# Thread: russian roulette

1. ## 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.

2. 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 $n$-th shot $p(n)= (5/6)^{n-1}(1/6)$, so the expected number of shots is:

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

If the cylinder is not spun between shots we have:

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

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

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

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

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

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

and the expected number of shots is:

$
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

3. 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?

4. 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.

5. 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.