Expectation of a die roll summed I know this problem involves conditional probability, but I'm confused as to how to tackle it.
Assume a die is rolled over and over, where the total is summed. If the die's roll is $\geq 3$ the game stops and the summed total is read out. What is the expectation of the total? What is the expected number of times the die was rolled?

You have $p=1/3$ to roll a die and get only $1$ or $2$.
So you have $P(n)=p^n q=p^n (1-p)$ to roll the die $n$ times getting less than $3$, and then more or equal $3$ at the $n+1$-th roll.
We include $n=0$, meaning that you get $\ge 3$ at the first roll.
The sum P(n)  over $0 \le n < \infty$ correctly gives $1$.
Now, the expected number of less than $3$ rolls will be
$$
\eqalign{
  & E(n) = \sum\limits_{0\; \le \,n\,} {n\,P(n)}  = (1 - p)\sum\limits_{0\; \le \,n\,} {n\,p^{\,n} }  =   \cr 
  &  = (1 - p)p{d \over {dp}}{1 \over {1 - p}} = {p \over {1 - p}} = {1 \over 2} \cr} 
$$
while the expected number of total rolls, of course is
$$
E(n + 1) = {3 \over 2}
$$
At each less than $3$ roll you can get,  with same probability, a 1 or a 2, thus in average $3/2$.
So espected sum of the rolls before stopping is $3/2E(n)=3/4$.
