This problem has been bothering me all day. Basically the idea is that you're playing russian roulette and you put 1 bullet in one of 5 chambers and you spin once in the beginning (and only once). you're playing by yourself. what's the probability you die in the first three pulls? my inner mathematician says it's not 60%, b/c the events are interrelated, but i cannot explain myself (been too long).
It is 60%, because you only spin once. Label the chamber it stops on as 1, then next one as 2, etc. You'll die in 3 pulls if and only if the bullet is in chamber 1, 2, or 3. If you respun after each pull, though, your probability of death = 1 - (4/5)^3, or 48.8%.
thanks guys, you are right. i realize what my fallacy was. basically, this is from a court case that ruled that the kid that fired the gun 3 times in a row put his friend in a 60% chance of death. while 60% is true before you even start playing... it is not true once you have the knowledge that the first shot failed and the second shot failed. by the time you shoot the third shot, there are 3 chambers left and a 1/3 chance the third pull will be fatal. so the court should've gone w/ the 1/3 chance in their reasoning.
the chance that the bullet is in the chamber is 48.8%. the chance of death is probably slightly less, if we factor in being left in a vegetative state.
Edit: NM, I read and did the problem wrong The set up of the problem was right but misread the number of chambers (aren't most guns 6 shooters?). 60% chance he dies in the first 3 pulls. 1 -(.8*.75*.666666)
