Hedge Funds puzzles, solved step by step

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  1. 043You may roll a fair die up to three times. After each roll you either stop and are paid the face in rupees, or throw that roll away and roll again; if you reach the third roll you must take it. What is the best stopping rule, and what is the game worth?Expected value and dice gamesHardSquarepoint CapitalLondon · 2026

    Try it first

    On the first roll you get a 4. What do you do?

    Show the worked solution

    Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.

    Why start from the last roll?

    Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.

    Solve from the last roll backwards: each value sets the next thresholdFirst roll3 rolls left123456keep 5 or 6reroll the restWorth4.67= 14/3Second roll2 rolls left123456keep 4, 5 or 6reroll the restWorth4.25= 17/4Last roll1 roll left123456must keep itWorth3.50= 7/2Keep a roll only if it beats what the remaining rolls are worth: 3.5, then 4.25Arrows run right to left: each stage uses the value of the stage after it
    With one roll left the game is worth 3.5, so with two left you keep 4 or more and the game is worth 4.25; with three left you keep only 5 or 6, and the whole game is worth 14/3, about 4.67.

    How do the thresholds come out?

    With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.

    The relationship
    Vn=16∑f=16max⁡(f, Vn−1),V1=3.5,  V2=4.25,  V3=143V_n = \frac{1}{6}\sum_{f=1}^{6} \max\left(f,\, V_{n-1}\right), \quad V_1 = 3.5,\; V_2 = 4.25,\; V_3 = \tfrac{14}{3}
    V_nthe value of the game with n rolls left
    fthe face you just rolled
    max(f, V_{n-1})keep the roll or throw it away, whichever is worth more
    What it says in wordsEach stage is worth the average, over the six faces, of the better of keeping the face or playing on.

    Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.

    Where candidates lose it

    The common loss is using 3.5 as the threshold at every stage, which keeps a 4 on the first roll. The fallback on the first roll is the two-roll game, 4.25, not a single roll.

    The second loss is solving forwards and getting lost. Say you will start from the last roll, compute 3.5, 4.25 and 14/3 in that order, and the thresholds fall out.

    What the interviewer asks next

    • What is the game worth with four rolls, and what is the first-roll threshold?
    • You must pay Rs 1 for every reroll. How do the thresholds change?
    • You are paid the square of the face instead. What is the optimal rule?

    Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Dynamic programming questions with focus on probability at the end.

  2. 068You and an opponent each secretly show heads or tails. If both show heads you win Rs 3, if both show tails you win Rs 1, and if they differ you pay Rs 2. The payoffs look balanced. What mix should each player use, and what is the game worth to you?Expected value and dice gamesHardCitadelsydney · 2025

    Try it first

    If both of you play well, what is the game worth to you per round?

    Show the worked solution

    Both players should show heads 3/8 of the time, and the game is worth minus Rs 0.125 a round to you. Choose your mix so the opponent gains nothing by switching: 3p - 2(1 - p) = -2p + (1 - p) gives p = 3/8. The opponent's mix solves the same balance from your side, also 3/8. At those mixes you lose an eighth of a rupee a round, although the payoffs look even.

    Why is a fair coin the wrong strategy?

    Think of a penalty taker and a goalkeeper. If the taker always shoots left, the keeper dives left; the taker's only defence is to mix so the keeper cannot profit from guessing. A mix is right only if it leaves the other side indifferent, and a fair coin here does not: against it the opponent's tails pays you 0.5 x (-2) + 0.5 x 1 = -0.50 a round. So the opponent always shows tails, and you lose Rs 0.50 a round rather than breaking even.

    How do you find the equilibrium mix?

    Set up your payoff for each of your choices as the opponent's chance of heads, q, varies. Showing heads pays 3q - 2(1 - q) = 5q - 2; showing tails pays -2q + (1 - q) = 1 - 3q; they are equal at q = 3/8, where both pay -1/8. The opponent plays 3/8 heads so that nothing you do beats -1/8. By the same balance from the opponent's side, you play 3/8 heads so that nothing the opponent does pushes you below -1/8. That pair is the Nash equilibriumA pair of strategies in which neither player can do better by changing only their own choice..

    The relationship
    5q−2=1−3q  ⇒  q=38,V=5⋅38−2=−185q - 2 = 1 - 3q \;\Rightarrow\; q = \tfrac{3}{8}, \qquad V = 5 \cdot \tfrac38 - 2 = -\tfrac18
    qthe opponent's chance of showing heads
    5q - 2your expected payoff if you show heads
    1 - 3qyour expected payoff if you show tails
    Vthe value of the game to you
    What it says in wordsThe opponent's mix makes your two choices pay the same, and that common payoff is what the game is worth.
    The opponent picks the mix that makes your two choices pay the same-2-1+1+2+3000.250.50.751Opponent's chance of showing heads, qyou show heads: 5q - 2you show tails: 1 - 3qq = 3/8: you get-1/8 either wayYour payoff, Rsopp. Hopp. Tyou H+3-2you T-2+1Both show heads 3/8Value to you: -Rs 0.125Fair coin vs best reply:you get -0.50 a round
    Your payoff from showing heads rises with the opponent's chance of heads and your payoff from tails falls; the lines cross at 3/8, where either choice pays minus Rs 0.125, so the opponent plays 3/8 heads and the game is worth minus Rs 0.125 a round to you.

    Why is the game negative when the payoffs look even?

    Because the balance is in the totals, not in the play. Your two winning cells need coordination the opponent will not give you, while their winning cells, the mismatches, pay the same 2 either way. The opponent can lean towards tails, starving your big Rs 3 cell, and the only price is feeding your small Rs 1 cell. Say what a desk would do with it: ask to be paid about 13 paise a round to play, or ask to swap sides.

    Where candidates lose it

    The trap is adding up the payoffs, 3 and 1 against 2 and 2, calling the game fair and playing a fair coin. That ignores that the opponent chooses too, and against a fair coin their best reply costs you Rs 0.50 a round.

    The second slip is solving for your own mix by making yourself indifferent. Each player's mix is chosen to make the other player indifferent; set up the equation from the opponent's payoffs.

    What the interviewer asks next

    • Change the tails-tails payoff to Rs 2. What are the mixes and the value now?
    • What would you pay per round to play this game from the opponent's side?
    • The opponent is known to show heads half the time. What do you do, and what do you earn?

    Asked at Citadel, Quantitative Trading, sydney, 2025 (Wall Street Oasis): many probability questions for OA. mix of prob and game theory for technical