strategy is to fire into the ground until one of them is dead, thus giving Brown first crack at the survivor.
Chances of survival:
If Smith goes first (1/2 a chance), he kills Jones with the first shot. Then Brown has half a chance of killing him: if he misses, Smith kills Brown with his second shot.
Total chances of survival if Smith goes first: Smith 25%, Jones 0%, Brown 25%
If Jones goes first (1/2 a chance), he shoots at Smith. He has an 80% chance of killing him. If he does not, Smith kills him and we jump to the above calculation. If he does, Brown fires and has half a chance of killing Jones. (Total chance of Brown surviving by killing Jones on his first shot: 20%. Chance that Jones has a chance to fire a second shot: 20%. ) Jones than has an 80% chance of killing Brown, who then has a 50% chance of killing Jones...(and on to infinity.)
Total survival chance for Jones: 8/45
Total for Smith: 30%
Total for Brown: 47/90