TITLE: Expectation of a die roll summed
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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}