Oct 08, 2026 by Offer

Citadel Dice Game 2027: Expected Value Solution (4.25)

The Citadel dice expected value question — commonly reported by candidates as "How much would you pay to play a game where you are paid the pips on a fair die, but may re-roll once?" — solves to 4.25. The optimal strategy: keep any first roll of 4, 5, or 6, and re-roll on 1, 2, or 3.

What This Citadel Dice Expected Value Question Assesses

This is an optimal-stopping problem disguised as a dice game. Citadel uses it to test whether you compare the value of acting now against the expected value of waiting — the core logic of every trading decision. Candidates who answer 3.5 (the plain die expectation) reveal they ignored the option value of the re-roll.

How to Answer This Citadel Dice Expected Value Question

Work it in three steps, narrating each:

Step 1 — Value the re-roll. A fresh roll of a fair die has expected value (1+2+3+4+5+6)/6 = 3.5. This is your fallback: any first roll below 3.5 should be discarded.
Step 2 — Set the threshold. Keep the first roll if it beats the re-roll's expectation; re-roll otherwise. Since die faces are integers, keep 4, 5, 6 and re-roll 1, 2, 3. State this rule explicitly — it's the optimal-stopping decision.
Step 3 — Compute the total expectation. Half the time (rolls 4–6) you keep (4+5+6)/6 = 2.5 of contribution; half the time you re-roll for 3.5, contributing 1.75. Total: 2.5 + 1.75 = 4.25.

Example line: "I'd re-roll anything below 4 since the re-roll is worth 3.5, giving an expected value of 4.25 — so I'd pay up to $4.25 to play."

Common Mistakes

Answering 3.5. Forgetting the re-roll option entirely is the most common error — the question is testing whether you value the option.
Wrong threshold. Re-rolling a 4 (above the 3.5 fallback) or keeping a 3 (below it) both destroy value; the cutoff logic must be exact.
No bidding logic. The question asks what you'd pay — always connect the expectation back to the price, since that's the trading decision.

Candidates commonly report follow-ups that extend this game: two re-rolls, a cost per re-roll, or a biased die. The method never changes — compare the current value against the expected value of continuing — so master the one-roll version and the extensions solve themselves.

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FAQ

What if there were two re-rolls allowed? Work backwards: the last re-roll is worth 3.5, so with one re-roll left you'd keep 4+ (value 4.25), making the two-roll game worth (4.25×3 + 5 + 6... ) — compute recursively. The threshold rises with more options.

What if each re-roll costs $1? Subtract the cost from the re-roll's value: re-roll only if the expected gain exceeds $1, i.e., keep anything above 2.5 — so keep 3 through 6.

Does the answer change with an unfair die? Only the numbers change. The method — threshold equals the expected value of the re-roll — is universal.

Why do trading firms ask this? Optimal stopping under uncertainty is the skeleton of market making and options pricing. They're testing the instinct, not the arithmetic.

Preparing for Citadel's interview? Our 2027 Citadel Online Assessment and Coding Challenge Tutorials has practice questions and answers — $79 one-time, instant download.