If you roll a 1 you roll again Suppose you decide to roll a single die $3$ times, and at any time you roll a $1$, you roll the die two more times. What's the expected number of rolls in this (rather contrived) game? If I change it to start with the intention of rolling once, and rolling a $1$ instead only adds one more roll, this becomes a relatively straightforward series and I think the answer comes out to be $6/5$ or thereabouts. But I don't know how to extend the solution to the $(3,2)$ case as I've asked it here, or the general $(m,n)$ case. I tried to define a random variable $X$ to count the number of rolls and then brute for the probabilities for each non-zero value of $X$, but it seemed too tedious to be the most efficient solution.

As was mentioned in a comment, by linearity of expectation the result is $3$ times the result for a single die with the same rules.

For a single die, you get $1$ roll no matter what you roll, and then additionally twice the expected value if you roll a $1$, so the expected value $E$ satisfies

$$ E=1+\frac16\cdot2E\;. $$

The solution is $E=\frac32$, and since you have $3$ of these, you expect to roll the die $\frac92$ times.