There are $n$ fair dice. They are all tossed every time except the dice that are removed. A dice is removed if $3$ is rolled. What is the expected number of rolls? Any help would be appreciated.
My attempt: Consider the case of two dice. The process ends on step one with probability $\frac{1}{36}$, if one of the dice rolls up $6$ and other doesn't the on average $\frac{10}{36}(1+6)$ more steps are needed. Finally if $6$ doesn't roll up on both of the rolls then $\frac{25}{36}(z+1)$ rolls will be needed where $z$ is the expected number of rolls until both dice are removed. Thus $$z= \frac{1}{36}+\frac{10}{36}(1+6)+\frac{25}{36}(z+1) $$ Is this correct? If it is I can extend it recursively for $n$ dice.