In the don't-spin scenario: Your argument that there are 5 chambers left and a bullet in one of them after your friend survives is actually an argument that $P(A|B)=1/5$, not that $P(A\cap B)=1/5$. In fact, $P(A\cap B)=2/3$, since there is a 4 in 6 chance that the bullet is not in either of the first two chambers. (You also define $A$ to be the event that you survive, but your calculation treats $A$ as the event that you die.)
Assuming no one ever respins the chambers (because otherwise the strategy is trivial—always pass the gun off):
Work backwards. Obviously, with only one chamber remaining, you would pass the gun off. Good luck captain!
With two chambers left, if you pass the gun, you have a $1/2$ chance of surviving, no matter what you do (pass the gun, or shoot again).
With three chambers left, if you pass the gun, your opponent has a $1/3$ chance of surviving that shot, after which your chances are both $1/2$. So your chances of survival by passing the gun off is $2/3$. So you pass the gun off.
With four chambers left, if you pass the gun, your opponent has a $1/4$ chance of surviving that shot, after which they will pass the gun off, and your chances of survival at that point are $1/3$. So the chances are $1/2$ for both, and it doesn't matter what you do.
With five chambers left, if you pass the gun, your opponent has a $1/5$ chance of surviving that shot, after which their chances of survival are $1/2$. So your chances of survival if you pass the gun off are $3/5$, so you pass the gun off.
You should be able to show by induction that passing the gun off is never a bad choice. With an even number of chambers left, it's always a $50$-$50$ proposition, and with an odd number of chambers left, it's strictly better to pass the gun off.
There might be an analysis that short-circuits the induction and makes the pattern obvious upfront, but I don't see it yet.
Best Answer
Let's model the gun has a list of 6 letters ABCDEF where EF have the bullets. The first person survives, so we must be in positions ABCD. Thus, there is a 1/4 chance of death if we pull the trigger because we can end up in positions BCDE and E is the only one that kills us. If we spin the wheel, then we have a 2/6=1/3 chance of death.
However, the strategic part (given that one of you must die, hurting your opponent is just as good as surviving for yourself) is that suppose I pull the trigger and survive. Now, I am in position CDE, so my opponent can now pick between a 1/3 chance (spinning) and a 1/3 chance (triggering). If he chooses to trigger again, the gun can only be in positions CD, so I have a 1/2 with trigger, and 1/3 with spinning.
Assuming optimal play for both parties, player one survives (otherwise, nothing interesting), player two triggers*, player one triggers (because spinning helps player two), player two spins (because 1/3 is better than 1/2), then player one triggers*. Note that the * parts are the same with opposite players. So basically the roles reverse. I am also of the opinion that player one has an advantage in long term, but is probably overshadowed by immediate chance to die. I will think about that some other time.