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?

Another way of looking at it:
If you roll a 3,4,5, or 6 on the first roll (1/6 chance each), then the expected total will be that number (3, 4, 5, or 6).
If you roll a 1 or a 2, (also 1/6 chance each), then the expected roll is 1 or 2, PLUS the total expected roll.
So the total expected roll $T$ should be:
\begin{aligned}
T &= \frac{1+T}{6} + \frac{2+T}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} \\
T &= \frac{1+2+3+4+5+6}{6} + \frac{2T}{6} \\
T &= \frac{7}{2} + \frac{T}{3}  & \text{(multiply by 3)} \\
3T &= \frac{21}{2} + T & \text{(subtract $T$)}\\
2T &= \frac{21}{2} & \text{(divide by 2)}\\
T &= \frac{21}{4} = 5.25
\end{aligned}
