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