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Four Fair Coins: Expected Payoff and Pricing Two Magic Wands

Last updated: Jul 2, 2026

Four Fair Coins: Expected Payoff and Pricing Two Magic Wands

Company: Jane Street

Role: Data Scientist

Category: Machine Learning

Difficulty: easy

Interview Round: Technical Screen

You play the following game: four fair coins are tossed at the same time. For every coin that lands heads, you are paid \$1. The interviewer walks you through the game in three stages, adding a "magic wand" that lets you modify the coins after you see the initial toss. ### Constraints & Assumptions - All four coins are fair and independent: each lands heads with probability $1/2$. - The payoff is exactly \$1 per head showing at the end; there are no other costs or payoffs. - In Parts 2 and 3 the wand is used **after** you observe the initial toss, you may use it an unlimited number of times, and each use is free. - "A pair" means any two distinct coins of your choice, and you may choose a different pair on every use (fully adaptive play). - You are risk-neutral, so the fair price of a wand is the increase in your expected payoff from owning it. ### Clarifying Questions to Ask - Are the coins fair, and are the four tosses independent? - Do I see the coins before deciding how to use the wand, and can I choose each pair adaptively based on the current state? - For the "turn over" wand, do both chosen coins deterministically flip to their opposite faces (heads becomes tails and vice versa)? - Is every use of the wand free, with no limit on the number of uses, and do I decide when to stop and collect the payoff? - Should I price the wand for a risk-neutral player, i.e., fair price equals the gain in expected payoff? ### Part 1 What is your expected payoff from the game (no wand)? ```hint One line, no enumeration Use linearity of expectation: the expected payoff is the sum of each coin's expected contribution. You never need to list all $2^4$ outcomes. ``` #### What This Part Should Cover - Applying linearity of expectation across the four coins instead of enumerating outcomes. - Stating the per-coin expected contribution and combining it into a clean numeric answer. ### Part 2 You are offered a magic wand. After seeing the toss, the wand lets you **turn over any pair of coins** (both chosen coins flip to their opposite faces). You may use it as many times as you like. What is a fair price for this wand? ```hint Look for an invariant Turning over two coins changes the number of heads by $-2$, $0$, or $+2$, depending on what the two coins were showing. What property of the head count can no sequence of pair-flips ever change? ``` ```hint Best reachable payoff Split the $16$ equally likely initial outcomes into classes the wand cannot move between. For each class, find the largest head count you can reach, and weight by the probability of the class. ``` #### What This Part Should Cover - Identifying what a pair-flip can and cannot change about the configuration of heads. - Determining the best reachable payoff from each class of starting outcomes, with a constructive argument that it is reachable and an argument that nothing better is. - Computing the expected payoff with the wand and pricing it as the difference versus Part 1. ### Part 3 The wand's power changes: now each use lets you **pick any pair of coins and re-toss both of them**. Again you may use it an unlimited number of times, free of charge, and you decide when to stop. What is a fair price for this wand? ```hint What changed versus Part 2 Re-tossing is random, so the constraint you found in Part 2 no longer binds. Notice that re-tossing two coins that both show tails can never make you worse off. Which pair should you re-toss whenever at least two tails are showing? ``` ```hint The only risky decision The interesting state is exactly three heads: any re-toss must now include a head, so you can lose ground. Either set up the optimal-stopping value equations for the states "two heads" and "three heads," or look for a strategy that only ever stops in one particular state. ``` #### What This Part Should Cover - Recognizing that re-tossing removes the structural restriction from Part 2 and injects fresh randomness with free retries. - A correct treatment of the stop-or-continue decision at three heads (value equations or a dominance/absorption argument). - Arguing termination: the proposed strategy reaches its target state with probability 1, so the claimed expected payoff is actually achieved. - Pricing the wand as the gain in expected payoff, and comparing it with the Part 2 wand. ### What a Strong Answer Covers Across all three parts, the interviewer is checking that the candidate: - Prices every option the same way: value of the tool = expected payoff with it minus expected payoff without it, under risk neutrality. - Reasons structurally about the allowed operation (invariants for the deterministic wand, monotone never-lose moves for the random one) instead of brute-forcing cases. - Is careful about what "unlimited free uses" buys, including why the two wands are not worth the same amount. - Communicates a clean final number for each part and sanity-checks it against simple bounds (the payoff can never exceed \$4). ### Follow-up Questions - Generalize Part 2 and Part 3 to $n$ fair coins. How do the two wand prices behave as $n$ grows? - Suppose each use of the Part 3 wand costs a fee $c > 0$. How does your strategy change, and for what values of $c$ is the wand worth buying at all? - What if the wand in Part 3 could only be used at most $k$ times? Set up (don't fully solve) the dynamic program for its value. - Redo Part 1 and Part 2 with biased coins that land heads with probability $p$. Which parts of your argument survive unchanged?

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|Home/Machine Learning/Jane Street

Four Fair Coins: Expected Payoff and Pricing Two Magic Wands

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Jane Street
Sep 11, 2025, 12:00 AM
easyData ScientistTechnical ScreenMachine Learning
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0

You play the following game: four fair coins are tossed at the same time. For every coin that lands heads, you are paid $1.

The interviewer walks you through the game in three stages, adding a "magic wand" that lets you modify the coins after you see the initial toss.

Constraints & Assumptions

  • All four coins are fair and independent: each lands heads with probability 1/21/21/2 .
  • The payoff is exactly $1 per head showing at the end; there are no other costs or payoffs.
  • In Parts 2 and 3 the wand is used after you observe the initial toss, you may use it an unlimited number of times, and each use is free.
  • "A pair" means any two distinct coins of your choice, and you may choose a different pair on every use (fully adaptive play).
  • You are risk-neutral, so the fair price of a wand is the increase in your expected payoff from owning it.

Clarifying Questions to Ask

  • Are the coins fair, and are the four tosses independent?
  • Do I see the coins before deciding how to use the wand, and can I choose each pair adaptively based on the current state?
  • For the "turn over" wand, do both chosen coins deterministically flip to their opposite faces (heads becomes tails and vice versa)?
  • Is every use of the wand free, with no limit on the number of uses, and do I decide when to stop and collect the payoff?
  • Should I price the wand for a risk-neutral player, i.e., fair price equals the gain in expected payoff?

Part 1

What is your expected payoff from the game (no wand)?

What This Part Should Cover

  • Applying linearity of expectation across the four coins instead of enumerating outcomes.
  • Stating the per-coin expected contribution and combining it into a clean numeric answer.

Part 2

You are offered a magic wand. After seeing the toss, the wand lets you turn over any pair of coins (both chosen coins flip to their opposite faces). You may use it as many times as you like. What is a fair price for this wand?

What This Part Should Cover

  • Identifying what a pair-flip can and cannot change about the configuration of heads.
  • Determining the best reachable payoff from each class of starting outcomes, with a constructive argument that it is reachable and an argument that nothing better is.
  • Computing the expected payoff with the wand and pricing it as the difference versus Part 1.

Part 3

The wand's power changes: now each use lets you pick any pair of coins and re-toss both of them. Again you may use it an unlimited number of times, free of charge, and you decide when to stop. What is a fair price for this wand?

What This Part Should Cover

  • Recognizing that re-tossing removes the structural restriction from Part 2 and injects fresh randomness with free retries.
  • A correct treatment of the stop-or-continue decision at three heads (value equations or a dominance/absorption argument).
  • Arguing termination: the proposed strategy reaches its target state with probability 1, so the claimed expected payoff is actually achieved.
  • Pricing the wand as the gain in expected payoff, and comparing it with the Part 2 wand.

What a Strong Answer Covers

Across all three parts, the interviewer is checking that the candidate:

  • Prices every option the same way: value of the tool = expected payoff with it minus expected payoff without it, under risk neutrality.
  • Reasons structurally about the allowed operation (invariants for the deterministic wand, monotone never-lose moves for the random one) instead of brute-forcing cases.
  • Is careful about what "unlimited free uses" buys, including why the two wands are not worth the same amount.
  • Communicates a clean final number for each part and sanity-checks it against simple bounds (the payoff can never exceed $4).

Follow-up Questions

  • Generalize Part 2 and Part 3 to nnn fair coins. How do the two wand prices behave as nnn grows?
  • Suppose each use of the Part 3 wand costs a fee c>0c > 0c>0 . How does your strategy change, and for what values of ccc is the wand worth buying at all?
  • What if the wand in Part 3 could only be used at most kkk times? Set up (don't fully solve) the dynamic program for its value.
  • Redo Part 1 and Part 2 with biased coins that land heads with probability ppp . Which parts of your argument survive unchanged?
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