This is the solution to Morris: Trial/Trual/Whatever. Please look there before reading the solution.
Jeff suggested that they might all choose not to shoot at each other and it would go on forever. This made me think about the logic. I’m going to say that someone shoots at someone at some point and everyone knows this. So everyone knows that if they shoot to miss then one of the others will shoot to hit at some point.
If anyone shoots to hit then they will shoot at the best shot. If they hit they will be in two-man duel and would prefer to be against the worst shot possible.
So François knows that no-one will shoot at him. Xavier knows that one of the others will shoot at him.
François has two options, he can shoot at Xavier or he can shoot to miss.
If he hits Xavier he will be in a two man dual with JC shooting first, he only has a one in five chance of surviving the first shot so his chance of survival is less than 1/5.
If François shoots to miss he knows that the other two will shoot at each other, if they choose to shoot to hit. So at some point JC or Xavier will kill the other. At this point François will be in a two man duel with himself shooting first. This gives him at least a fifity/fifty chance.
So François shoots to miss.
Xavier and JC know this. They know they are effectively in a two man duel with each other. They therefore shoot at each other.
That sorts out the strategy, what about the probabilities?
Let’s consider JC first. There is a fifty/fifty chance that Xavier or JC will shoot first. JC’s only chance is to shoot first and hit. The chance is 1/2 x 4/5 = 2/5.
So 2/5 of the time JC will face François with Francois shooting first.
Let’s say that the chance that François wins is f. He has a 1/2 chance of killing JC with his first shot. There is a 1/2 x 1/5 = 1/10 chance that both miss with their first shot, we are back to the starting position so the probability of François surviving is now f again.
The chance of François surviving is f = 1/2 + 1/10 f. This gives 9/10 f = 1/2 and f = 5/9. The chance of François surviving is 5/9. The chance of JC surviving is 4/9.
This gives JC a total survival chance of 2/5 x 4/9 = 8/45.
Now consider Xavier. He has a 3/5 chance of facing François with François shooting first.
His chance of surviving is 1/2, the chance that François misses.
His total survival chance is 3/10.
Finally Francois’ chance of surviving is 2/5 x 5/9 + 3/5 x 1/2 = 2/9 + 3/10 = 47/90.
So the probabilities of being the last man standing are: François 47/90; Xavier 3/10 and JC 8/45.
Amazingly the worst shot has the best chance of survival!