Value a Die Game with Optional Continuation

You roll a fair six-sided die repeatedly. Each rolled value is added to your running total. If the roll is 1 or 2, the game stops immediately. If the roll is 3, 4, 5, or 6, you may either stop with your accumulated winnings or keep rolling. Assuming risk-neutral preferences and no discounting, what is the fair price of this game, and what is the optimal stopping policy? Derive the value rather than estimating it through simulation.

Constraints & Assumptions

• Winnings accumulate across rolls.
• The choice to continue is made after collecting a 3, 4, 5, or 6.
• There is no fixed horizon, transaction cost, or loss on a later roll.
• “Fair price” means expected monetary value to a risk-neutral player.

Clarifying Questions to Ask

• Do the amounts accumulate, or does only the final roll pay?
• Is stopping allowed after a 1 or 2, or is termination forced?
• Does continuing ever forfeit prior winnings?
• Is the player risk neutral?

Hint: Write a continuation-value equation. After any optional stopping point, the future game has the same value as it did before the first roll.

What a Strong Answer Covers

• A Bellman or first-step equation for the continuation value.
• Justification that continuing after 3 through 6 is optimal.
• The numerical expected value and a distinction between fair value and personal willingness to pay.
• A check that the expected game length is finite.

Follow-up Questions

• How would a per-roll fee change the stopping decision?
• What if a later roll of 1 or 2 erased all accumulated winnings?
• How would risk aversion affect the maximum entry price?

Overview: Value an infinite-horizon die game with optional continuation. Derive the Bellman equation, optimal always-continue policy, expected entry value, and the role of risk preferences.

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Community answers

Answer by iyad.e.k2005

One should continue playing as long as the expected gain $g_t$ at time $t$ is at least 0; since the player is risk neutral, he will play even when the expected value is 0. The expected gain is computed as $1/3$ times the expected gain conditional on rolling 1 or 2 plus $2/3$ times the expected gain conditional on rolling 3, 4, 5, or 6. The first term is $1.5 - x$, where $x$ denotes the total accumulated winnings, and the second is $4.5$ (assuming that if you roll 1 or 2, you lose everything already accumulated and leave only with $1 or $2). Solving for $x$ such that the gain is at least zero shows that one should continue as long as $x$ is smaller than $10.5$.