1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – MCT Mar 8 '18 at 1:34
  • $\begingroup$ @MCT, well it is not written in the problem, so I assumed this was the case and was solving the problem under this assumption. $\endgroup$ – user1620696 Mar 8 '18 at 1:37
  • $\begingroup$ Okay. I'm also not sure what you mean that $X$ has values all one. $\endgroup$ – MCT Mar 8 '18 at 1:39
1
$\begingroup$

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 $$E[X] = \frac1m \sum_{n=1}^\infty np^{n-1} = \frac1m\frac1{(1-p)^2} = \frac{m^2}m=m.$$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user1620696 Mar 8 '18 at 2:24
  • $\begingroup$ That’s it exactly. $\endgroup$ – amd Mar 8 '18 at 2:25
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.