Sum of Rolls Until Threshold

A fair 6-sided die is rolled repeatedly until a roll of 3 or greater appears. All rolls are summed. What is the expected value of this sum?

Answer

Let $$p$$ be the probability of rolling 3 or greater on a fair 6-sided die. Then

p = \frac{\text{rolls }3,4,5,6}{6} = \frac{4}{6} = \frac{2}{3}.

Hence, the expected number of rolls needed to reach this threshold (3 or greater) is $\frac{1}{p} = \frac{3}{2}$ . Among these $\tfrac{3}{2}$ rolls on average, exactly 1 of them is the final roll (which is $\ge3$ ), so the remaining $\tfrac{3}{2}-1 = \tfrac12$ rolls on average must be rolls of 1 or 2. The average of $\{1,2\}$ is $\tfrac{1+2}{2} = 1.5$ . Therefore, the contribution to the sum from these “non-threshold” rolls is

\Bigl(\tfrac{3}{2} - 1\Bigr)\times 1.5 = \tfrac12 \times 1.5 = \tfrac{3}{4}.

For the final roll, which is at least 3, the possible outcomes are $\{3,4,5,6\}$ , whose average is $\tfrac{3+4+5+6}{4} = 4.5$ . Adding these contributions gives

\mathbb{E}[\text{sum}] = \tfrac{3}{4} \;+\; 4.5 = \tfrac{3}{4} + \tfrac{9}{2} = \tfrac{3}{4} + \tfrac{18}{4} = \tfrac{21}{4} = 5.25.