Derivatives Foundation puzzles, solved step by step

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.

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.

86/99, about 86.9%. Per game I win with probability 1/3, you win with 1/6 and we draw with 1/2. The match has five live states: no streak, my streak of one or two, your streak of one or two. Writing my chance of winning the match from each state as an unknown, each state's equation is a weighted average of its neighbours, and solving the five equations gives 86/99 from the start. The naive ratio of (1/3)^3 to (1/6)^3 gives 89% and is wrong.

Why does the match need states rather than a single formula?

A tennis game at deuce is the everyday version: whoever is a point ahead is in a different position from level, and the chance of winning the game from deuce is best found by naming the positions and linking them. What matters here is not the game count but the current streak, and only five positions are possible before the match ends: no streak, me on one, me on two, you on one, you on two. Every game moves the match from one of those positions to another, with the same three probabilities each time, so the match is a Markov chain and the answer is a small linear system rather than a series. Three in a row sounds like it needs a long sum over all the ways the match can go; the states collapse that sum into five unknowns.

How do you write and solve the equations?

Call my winning chance x from no streak, a1 and a2 from my streaks, b1 and b2 from yours. From any state a draw, probability 1/2, takes you to x. From no streak a win takes you to a1 and a loss to b1, so x = x/2 + a1/3 + b1/6. From a1 a win takes you to a2 and a loss to b1. From a2 a win ends the match in my favour, worth 1. From b1 a loss takes you to b2 and a win takes you to a1; from b2 a loss ends it, worth 0. Five equations in five unknowns, and the structure is friendly: substitute the draw term first, since x/2 appears everywhere, and the system reduces by hand in a few lines. The solution is x = 86/99, a1 = 29/33, a2 = 10/11, b1 = 28/33, b2 = 8/11. A simulation of 200,000 matches gives 0.869, which confirms the fraction.

Why is the naive ratio wrong, and in which direction?

The tempting shortcut compares the chance of three straight wins for me, (1/3)^3, with three straight for you, (1/6)^3, and takes my share: 8 over 9, 88.9%. That treats the match as a single race from scratch, but a broken streak is not a reset to equal footing: when you beat me on my streak of two, you start a streak of one, and the shortcut ignores every such hand-over. Those hand-overs favour the weaker player a little, which is why the true 86.9% sits below 88.9%. It is also worth saying that the draws change nothing about who wins: they only lengthen the match, which lasts about 33.9 games on average, because every draw sends both streaks back to zero.

135/211, about 64.0%. Let r be q over p, which is 0.4 / 0.6 = 2/3. The probability of reaching N from a stake of i is (1 - r^i) / (1 - r^N). With i = 2 and N = 5 that is (1 - 4/9) / (1 - 32/243) = (5/9) x (243/211) = 135/211. A fair coin would give 2/5 = 40%; the 60% edge lifts it to 64.0%. The game lasts about 6.0 bets on average.

Why is the answer not simply 2 out of 5?

With a fair coin the answer is 2/5, because a fair game cannot create or destroy expected money: you start with Rs 2, you finish with Rs 5 or Rs 0, so the chance of Rs 5 must be 2/5 to keep the average at 2. With a 60% coin each bet gains you Rs 0.20 on average, so the walk drifts upward and the chance of hitting the top is higher than the fair-coin fraction; what you need is a quantity that is still conserved under the biased coin. That quantity is (q/p) to the power of your stake. A win multiplies it by q/p, a loss by p/q, and weighted by their probabilities the two moves cancel: p x (q/p) + q x (p/q) = q + p = 1. Because that quantity is conserved, its starting value must equal its average finishing value, and that one line gives the formula.

How do you derive it from the states if you forget the formula?

Write P_i for the chance of reaching 5 from a stake of i. Then P_0 = 0, P_5 = 1, and in between P_i = 0.6 P_(i+1) + 0.4 P_(i-1), one equation per state. That is a second-order linear recurrence whose solutions are of the form A + B r^i with r = q/p, and the two boundary conditions fix A and B. Solving the five equations directly gives P_1 = 81/211, P_2 = 135/211, P_3 = 171/211 and P_4 = 195/211, and the recurrence is the thing to write on the whiteboard first, because it works for any rule change. A simulation of 200,000 games gives 0.639, agreeing with the fraction to three places. The same system with a 1 on the right-hand side of each interior equation gives the expected duration, about 6.0 bets from Rs 2.

What does the biased formula tell you about trading with an edge?

Let the target go to infinity. With a fair coin the chance of never going broke is zero: any finite stake is eventually lost. With the 60% coin it is 1 minus r to the stake, which from Rs 2 is 1 - 4/9 = 5/9, about 56%, and from Rs 10 it is above 98%. An edge does not protect a thin stake: with Rs 2 behind a 60% coin you still go broke 44% of the time, and the cure is not a better coin but a bigger stake relative to the bet. That is why a desk with a genuine edge still caps position size, and why the question sits next to the Kelly one. The limitation is that the bets here are of fixed size; once you can resize the bet with your capital, the ruin arithmetic changes completely.

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.

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.

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.

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.

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.

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.

232 of the 1,024 paths touch 104, about 22.7%. A path ending at or above 104 must have touched it: that needs at least 7 up days, 176 paths. A path that touches 104 and ends below it can be reflected in the 104 line after its first touch, and the reflection ends above 104, needing at least 8 up days: 56 paths. The pairing is one to one, so the count is 176 + 56 = 232.

Why is the reflection a fair way to count?

Imagine a walker who reaches a wall at some point and then wanders on. Flip every step after the first touch, up for down, and you get another walker who also touched the wall at the same moment and is now the same distance on the other side of it. Each flipped step is as likely as the original, so the two walks have equal probability. Reflection pairs every path that touches 104 and finishes below it with exactly one path that finishes above 104, so the hard count becomes an easy count of endpoints. Paths that finish at exactly 104 are their own partners and are counted once.

How do the binomial counts fall out?

After 10 days of plus or minus 1, the position is 100 plus ups minus downs, which is 100 + 2 x ups - 10. Ending at or above 104 needs ups of at least 7: C(10,7) + C(10,8) + C(10,9) + C(10,10) = 120 + 45 + 10 + 1 = 176. Ending strictly above 104 means at or above 106, since the endpoint is always even, so ups of at least 8: 45 + 10 + 1 = 56. The probability of touching is (176 + 56)/1,024 = 232/1,024, about 22.7%, against only 176/1,024 = 17.2% for ending at or above 104. The gap is the 56 paths that visited the barrier and came back.

Say why a derivatives desk cares. A knock-out or one-touch option pays on exactly this event, and the reflection principle is how the closed-form barrier prices are derived in continuous time. The limitation is in the symmetry: reflection needs the up and down steps to be equally likely and the barrier to sit on the lattice. With a drift, or in continuous time with a non-zero rate, the reflected path is not equally likely and a correction factor appears, which is why barrier formulas carry that extra power of the barrier over the spot.

The queue is empty one third of the time and holds 2 orders on average. With arrivals at 2 a minute and fills at 3, the fill process is busy a fraction 2/3 of the time, so the queue is empty 1/3 of the time. Balancing flow across each arrow of the chain gives share(n + 1) = (2/3) share(n), a geometric ladder starting at 1/3, whose mean is (2/3)/(1/3) = 2.

Why does balancing flow across one arrow solve the whole chain?

Picture a ticket counter with one clerk. Over a long day, the number of times the queue grows from 3 people to 4 must equal the number of times it shrinks from 4 to 3, because you cannot go up through that boundary twice without coming back down through it once. In the long run the flow of arrivals across each boundary equals the flow of fills back across it, which pins the share of time in each state to the share in the state below. Arrivals push at 2 a minute times the share of time at n; fills pull at 3 a minute times the share at n + 1. Equal flows mean share(n + 1) = (2/3) share(n).

How do the shares add up to one, and what is the mean?

The shares form a geometric series: share(0), share(0) x 2/3, share(0) x 4/9 and so on. They must sum to 1, and a geometric series with ratio 2/3 sums to 3 times its first term, so share(0) = 1/3. The mean number in the queue is the sum of n times share(n), which for a geometric ladder is the ratio over one minus the ratio: (2/3)/(1/3) = 2. A queue fed at two thirds of its capacity holds two orders on average, and because the mean is rho/(1 - rho) it explodes as arrivals approach fills. At 2.7 arrivals a minute the average would be 9; at 3 it is unbounded.

Two sentences that show you can use it. Little's law says time in the system equals mean number over arrival rate, 2/2 = 1 minute, so an order joining this queue expects to wait a minute to be filled, which is what a market maker deciding whether to post at this price wants to know. The limitation: the chain assumes memoryless arrivals and fills and no cancellations, and real order books have cancellations that depend on queue position, so the geometric shape is a first model, not a description.