Thread: Russian Roulette Statistics

I have a problem concerning Russian Roulette. Not a personal problem, mind you, but a statistical one. The game is a sole player with 1 gun, 1 bullet and the chamber is spun each time. I believe I have MOST of what I need to solve the problem, but am unsure.

Starting with this probability eq:


I think I have the constants set out to be
N=# of rounds played
p= probability of getting the bullet=1/6
q=probability of getting empty round=5/6

If the question is what is the probability of still being alive after N rounds, I can't figure what n should be. I think with that, the problem should be solved, but I can't wrap my mind around what it should be.

ps. that's q to the (N-n) power, btw, I don't think it looks quite right

Originally Posted by pentaquark

I have a problem concerning Russian Roulette. Not a personal problem, mind you, but a statistical one. The game is a sole player with 1 gun, 1 bullet and the chamber is spun each time. I believe I have MOST of what I need to solve the problem, but am unsure.

Starting with this probability eq:


I think I have the constants set out to be
N=# of rounds played
p= probability of getting the bullet=1/6
q=probability of getting empty round=5/6

If the question is what is the probability of still being alive after N rounds, I can't figure what n should be. I think with that, the problem should be solved, but I can't wrap my mind around what it should be.

ps. that's q to the (N-n) power, btw, I don't think it looks quite right

The equation should be expressed as follows.

If you're alive afer N rounds then the bullet hasn't been shot after N trials. The probability of beng alive is therefore .

By the way, I don't think the binomial distribution appropriate here. Does it make sense to have more than zero failures in N trials ....?

Ok, I see. I was way over thinking the problem.

Yeah, I see why the BD wouldn't work now. Thanks!