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You may re-roll a fair die as often as you like, free of charge, and are paid the face value you accept. What is the game worth?

Answer with a number. Fractions, powers and expressions like 23/6 or C(52,5) are read correctly in practice.

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Answer

6

With unlimited re-rolls every roll presents the same decision, so the value is a fixed point: $V = \frac{1}{6}\sum_f \max(f, V)$ . Guess the threshold sits between $5$ and $6$ , meaning you keep only a $6$ : then $V = (5V + 6)/6$ , so $6V = 5V + 6$ and $V = 6$ . Check the guess against the answer, which is the step people skip: $V = 6$ says re-roll anything below $6$ , exactly the policy the equation assumed. Intuitively a free retry has no downside, so you simply wait for the maximum.

Worked solution

Formula $V = \frac{1}{6}\sum_{f=1}^{6}\max(f,\ V)$
Substitute $\text{if } 5 < V < 6:\ V = \frac{5V + 6}{6}$
Solve $6V = 5V + 6$
Answer $V = 6$

Sanity check. Consistent: $V = 6$ means you re-roll anything below 6, which is the policy the equation assumed.

Takeaway: Unlimited free re-rolls means you simply wait for the best outcome.

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Pricing a game: EV, re-rolls and when to stop

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You may re-roll a fair die as often as you like, free of charge, and are paid the face value you accept. What is the game worth?

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