Die roll, forced re-roll on anything below four

Question

You roll a fair six-sided die. If it shows 1, 2, or 3, you must re-roll once and keep that second roll. If it shows 4, 5, or 6, you keep the first roll.

On average, what is the value of the roll you keep, in dollars?

Accepted Answer

Break into the two cases and blend them by their probabilities. Case A: you keep the first roll. This happens for 4, 5, or 6, which is 3 of the 6 faces, probability 3/6. Given you kept it, the value averages: \frac{4+5+6}{3} = \frac{15}{3} = 5. Case B: you re-roll. This happens for 1, 2, or 3, probability 3/6. The kept value is a fresh die, averaging 3.5. Blend by probability: EV = \frac{3}{6}(5) + \frac{3}{6}(3.5) = \frac{5 + 3.5}{2} = \frac{8.5}{2} = 4.25. Because the split is exactly half and half here, the answer is simply the midpoint of the two case averages, 5 and 3.5. Re-rolling the b…

Die roll, forced re-roll on anything below four

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