Let’s take this problem. $A$,$B$ and $C$ take a revolver (6 slots) and load just one bullet. Then $A$ tries to shoot a target. If it shoots, then the game is over, if not he gives the gun to $B$ who tilts the wheel and tries to shoot, too. It goes on until someone shoots. What’s the probability that $C$ wins?

If I consider the shot as a geometric variable $X$ of parameter $1/6$ then it is easy to see that the probability is $\sum_{j=1}^{\infty} P(X=3j)=\sum_{j=1}^{\infty} (5/6)^{3n-1} (1/6)=25/91$.

But let’s take another road. I consider the events $K$ that is “$C$ wins” and the events $A_i$ that are “ the bullet is shot at the i-th round”. Then I have that $P(K)=\sum_{i=1}^{\infty} P(K|A_i)P(A_i)$.

Now, $P(A_i)=(5/6)^{3i-3} (1/6) (1+ 5/6 + 25/36)=(5/6)^{3i-3} (1/6)(91/36)$ because I don’t shoot the previous rounds and one of the three shoots at the i-th round. Now I calculate $P(K|A_i)$. It can be shown manually that it doesn’t depend on $i$, and using the definition one can see that it actually is equal to $(5/6)^2 (36/91)$ for every $i$. This way the answer is the same obtained above with the other method.

But from an intuitive point of view I could say: ok, I want to know the probability of $K$ knowing (supposing) that $A_i$ is verified, that is knowing that the gun shoots at the i-th round. This means that the $i$ doesn’t matter, and I will only start calculating once the i-th round begins (because I know the gun won’t shoot before that). Then I say that if the first two people don’t shoot, then the third is bound to shoot, since I know the gun must shoot in this round. But with this intuitive way of thinking I would get a probability of $(5/6)^2$, and $36/91$ is missing.

What am I getting wrong in the intuitive approach?

  • 1
    $\begingroup$ The mistake is that, conditional on $A_i$, the probability that player A does not shoot in round $i$ is not 5/6. $\endgroup$ Sep 25, 2018 at 22:42

1 Answer 1


The crux of your mistake is in the following false assertion: "Given that the bullet will be fired in round $i$, the probability that it is fired by the first shooter is $5/6$."

If you actually wanted to compute the conditional probability of this occurring, you'd need to do it carefully. Let $A_i$ denote the event that the gun is fired during round $i$, and let $B_i$ denote the event that the gun is fired by the first shooter within round $i$. (Clearly, $B_i \subset A_i$.) Then $$\mathbb P(B_i \mid A_i) = \frac{\mathbb P(B_i)}{\mathbb P(A_i)} = \frac{(5/6)^{3i-3}(1/6)}{(5/6)^{3i-3}(1/6)(1 + 5/6 + 25/36)} = 36/91.$$

That calculation isn't terribly relevant to what you actually want, but hopefully it's instructive about where the error is. To get the actual probability you want as a conditional probability, let $C_i$ denote the event that the gun is fired by the third shooter in round $i$: $$\mathbb P(C_i \mid A_i) = \frac{\mathbb P(C_i)}{\mathbb P(A_i)} = \frac{(5/6)^{3i-3}(1/6)(25/36)}{(5/6)^{3i-3}(1/6)(1 + 5/6 + 25/36)} = 25/91.$$

The following is true, though: "Given that the bullet has not yet been fired by round $i$, the probability that it is fired by the first shooter is $5/6$." But that's a rather different statement than the earlier one.

  • $\begingroup$ So the point is that when I suppose that the bullet will be fired in the i-th round, I automatically get a variable that is non geometric and whose probability must be obtained through calculations. Actually this makes sense, a geometric variable has no memory, while this one has. However, I guess there are not intuitively obvious reasons for which the answer is $36/91$? I mean something obvious like “1 bullet, 6 holes, the probability is 1/6”, that does not even require the definition of conditional probability. $\endgroup$
    – tommy1996q
    Sep 26, 2018 at 8:41
  • $\begingroup$ Yeah, I don't think I have an intuitive explanation for why it's $36/91$ like you're hoping for. At least, nothing that wouldn't boil down to just doing the calculations anyway. $\endgroup$ Sep 26, 2018 at 17:25

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