Derivatives Foundation puzzles, solved step by step

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  1. 002Every market day is either trending or choppy. A trending day is followed by another trending day 70% of the time, and a choppy day by another choppy day 60% of the time. In the long run, what fraction of days are trending? And when a trend starts, how many days does it last on average?Random walks and Markov chainsCoreQuant tradingHedge funds

    Try it first

    Gut call before the algebra: which is larger, the share of trending days or the share of choppy days?

    Show the worked solution

    4/7 of days trend, about 57.1%, and a trend lasts 3.3 days on average. In the long run the flow out of trending must equal the flow in: 0.3 times the trending share equals 0.4 times the choppy share, so the shares stand 4 to 3. A trend ends on any given day with probability 0.3, so its expected length is 1/0.3, and a choppy spell lasts 1/0.4 = 2.5 days.

    Why must the two flows balance?

    Picture two rooms at a party with a door between them. Each minute, 30% of the people in room A wander into B and 40% of those in B wander into A. The crowd settles when the two queues through the door carry the same number of people; otherwise one room keeps filling. A stationary split is one where the number of days leaving each state equals the number entering it, and that single equation fixes the split. Days leaving trending: 0.3 times the trending share. Days entering it from choppy: 0.4 times the choppy share. Set them equal and the ratio is 4 to 3.

    Two states, four arrows: in the long run the two crossing flows must balanceTrendingdayChoppydaystays trending 0.7stays choppy 0.60.3 of trending days flip0.4 of choppy days flipBalance: 0.3 x (trending share) = 0.4 x (choppy share)so trending : choppy = 4 : 3Long-run share of daysTrending: 57.1% of days4/7average streak 1/0.3 = 3.3 daysChoppy: 42.9% of days3/7average streak 1/0.4 = 2.5 daysCheck: 4/7 x 3.3 against 3/7 x 2.5share = streak x how often a streak starts
    Trending days keep 0.7 of their successors and lose 0.3 to choppy, while choppy days keep 0.6 and lose 0.4 back, so the long-run shares stand 4 to 3, 57.1% trending and 42.9% choppy, with trends lasting 3.3 days and choppy spells 2.5 days on average.
    The relationship
    0.3 πT=0.4 πC,πT+πC=1  ⇒  πT=0.40.3+0.4=47E[streak]=10.3=3.330.3\,\pi_T = 0.4\,\pi_C,\quad \pi_T + \pi_C = 1 \;\Rightarrow\; \pi_T = \frac{0.4}{0.3 + 0.4} = \frac{4}{7} \qquad E[\text{streak}] = \frac{1}{0.3} = 3.33
    pi_T, pi_Cthe long-run shares of trending and choppy days
    0.3, 0.4the chance a trending day flips to choppy, and a choppy day flips to trending
    1/0.3expected length of a run that ends with probability 0.3 each day
    What it says in wordsEach state's share is the other state's flip rate divided by the sum of the two flip rates, and a run's length is one over its own flip rate.

    Why is the average streak 1/0.3 and not something longer?

    A trend that has lasted five days is no more likely to end tomorrow than one that started today: the chain has no memory beyond yesterday. Each trending day ends the run with probability 0.3, independent of its age, so the run length is a geometric count with mean 1/0.3 = 3.33 days. That is the same reason the expected number of rolls to a six is 6. The two answers also check each other: the share of a state equals how often a run of it starts, times how long it lasts, and 4/7 against 3/7 is exactly 3.33 against 2.5 scaled by the same start rate.

    What does the chain say about tomorrow, given today?

    This is where the puzzle connects to trading. Today's state carries real information: after a trending day, tomorrow trends with probability 0.7, well above the unconditional 57%. After a choppy day it is only 0.4. The long-run split tells you nothing about tomorrow; the transition row for today's state does. Say that distinction out loud, because an interviewer who hears 57% quoted as a one-day forecast knows you have confused the stationary distribution with a conditional one. The limitation to add: a two-state chain with fixed probabilities is a toy, and real regime persistence drifts over time.

    Where candidates lose it

    The first wrong answer is 50%, on the grounds that each state has one way in and one way out. The flows are not equal in rate: trending leaks at 0.3, choppy at 0.4, and the slower leak wins more of the time.

    The second loss is writing out eigenvectors of a two-by-two matrix under time pressure. The balance equation, flow out equals flow in, takes one line and is what the interviewer wants to hear.

    What the interviewer asks next

    • Starting from a choppy day, what is the chance that the day after tomorrow is trending?
    • Add a third state, a crash day, that follows a choppy day 5% of the time. How does the method change?
    • How would you estimate these transition probabilities from a year of daily data, and how noisy would they be?
  2. 055A token sits on one corner of a square. Every second it moves to one of the two neighbouring corners, chosen by a fair coin. What is the expected number of seconds until it first reaches the opposite corner?Random walks and Markov chainsCoreQuant trading

    Try it first

    The opposite corner is two steps away. Expected time to get there?

    Show the worked solution

    4 seconds. By symmetry, the two corners next to the start are the same state, call it adjacent. From the start you always move to adjacent in one step. From adjacent, half the time you reach the target and half the time you return to the start. So E(start) = 1 + E(adjacent) and E(adjacent) = 1 + E(start)/2, giving E(adjacent) = 3 and E(start) = 4.

    Why collapse four corners into three states?

    If you are lost in a town with a river on one side, what matters is how far you are from the river, not which street you are on. The square is the same: standing at either corner next to the start, the token's future looks identical, one coin flip from the target and one from the start. Grouping corners by their distance from the target turns a four-state chain into a three-state line, and a line is solved with one equation per unknown. You have two unknowns, the expected time from the start and from an adjacent corner, because the target itself takes zero time.

    Collapse four corners into three distances, then solve two equationsstarttarget1 step away1 step awayeach move: a coin flipbetween the two neighboursStartdistance 0Adjacentdistance 1Targetdistance 211/21/2 back to startE(start) = 1 + E(adjacent)E(adjacent) = 1 + 1/2 x 0 + 1/2 x E(start)E(adjacent) = 1 + 1/2 (1 + E(adjacent)) so E(adjacent) = 3E(start) = 1 + 3 = 4 steps on average
    Grouping the two corners next to the start into one state gives a three-state chain in which the start always moves to adjacent, and adjacent finishes or returns to the start with equal chance, so E(adjacent) = 3 and E(start) = 4 seconds.

    How do you set up and solve the equations in the room?

    Each equation says the same sentence: one step, plus the average of what is left from where you land. From the start, every step lands on an adjacent corner, so E(start) = 1 + E(adjacent); from an adjacent corner, half the steps finish and half return, so E(adjacent) = 1 + (1/2) x 0 + (1/2) x E(start). Substitute the first into the second: E(adjacent) = 1 + (1/2)(1 + E(adjacent)), so E(adjacent)/2 = 3/2, E(adjacent) = 3, and E(start) = 4. Saying the sentence before the symbols is what keeps the equations honest.

    The relationship
    E0=1+E1,E1=1+12E0  ⇒  E1=3,  E0=4E_0 = 1 + E_1, \qquad E_1 = 1 + \tfrac{1}{2}E_0 \;\Rightarrow\; E_1 = 3,\; E_0 = 4
    E_0the expected steps to the target from the starting corner
    E_1the expected steps from either corner next to the start
    1/2the chance a step from an adjacent corner lands on the target
    What it says in wordsFrom the start you always move one step closer; from there a coin flip either finishes or sends you back, and the two equations give four steps on average.

    What is the check, and what does the general case look like?

    Check it with the geometric picture. After the first step you are adjacent, and from there each attempt either finishes in one step or costs two steps, back to the start and out again, before you are adjacent once more. Each attempt succeeds with chance 1/2, so E(adjacent) = (1/2) x 1 + (1/2) x (2 + E(adjacent)), which again gives 3, and the first step makes it 4. On a cube the same method with four distance classes gives 10 steps to the opposite vertex; on a general graph the method is the same, but the number of distance classes grows and the arithmetic stops being mental.

    Where candidates lose it

    The fast wrong answer is 2, the length of the shortest path. The interviewer is checking whether you see that the token can bounce back, and whether you reach for the first-step equations rather than trying to sum a series.

    The second loss is writing four equations, one per corner. Say the symmetry out loud, collapse to three states, and the whole thing is two lines.

    What the interviewer asks next

    • What is the expected time to return to the starting corner for the first time?
    • Same walk on the eight corners of a cube. Expected time to the opposite vertex?
    • What is the probability the token reaches the opposite corner within 4 steps?
    • The coin is biased: it moves clockwise with chance 0.7. Does the expected time change?
  3. 074What is the expected number of fair coin flips needed to see two heads in a row? And how many to see a head followed by a tail?Random walks and Markov chainsCoreSquarepoint CapitalLondon · 2025

    Try it first

    HH and HT each have probability 1/4 on any two flips. Do they take the same expected time to appear?

    Show the worked solution

    6 flips for HH and 4 flips for HT. Track one state: whether the last flip was a head. For HH, a tail after a head sends you back to the start, so E(start) = 1 + E(one H)/2 + E(start)/2 and E(one H) = 1 + E(start)/2, giving 6. For HT, a head after a head leaves you still holding a head, so E(one H) = 1 + E(one H)/2 = 2 and E(start) = 2 + 2 = 4.

    Why do two patterns with the same probability take different times?

    Two queues at a counter: in one, a mistake sends you to the back; in the other, a mistake keeps your place. Both queues move at the same speed, but one is a much longer wait. Waiting for a pattern is a race with restarts, and what matters is how much progress a failure destroys: a tail after a head destroys everything for HH, while a head after a head destroys nothing for HT. That asymmetry, not the probability of the pattern, sets the expected time. It is also why HT and TH take 4 while HH and TT take 6.

    A failed HH sends you back to the start; a failed HT does notWaiting for H then HStartnothing useful yetOne Hfirst head seenDoneHTT: stayT: back to the startE(start) = 1 + 1/2 E(one H) + 1/2 E(start)E(one H) = 1 + 1/2 x 0 + 1/2 E(start)E(one H) = 4E(start) = 6 flipsWaiting for H then TStartnothing useful yetOne Hfirst head seenDoneHHT: stayH: stay, still one headE(start) = 1 + 1/2 E(one H) + 1/2 E(start)E(one H) = 1 + 1/2 x 0 + 1/2 E(one H)E(one H) = 2E(start) = 4 flips
    Waiting for HH takes 6 flips because a tail after a head sends you back to the start, while waiting for HT takes 4 because a second head after a head still counts as a first head, so no progress is lost.

    How do you set up and solve the equations in the room?

    Two states, two equations, each saying one flip plus the average of what remains. For HH: from the start, a head takes you to one H and a tail keeps you at the start, so E0 = 1 + E1/2 + E0/2; from one H, a head finishes and a tail returns you to the start, so E1 = 1 + E0/2; substitute to get E1 = 4 and E0 = 6. For HT the second equation changes to E1 = 1 + E1/2, since a head keeps you at one H, giving E1 = 2, and the first equation gives E0 = 2 + E1 = 4. The first equation is the same in both problems; only the failure branch differs.

    The relationship
    HH: E0=1+12E1+12E0, E1=1+12E0⇒E0=6HT: E1=1+12E1⇒E1=2, E0=4\text{HH: } E_0 = 1 + \tfrac{1}{2}E_1 + \tfrac{1}{2}E_0,\ E_1 = 1 + \tfrac{1}{2}E_0 \Rightarrow E_0 = 6 \qquad \text{HT: } E_1 = 1 + \tfrac{1}{2}E_1 \Rightarrow E_1 = 2,\ E_0 = 4
    E_0expected flips remaining from the start, no useful progress
    E_1expected flips remaining once the last flip was a head
    1/2the chance of a head or a tail on each flip
    What it says in wordsThe pattern whose failures throw away progress takes six flips on average, and the pattern whose failures keep progress takes four.

    What is the quick check, and where does this matter beyond coins?

    There is a general rule: the expected time to a pattern is the sum of 2 to the k over every length k at which the pattern overlaps itself. HH overlaps itself at lengths 1 and 2, so 2 + 4 = 6; HT overlaps only at its full length 2, so 4. The same arithmetic prices a bet on which of two patterns appears first, and it shows up on a desk whenever a signal needs a run of confirmations: a rule that resets on any contradicting tick waits far longer than one that keeps partial progress. The limitation is the fair coin; with a biased coin the overlap rule still holds but the powers of two become products of the relevant probabilities.

    Where candidates lose it

    The common answer is that HH and HT take the same time because each has probability 1/4. The interviewer is testing whether you see that waiting time depends on what a failure costs, not on the pattern's probability.

    The second loss is setting up the HT chain with a return to the start after a second head. A head after a head is still one head; the state does not change.

    What the interviewer asks next

    • What is the expected number of flips to see three heads in a row?
    • Which appears first on average, HHT or HTH, and why are their waiting times different?
    • The coin lands heads with probability 0.6. Expected time to HH?
    • You flip until you see HT and I flip until I see HH. What is the chance you finish first?

    Asked at Squarepoint Capital, Quantitative Research, London, 2025 (Wall Street Oasis): a few siimple questions on statistical problems e.g. # of throws expected to get 2 heads in a row