← Jane Street Interview Insights
My first instinct was 'accept anything >= 0' and I was pretty confident about it for an embarrassingly long time.
Model the game as a Markov decision process with a value function representing the expected payoff to the current roller. Use dynamic programming to find the optimal stopping threshold: the roller accepts if the roll exceeds the continuation value, otherwise passes. Then solve for the fixed point where the value function equals the expected payoff under this threshold strategy.
Pro tip: Frame the problem as an optimal stopping problem and emphasize that the threshold is determined by the value of passing, which itself depends on the threshold. This shows you understand the recursive nature and can solve it analytically or numerically.
Let V be the expected payoff to the current roller when both play optimally. By symmetry, the game is zero-sum and the value is the same for both players when it's their turn.
The roller will accept a roll x if x >= V (since passing gives expected payoff V). Thus the optimal strategy is a threshold strategy: accept if x >= t, where t is the threshold.
Given threshold t, the expected payoff is V = (1/21) * sum_{x >= t} x + (1/21) * sum_{x < t} V. Solve for V as a function of t.
The optimal threshold t* satisfies t* = V(t*). Solve for t* and compute the corresponding V. Since the die is discrete, check integer thresholds around the continuous solution.
Confirm that the threshold strategy is indeed optimal (e.g., by checking that accepting at t* is better than passing). State the expected payoff to the first player.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.