Mean value in a russian roulette game

Consider a russian roulette game and suppose that there are $m$ chambers and $1$ bullet. In a problem one is asked to find the mean value of the number of times that a player has the chance of pulling the trigger.

I did the following: since there are $m$ chambers, the probability in one turn that the player will be alive is $p = \frac{m-1}{m}$.

In that case, consider the random variable $X$ taking values in $\mathbb{N}$ whose value is the number of times a player has played.

The probability of being alive after $n$ turns (assumed to be indepedent) can easily be found to be $p^n$ which is $p^n=\frac{(m-1)^n}{m^n}$.

Thus, the probability distribution $P(X=n)$ can be found considering that: the player having played $n$ turns means that he survided this $n$ turns. This is the same probability as above and hence $P(X=n)=p^n$.

The mean value then becomes

$$\langle X\rangle=\sum_{n=1}^\infty n P(X=n)=\sum_{n=0}^\infty n p^n=p\dfrac{d}{dp}\sum_{n=0}^\infty p^n=p\dfrac{d}{dp}\dfrac{1}{1-p}=p\dfrac{1}{(1-p)^2}.$$

But $1-p = \frac{1}{m}$ and hence $(1-p)^{-2}=m^2$, from where we obtain

$$\langle X \rangle = \frac{m-1}{m} m^2=m^2-m.$$

I believe this is wrong. In one book with this problem, considering $m=6$ the solutions says that $\langle X \rangle = 6$.

This would be the case if we had

$$\langle X\rangle = \sum_{n=0}^\infty p^n=\dfrac{1}{1-p}=m,$$

but this doesn't seem to make much sense, after all this would be the mean value of a random variable whose values are all $1$ with probabilities $p^n$ each.

So: is my approach correct? If not, where am I getting this wrong?

• I assume that, if a player shoots a blank, the barrel of the gun is re-randomized? (Is this always the assumption? I'm not much of a russian roulette player) – MCT Mar 8 '18 at 1:34
• @MCT, well it is not written in the problem, so I assumed this was the case and was solving the problem under this assumption. – user1620696 Mar 8 '18 at 1:37
• Okay. I'm also not sure what you mean that $X$ has values all one. – MCT Mar 8 '18 at 1:39

To have $X=n$, the player must have clicked on empty chambers for the first $n-1$ turns and then had the ill fortune of finding the loaded chamber on the $n$th go. Therefore, $$\Pr(X=n) = \left({m-1 \over m}\right)^{n-1}\left(\frac1m\right) = \frac1m p^{n-1},$$ and so $$E[X] = \frac1m \sum_{n=1}^\infty np^{n-1} = \frac1m\frac1{(1-p)^2} = \frac{m^2}m=m.$$
• so the method is the same I tried, but what I got wrong was how to interpret $X = n$, which affected the probability? So the point is that if the player played $n$ times, this means that he didn't play anymore which in turn has to be taken as meaning that on the $n$-th turn he found the chamber loaded? – user1620696 Mar 8 '18 at 2:24
Generally, if there is an event which happens with probability $p$, then the expected number of tries one must have until it happens is $1/p$. (For example, you are expected to roll a die six times before rolling a three).
Thus, it takes on average $m$ tries until someone successfully (or unsuccessfully?) pulls the trigger. Which means that each player, on average, has $m/m = 1$ try (assuming the players, say, go in a circle and the first person is random).