When the biggest face is the trap

Question

You roll a fair die, adding each face value to a running score. But rolling a $\mathbf{6}$ wipes the whole accumulated score and ends the round. At any point you may stop and bank your current score.

What is your stopping rule, and roughly what is the game worth?

Accepted Answer

One-step reasoning. With score s, roll once more: - With probability 1/6 you roll a 6 and lose s (change = -s). - With probability 1/6 each you roll 1, 2, 3, 4, 5, gaining that amount. Expected change: E[\Delta] = \frac{1}{6}(-s) + \frac{1}{6}(1+2+3+4+5) = \frac{-s + 15}{6} = \frac{15 - s}{6}. This is positive exactly when s < 15, so keep rolling while your score is below 15; stop once you reach 15 or more. The wipeout probability (1/6) matches the original game, but now the killer face is the biggest one, so the safe faces only sum to 15 instead of 20, and the threshold drops from 20 to 15. …

When the biggest face is the trap

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