Probability of there being a bullet in the chamber Suppose we have three $6$-chambered guns: * The first has no bullets. * The second has one bullet. * The third has two bullets in consecutive chambers. The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, 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 you are shot?

**Setup** :

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $\left\\{1,2,3,4,5,6,\infty\right\\}$

Note that there are $18$ chambers in total having equal probability to be chosen for the first shot, and it is not really difficult to find that:

$P\left(X=1\right)=\frac{3}{18}$

$P\left(X=2\right)=P\left(X=3\right)=P\left(X=4\right)=P\left(X=5\right)=\frac{2}{18}$

$P\left(X=6\right)=\frac{1}{18}$

$P\left(X=\infty\right)=\frac{6}{18}$

Based on this you can find $P(X=t+1\mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $t\in\\{0,1,2,3,4,5\\}$.

I leave the rest to you.