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.
Best Answer
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 $t=1$:
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 $0$.