When people (like me) ask probability problems you can be sure of two things: (1) the least likely solution at the first glance is correct and (2) there is a slightly unfair trick hidden in the wording of the task.
Let's start with Andy's notation (for clarity and brevity). I'm convinced Andy's solution is correct starting with his assumptions (at least it looks very good to me without doing all the calculations). But there is a way, in which C (Joe) can push his chances of survival well above 50%. To do this, C (Joe) has not to follow Andy's assumption/advice three (If C shoots at B first and hits he faces certain death (from A) so he too must shoot A first), but whenever elected to shoot, as long as both opponents are still alive, he shoots into the air (a shooter with a 50/50 chance of hitting should be able to do that without fail). C only starts shooting with the aim to hit when the first of the two opponets is dead. He'll always be the first to shoot (in the remaining duell) this way.
With A remaining, C has a 50% chance to survive if he hits with the first shot, otherwise it's B surviving. With B remaining opponent, C has an above 50% chance to survive, for even if he doesn't kill with the first aimed shot, he has a 20% chance for a second, and so on.
With this reasoning, B has the worst chance of surviving (other than in Andy's scenario), for the possibility that C shoots first and hits A is taken away by C's reasoning. So without the exact probabilities (I didn't ask for them), C has the highest chance of surviving (above 50%), A has the second highest, and C the lowest.
Yes, it's a bit unfair, I know, but nowhere I had said explicitely that they may not shoot into the air if it is to their advantage.

By the way, with this thinking the truell is reduced to two successing duells, the first is A against B (A's chances higher) and then the winner shoots against C with the 50% shooter C starting first (C's chances higher)

Wolfgang