For the dice problem, the problem is now: every round we draw nice, if you get to stay, you get $4. If you is kicked out, you get nothing. If you choose to quit, you get $10. We will now use a mixed strategy: "I want to first take the risk and earn at least X dollars before I quit and take my $10". What's the optimal X?

Implement policy iteration to find out X, suppose the max money you can get is $100. Define the state space to include the money you got so far.

https://colab.research.google.com/drive/13VwGV6JRm5_mwuKb2mtX6XE45cKC8t14?usp=sharing
Links to an external site.

The optimal X is:

Question

For the dice problem, the problem is now: every round we draw nice, if you get to stay, you get $4. If you is kicked out, you get nothing. If you choose to quit, you get $10. We will now use a mixed strategy: "I want to first take the risk and earn at least X dollars before I quit and take my $10". What's the optimal X?

Implement policy iteration to find out X, suppose the max money you can get is $100. Define the state space to include the money you got so far.

https://colab.research.google.com/drive/13VwGV6JRm5_mwuKb2mtX6XE45cKC8t14?usp=sharing
Links to an external site.

The optimal X is:

BlackTom AI · Accepted Answer

We start by restating the problem setup and recognizing what the decision-maker is optimizing. In this dice problem, each round you either stay and earn $4, are kicked out and earn $0, or quit and lock in $10. With a mixed policy that attempts to maximize the expected payoff, you plan to continue playing until your accumulated money reaches a threshold X, at which point you quit and take the $10 instead of risking further rounds. The question effectively asks for the optimal stopping threshold under a policy iteration framework, given a maximum possible total of $100 and a state space that includes the current amount of money earned so far.

First, consider the dynamic programming structure. The state is the current total amount you have accumulated. From any non-terminal state s < X, you have an action to continue playing, which leads to a probabilistic next state: you might stay and gain some additional amount, be kicked out and end with s, or eventually reach or exceed the threshold where quitting yields $10 more than what you would have earned had you kept going without hitting the threshold. In a simplified formulation, the decision at state s is whether the expected value of continuing exceeds the sure payoff if you quit at that moment. This is the core of the Bellman equation for optimal stopping problems: V(s) = max{ Q_continue(s), 10 }, where Q_continue(s) encodes the expected value of continuing one more round given the current s and the policy for subsequent rounds.

Next, think about how policy iteration would proceed. You start with an initial policy that specifies, for each state s, whether you continue or quit. Then you evaluate the value function V under that policy by solving a system of linear equations implied by the Bellman equations for the states under that policy. After evaluation, you improve the policy by comparing, for each non-terminal state s, the value of continuing versus quitting with the current V, and update the policy to the action that yields the higher value. This process repeats until the policy stabilizes and you have an optimal policy.

To reason about the threshold X conceptually, imagine the trade-off at state s where you are considering whether to continue or quit. If you continue, you risk ending up with a final payoff lower than what you would get by quitting now, depending on your current total and the probabilities of different outcomes in future rounds. If the expected gain from one more round, plus the expected value of future optimal play from the new state, is greater than quitting for $10, you should continue; otherwise, you should quit, locking in $10. The optimal threshold X is the smallest s such that quitting becomes at least as good as continuing under the optimal continuation values.

In the policy-iteration cycle, you would compute V(s) for all s with the current policy, then update the policy by choosing the action that yields the higher value at each s. This update typically pushes the threshold upward: as you become more confident that continuing can outperform quitting at higher totals, the policy prefers to keep playing longer and only quit later, increasing X until the policy no longer changes. The precise numerical determination of X depends on the exact transition probabilities and payoffs used in the model for the next-state values after a continue action, as well as the cap of $100 in the state space.

Intuitively, the threshold X is the decision point where the expected value of continuing from s is equal to or just below the sure payoff of quitting for $10. If you imagine stepping through states from low to high money amounts, early on continuing generally has a higher chance to improve your total, so the policy favors continuing; as s grows, the advantage of risking further rounds diminishes, and quitting becomes optimal.

If you are following a concrete implementation in the linked Colab, the final X would be found by iterating policy evaluation and improvement until the action at every s is stable, and the point at which the continue option ceases to be optimal matches the stopping threshold. The numerical value of X emerges from that convergence process rather than from a closed-form formula, since it depends on the exact transition dynamics and the truncation at $100.

Therefore, the optimal X is the threshold that satisfies the condition where the expected return from continuing, given the current value function, equals the alternative of taking the fixed $10 payoff, with policy iteration guiding the updates until no further changes occur in the policy across all states.

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