The die that explodes on a six

Question

You roll a fair six-sided die and add up what you roll. There's a twist: every time you roll a 6, you keep the 6 and roll again, adding the new roll too. You keep going as long as you keep rolling sixes, stopping the first time you roll something other than a 6.

What is the expected value of your total?

Accepted Answer

Let E be the expected total. Condition on the first roll. With probability 	frac16 each, a roll of 1 through 5 ends the game at that value. A roll of 6 (probability 	frac16) contributes 6 and restarts the process, adding another E in expectation: E = \underbrace{\frac{1+2+3+4+5}{6}}{	ext{stop on 1–5}} + \frac{1}{6}\bigl(6 + E\bigr). Simplify: E = \frac{15}{6} + 1 + \frac{E}{6} = 3.5 + \frac{E}{6}. Solve for E: E - \frac{E}{6} = 3.5 \;\Rightarrow\; \frac{5}{6}E = 3.5 \;\Rightarrow\; E = \frac{3.5	imes 6}{5} = 4.2. The expected total is 4.2. Takeaway. Self-referential (recursive) expectatio…

The die that explodes on a six

Your answer

More Probability questions