# russian roulette

• Aug 2nd 2008, 12:10 PM
frankdent1
russian roulette
I was just wondering how to calculate the expected number of trigger pulls required to shoot a single bullet out of a six-shot revolver.
• Aug 2nd 2008, 12:33 PM
CaptainBlack
Quote:

Originally Posted by frankdent1
I was just wondering how to calculate the expected number of trigger pulls required to shoot a single bullet out of a six-shot revolver.

Spinning the cylinder between shots:

Probability that you die on the $\displaystyle n$-th shot $\displaystyle p(n)= (5/6)^{n-1}(1/6)$, so the expected number of shots is:

$\displaystyle E(n)=\sum_{n=1}^{\infty} n(5/6)^{n-1}(1/6)=6$

If the cylinder is not spun between shots we have:

prob of dying on first shot $\displaystyle p(1)=1/6$

$\displaystyle p(2)=(5/6)(1/5)$

$\displaystyle p(3)=(5/6)(4/5)(1/4)$

$\displaystyle p(4)=(5/6)(4/5)(3/4)(1/3)$

$\displaystyle p(5)=(5/6)(4/5)(3/4)(2/3)(1/2)$

$\displaystyle p(6)=(5/6)(4/5)(3/4)(2/3)(1/2)(1)$

and the expected number of shots is:

$\displaystyle E(n)=1 \times p(1)+2 \times p(2) + 3 \times p(3) +4 \times p(4)+5 \times p(5) + 6 \times p(6) = 3.5$

RonL
• Aug 9th 2008, 12:18 PM
frankdent1
Another related question.

In 2 player russian roulette, alternating and not spinning the cylinder between shots, what's the probability the first guy will die?
• Aug 9th 2008, 12:47 PM
frankdent1
Nevermind it's 50/50 because it's 3.5, which is between 3rd shot (player 1) and 4th shot (player 2). Although you'd think the guy who goes first would have a disadvantage.
• Aug 9th 2008, 06:53 PM
Quick
Quote:

Originally Posted by frankdent1
Although you'd think the guy who goes first would have a disadvantage.

He would if it you spin the wheel each time. Otherwise the loser is predetermined and there would be a 50-50 chance because there is a 50% chance it's in an odd round, and a 50% chance it's in an even round.