Two faces bust you now

Question

You roll a fair die, adding each face value to a running score. But if you roll a $\mathbf{1}$ or a $\mathbf{2}$ , you lose the whole accumulated score and the round ends. At any point you may instead 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 2/6 you roll a 1 or 2 and lose s (change = -s). - With probability 1/6 each you roll 3, 4, 5, 6, gaining that amount. Expected change from one more roll: E[\Delta] = \frac{2}{6}(-s) + \frac{1}{6}(3+4+5+6) = \frac{-2s + 18}{6} = \frac{18 - 2s}{6}. This is positive exactly when s < 9. So keep rolling while your score is below 9; stop once you reach 9 or more. Two losing faces double the per-roll wipeout risk, so the safe threshold drops well below the original game's 20. Because the downside grows with s while the fixed upside…

Two faces bust you now

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