Answer to old chestnut: truel
C's best strategy is to intentionally shoot his first shot into the ground away from the other two men. Consider the following:
At any time that there are two men left, of course, they shoot at each other. So what remains to be determined is the strategy for each man on a turn when all three are still in the truel.
If A gets such a turn, he kills B. The results of this is that C shoots at A, killing him with probability 30%, and otherwise A kills C and wins with probability 70%. If A kills C first instead, then B kills A 50% of the time on the next shot.
If B gets such a turn, he shoots at A. If he kills him, then C and B have a shoot-out, with C winning on the first shot 30% of the time, B on the next shot 50% of the remaining part of the time (35% of the time that this shootout begins), and so forth, resulting in B having a slightly better than 50% chance of winning if he kills A (the probability in the 35% of cases where both miss their first shots at each other is the same as the initial probability, so B wins 7/13 of the time and C wins 6/13 of the time. On the other hand, if he shoots at C and kills him, then A kills B on the next turn and B has no chance of winning. In either case if B misses, it becomes another shooter's turn, regardless of who B shot at, so B's chances of winning are better if he shoots at A.
On C's turn, if he shoots at A and kills him, the B-C shootout starts but with B having the first shot. In this case, B wins 1/2 of the time, and then C wins 6/13 of the remaining time (3/13 overall). If he shoots at B and kills him, A kills C the next turn and C has no chance of winning. On the other hand, if C misses, he has a 3/10 chance of winning if A has the next shot, and the average of a 3/10 and 6/13 chance of winning if B has the next shot. In either case, he has a better chance of winning if he misses than if he hits, so he intentionally misses with his first shot.
Note that this strategy of intentionally missing does not apply to the other players. A wins 70% of the time by killing B, but expects(*) B to kill him 50% of the time if he does anything else. B wins 7/13 of the time if he kills A on the first shot, but expects not to live at all otherwise (since he knows C doesn't shoot in the 3-player situation, A has the next shot in all cases and shoots B (*).
(*) If A assumes B chooses the intentional miss strategy, or vice versa, then the chance of winning with the intentional miss strategy becomes undefined, as everybody is intentionally missing. However, the players have in mind the goal of winning the duel, and nobody ever will when playing this way, so it is safe to assume that if they start out this way, somebody will break the cycle by shooting st somebody else.