Then P1 = Pr[injured & survive] and P2 = Pr[injured | survive] = P1 / Pr[survive]. But Pr[survive] is always at least 1/2: if your opponent shoots himself before you do, then you definitely survive. Therefore P2 is at most twice P1, and is negligible whenever P1 is negligible.

I see how you calculated that, but I think you're looking at the wrong pieces of evidence, and I agree with TheOtherDave.

You have an even split chance of getting the real bullet in play, so let's put that down:

P[bullet] = 0.5 P[¬bullet] = 0.5

Then, given that you DO get the bullet, you have a very high chance of being dead if you don't know how you will die:

P[die | bullet] = 0.99 P[¬die | bullet] = 0.01

Of course, this means that overall, P[die] = 0.495, and P[injury] = 0.005. However, if you already also know that P[¬die] = 1, then...

P[bullet & ¬die] = 0.5 P[¬bullet & ¬die] = 0.5

...because P[bullet] is computed before its causal effects (death or injury) can enter the picture, which means you're left with P[injury] = 0.5 (a hundred times larger than P1!).

Thus, while the first chance of injury is negligible, the chance of injury once you already know that you won't die is massively larger, given that P[injury XOR death | bullet] = 1 (which is implied in the problem statement, I would assume).

Edit: I realize that this makes the assumption that your chances of getting the bullet doesn't correlate with knowing how you will die, but it most clearly exposes the difference between your calculation and other possible calculations. This is not the correct way to calculate the probabilities in real life, since it's much more likely that non-death is achieved by not having the bullet in the first place (or by failing to play Russian Roulette at all), but there's all kinds of parameters you can play with here. All I'm saying is that P2 isn't necessarily at most twice P1, it all depends on the other implicit conditions and priors.