Probability of survival without rolling chamber (russian roulette)

Question

Say we are playing russian roulette with a pistol with $6$ slots, $1$ with a bullet. Then what's the probability that you'll survive if you've shot the gun $n$ times without rolling the chamber after each shot?

First of all, note that for $n\geq 6$, $P(A_n)=0$, where $A_n\equiv$ surviving $n$th shot.

The problem arises when $n\leq6$:

Solution says $P(A_n)=\dfrac{6-n}{6}$, but I got something different which makes sense too.

What I thought was that the probability of surviving each shot would be the number of empty slots left over all the slots left to shoot which are the empty ones plus the one with the bullet. Namely,

$$P(A_n)=\dfrac{6-n}{6-n+1}.$$

Thus, $$\{P(A_i)\}_{i=1}^6=\left\{\dfrac{5}{6},\dfrac{4}{5},\dfrac{3}{4},\dfrac{2}{3},\dfrac{1}{2},0\right\}.$$

My question is: which of the two is the correct? Mine or the solution sheet's?

Answer 1

As discussed in the comment section, what I actually computed was $P(A_n|\displaystyle{\cap}_{i=1}^{n-1}A_i)$ instead of $P(\displaystyle{\cap}_{i=1}^{n}A_i)$.

Employing multiplication rule we can find the actual answer: $$ \begin{aligned} P(\displaystyle{\cap}_{i=1}^{n}A_i)&=P(A_1)\prod_{i=2}^nP(A_i|\displaystyle{\cap}_{j=1}^{i-1}A_j)\\ &=P(A_1)\prod_{i=2}^n\dfrac{6-i}{6-i+1}\\ &=P(A_1)\dfrac{6-n}{5}=\dfrac{5}{6}\dfrac{6-n}{5}=\dfrac{6-n}{6}. \end{aligned}$$

Answer 2

Although the question has been answered in comments, I present an alternative view. Suppose the consecutive bullet chambers are named $A,B,C,D,E,F$ with the bullet in chamber $F$, wlog. On a diagram, letters $A,B,C,D,E,F$ are written in a circle in this order.

To survive the second shot means to survive two consecutive shots, which is same as choosing a pair of consecutive letters excluding $F$. $AB, BC, CD, DE$ are favorable, $EF, FA$ fail. So $P(A_2)=4/6$.

Similarly to survive third shot means to select three consecutive letters while excluding $F$. $ABC, BCD, CDE$ are favorable, $DEF$, $EFA$, $FAB$ fail. So $P(A_3)=3/6$.

In general there are $6$ groups of consecutive letters of length $n$ and $F$ is included in $n$ of these. It is excluded in $6-n$ of these. Hence for $n \le 6$, $P(A_n)=\dfrac{6-n}{6}$.