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This is the classic secretary-ish stopping problem and I'd seen something like it before, so I wasn't totally lost.
Solve the problem using backward induction: first determine the optimal threshold for the final roll, then use that to set the threshold for the second roll, and finally for the first roll. Compute the expected payout under this optimal strategy and clearly state the stopping rule.
Pro tip: After presenting the solution, mention that this is a classic optimal stopping problem and that similar logic applies to real-world decisions like when to accept a job offer or sell a stock. This shows you can connect technical concepts to business contexts.
Clarify that you want to maximize expected payout by choosing when to stop. The decision is based on the current roll and the number of rolls remaining.
On the third roll, you must accept whatever you get. The expected value is the average of a fair six-sided die: (1+2+3+4+5+6)/6 = 3.5.
On the second roll, you can either stop or take the third roll. The expected value of continuing is 3.5, so you should stop if your current roll is greater than 3.5, i.e., 4, 5, or 6. If you roll 1, 2, or 3, you should continue. Compute the expected value of this strategy.
On the first roll, you can stop or continue to the second roll. The expected value of continuing is the value computed in step 3. Stop if your current roll exceeds that expected value; otherwise continue.
Using the optimal thresholds, calculate the expected payout from the first roll. This is the maximum expected value you can achieve.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This tripped me up more than it should have.
Use backward induction to compute the optimal stopping thresholds and expected payoff for a three-roll die game with a $1 continuation cost. Start from the final roll, then work backwards to determine the minimum roll value that justifies paying $1 to continue at each stage.
Pro tip: Clearly state that the thresholds are the minimum die values for which continuing is optimal, and emphasize that the expected payoff is the net value after subtracting continuation costs. This shows you understand the cost structure and can communicate results precisely.
Clarify that there are up to three rolls, and after each of the first two rolls, you may pay $1 to continue to the next roll. The goal is to maximize expected net payoff.
On the third roll, you must accept the outcome, so the expected payoff is the average of a fair six-sided die: (1+2+3+4+5+6)/6 = 3.5.
On the second roll, you can either stop with the current value x or pay $1 to continue to the third roll, which has expected value 3.5. Continuing is optimal if x < 3.5 - 1 = 2.5, so you continue if x ≤ 2 and stop if x ≥ 3. The expected value of having a second roll is then (2/6)*3.5 + (4/6)*average(3,4,5,6) = 3.5? Wait, compute correctly: average of stopping values for x=3,4,5,6 is (3+4+5+6)/4=4.5. So EV = (2/6)*3.5 + (4/6)*4.5 = 1.1667 + 3 = 4.1667. But subtract? No, the $1 is paid only if you continue, so the EV already accounts for that: if you roll 1 or 2, you pay $1 and then get 3.5, net 2.5; if you roll 3-6, you stop and get x. So EV = (2/6)*2.5 + (4/6)*4.5 = 0.8333 + 3 = 3.8333. Actually, careful: The expected value of continuing is 3.5 - 1 = 2.5. So you continue if x < 2.5, i.e., x=1,2. Then EV = (2/6)*2.5 + (4/6)*4.5 = 0.8333 + 3 = 3.8333.
On the first roll, you can stop with x or pay $1 to continue to the second roll, which has expected value 3.8333. Continuing is optimal if x < 3.8333 - 1 = 2.8333, so you continue if x ≤ 2 and stop if x ≥ 3. The expected payoff of the game is then (2/6)*(3.8333 - 1) + (4/6)*average(3,4,5,6) = (2/6)*2.8333 + (4/6)*4.5 = 0.9444 + 3 = 3.9444.
State the optimal stopping thresholds: on the first roll, continue if you roll 1 or 2, stop if 3-6; on the second roll, continue if you roll 1 or 2, stop if 3-6. The expected net payoff is approximately $3.94.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Subtle difference from the previous version and I almost missed it.
Solve the problem using backward induction: start from the final roll (roll 3) and determine the optimal decision rule based on the expected value of continuing versus stopping. Then move to roll 2, incorporating the cost of continuing and the optimal policy from roll 3, and finally to roll 1. Compute the expected payoff under the optimal policy.
Pro tip: Clearly state that the optimal policy is a threshold rule: continue only if the current roll is below a certain cutoff. This demonstrates understanding of dynamic programming and makes the solution easy to follow.
Clarify that there are three rolls of a fair die, you may stop after any roll and receive the face value, but you must pay $1 to take each subsequent roll (roll 2 and roll 3). The goal is to maximize expected net payoff.
On roll 3, you must accept the outcome (no further rolls). The expected payoff if you reach roll 3 is the expected value of a fair die, which is 3.5.
At roll 2, you can stop and take the current value, or pay $1 to continue to roll 3. The expected value of continuing is 3.5 - 1 = 2.5. So you should continue if the current roll is less than 2.5, i.e., if you roll 1 or 2. Otherwise stop. Compute the expected value at roll 2 under this policy.
At roll 1, you can stop and take the current value, or pay $1 to continue to roll 2. The expected value of continuing is the expected value at roll 2 (computed in step 3) minus $1. Compare this to the current roll and decide to continue if the roll is below the threshold. Compute the overall expected payoff.
Calculate the expected value at roll 1 under the optimal policy, which is the answer. Clearly state the optimal policy thresholds and the expected payoff.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Bonus question and I was already a bit fried.
Model the problem as an optimal stopping problem where the value function V satisfies a Bellman equation. Derive the threshold by comparing the expected value of continuing versus stopping, and solve for the threshold as a function of n and c. Then express the threshold in closed form or as an inequality.
Pro tip: Always verify your threshold with small n (e.g., n=2,3) and check edge cases like c=0 (no cost) and c large (always stop). This demonstrates rigor and catches errors.
Let V be the maximum expected net payoff from a roll. Set up the Bellman equation: V = max(0, (1/n) * sum_{i=1}^n max(i, V) - c).
Assume a threshold t such that you stop if the roll is >= t and continue if it is < t. Then V = (1/n) * [sum_{i=t}^n i + (t-1) * V] - c.
Rearrange the equation to solve for V in terms of t, then use the indifference condition at the threshold: V = t - 1 (or t, depending on convention). Solve for t as a function of n and c.
Express t as the smallest integer satisfying a certain inequality, or solve the quadratic equation for t. Present the result clearly.
Check that t is between 1 and n, and that as c increases, t decreases (stop earlier). Discuss limiting cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.