The chances of injury are then not quite double what they would have been without death-immunity. This is reasonably low, because the prior odds of survival at all are reasonably high (0.505) - had the experiment been riskier, such that there was only a 0.01% chance of survival overall, then the chance of injury in the death-immunity case would be correspondingly higher.
(We also have not yet taken into account the effect of the first player—in such a game of Russian Roulette, he who shoots first has a higher prior probability of death).
I agree that this is how it should be reasoned ideally, which I only realized after first posting the grandparent. See other comments and the edit for how I arrived at the 50⁄50 reasoning. If you know the answer for the bottom/last question in this comment, I’d be interested to know.
No, that’s not right. What we’re interested in here, is P[injury|¬die]. Using Bayes’ Theroem:
P[injury|¬die] = {P[¬die|injury]*P[injury]}/P[¬die]
Using the figures you assume, and recalling that “injury” refers only to non-fatal injury (hence P[¬die|injury]==1):
P[injury|¬die] = {1*0.005}/0.505 = 1⁄101 = approx. 0.00990099
The chances of injury are then not quite double what they would have been without death-immunity. This is reasonably low, because the prior odds of survival at all are reasonably high (0.505) - had the experiment been riskier, such that there was only a 0.01% chance of survival overall, then the chance of injury in the death-immunity case would be correspondingly higher.
(We also have not yet taken into account the effect of the first player—in such a game of Russian Roulette, he who shoots first has a higher prior probability of death).
Thanks for the full Bayes Theorem breakdown.
I agree that this is how it should be reasoned ideally, which I only realized after first posting the grandparent. See other comments and the edit for how I arrived at the 50⁄50 reasoning. If you know the answer for the bottom/last question in this comment, I’d be interested to know.