Die roll, forced re-roll on any odd number

Question

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

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

Accepted Answer

Two cases, blended by their probabilities. Case A: you keep the first roll. This happens for even faces 2, 4, 6, which is 3 of 6, probability 3/6. Given you kept it, the value averages: \frac{2+4+6}{3} = \frac{12}{3} = 4. Case B: you re-roll. This happens for odd faces 1, 3, 5, probability 3/6. The kept value is a fresh die, averaging 3.5. Blend by probability: EV = \frac{3}{6}(4) + \frac{3}{6}(3.5) = \frac{4 + 3.5}{2} = \frac{7.5}{2} = 3.75. Notice the boost above 3.5 is small here. Odd faces include the high value 5, so re-rolling them throws away a decent roll sometimes, which is why keepi…

Die roll, forced re-roll on any odd number

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