# Intuitive conditional probability seemingly not working

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?

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

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.
• 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. Sep 26, 2018 at 8:41
• 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. Sep 26, 2018 at 17:25