Conditional Probability With a Madman

Question

There are three six-chambered revolvers. The first has no bullets; the second has one bullet; and the third has two bullets in consecutive chambers. The cylinder advances automatically as the trigger is pulled. A madman grabs a revolver at random, aims it at his own head, pulls the trigger t times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability that you are shot given t?

The provided answers are $[0.167 (t = 0), 0.133 (t = 1), 0.154 (t = 2), 0.182 (t = 3), 0.222 (t = 4), 0.143 (t = 5), 0 (t ≥ 6)]$

I can't seem to match it. Can someone tell me where I'm wrong?

My attempt:

$\Pr(S|t = 0) = \dfrac{1}{3}[0 + 1/6 + 2/6] = 0.167$

$\Pr(S|t = 1) = \dfrac{1}{3}[0 + (5/6)(1/6) + (4/6)(2/6)] = 0.144..$

$\Pr(S|t = 2) = \dfrac{1}{3}[0 + (5/6)(4/5)(1/4) + (4/6)(3/5)(2/4)] = 0.122..$

I stopped here cause my answers weren't matching.

Answer 1

The thing you're missing is that after the trigger is pulled once, the revolver can only actually shoot you in a particular state - the madman spun it to exactly one before a bullet (note, per @Dan Uznanski's answer, the madman doesn't spin the chamber again). You're also missing the conditional normalization.

Remember $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$$ where $A$ is "madman fires" and $B$ is "madman pulls the trigger", and count out all 18 equally likely cases for $t=1$:

1. In 6 cases, 0-bullets is chosen. All 6 cases have the madman pulling the trigger and not firing.
2. In 6 cases, 1-bullet is chosen. 5 of the cases have the madman pulling the trigger (the last case, he already killed himself). Only 1 case - of these 5 - has him firing at you.
3. In 6 cases, 2-bullets is chosen. 2 cases don't get to this point. Only 1 case out of 4 has him firing (and now the fact that both bullets are consecutive is relevant).

Total: 15/18 worlds have the madman pulling the trigger. 2/18 cases have him firing. $2/15=0.133$.

For $t=2$, the answer is $(0+1+1)/(6+4+3)=0.154$.

For $t=3$, it's $(0+1+1)/(6+3+2)=0.182$.

For $t=4$, it's $(0+1+1)/(6+2+1)=0.222$.

For $t=5$, something interesting happens. It is now impossible for 2-bullets to have been chosen originally. The answer becomes $(0+1+0)/(6+1+0)=0.143$.

For $t\geq6$, even 1-bullet cannot be the chosen revolver. The chance of him firing is obviously $0$.

Answer 2

The most important thing is: he's not spinning the cylinder each time: the probability of each trigger pull having a bullet in the chamber is not independent of the previous trigger pulls, and instead depends on where the cylinder started and how many times the trigger has been pulled, advancing to the next chamber.

There are 18 possible starting positions for the revolver: six chambers in three weapons. After each trigger pull at his own head, some of these starting positions are made impossible: the first means that the three chambers with bullets in them were not the ones that the gun started in, the second disqualifies the two chambers immediately before a bullet, etc.

Meanwhile, there are only a few starting positions in which a bullet is now in the barrel after the madman is done: if he didn't pull the trigger at all, three, if he pulled it 1-4 times, two (since the second bullet would only show up if the first bullet didn't get fired), only one for 5 times (since the gun with two bullets is now disqualified), and none for six or more times.

So:

Pulls	Starts	Bullets	Probability
0	18	3	$1/6 ≈ 0.167$
1	15	2	$2/15 ≈ 0.133$
2	13	2	$2/13 ≈ 0.154$
3	11	2	$2/11 ≈ 0.182$
4	9	2	$2/9 ≈ 0.222$
5	7	1	$1/7 ≈ 0.143$
6+	6	0	$0$