Common trading-interview formats

Any trading role · Market making · Market making · practice question on a reported topic

"I flip a fair coin until it lands heads and pay you the number of flips. Make me a market."

1. What fair value should your market be centred on?
   Answer and worked solution

   Answer: 2

   Key idea. Centre your market on the expected payout.

   1. The number of flips until the first head is geometric with p = ½.
   2. E = 1/p = 2.

   Check. A first market like 1.5 @ 2.5, then adjust as you're traded against.

2. Follow-up (ours): what is the standard deviation of the payout? (This guides your spread width.)
   Answer and worked solution

   Answer: 1.4142

   Key idea. The spread of the payout guides how wide to quote.

   1. Var = (1 − p)/p² = 0.5/0.25 = 2.
   2. SD = √2 ≈ 1.41.

   Check. The distribution is skewed (occasionally 5+ flips), so be careful about selling it cheaply.

Interviewers may push further: The interviewer buys at your offer three times. What do you do? (Raise your market; they may know something.)

Source: Quantt: quant trader interview questions · Try it in the app →

Any trading role · Market making · Market making · practice question on a reported topic

"I roll five dice and pay you the median. Make me a market."

1. Where should the market be centred?
   Answer and worked solution

   Answer: 3.5

   Key idea. Use symmetry: swap every face k for 7 − k.

   1. That swap mirrors the dice, and so mirrors their median around 3.5.
   2. A symmetric distribution's mean is its centre: 3.5.

   Check. The median of five dice is more concentrated than one die, so you can quote tighter than for a single die.

Interviewers may push further: What is P(median = 1)?

Source: Quantt: quant trader interview questions · Try it in the app →

Options trader · Technical · Options · practice question on a reported topic

A stock is at $100. The 1-year at-the-money call is worth $10. Rates are 5% (continuously compounded), no dividends.

1. (a) What is the 1-year ATM put worth?
   Answer and worked solution

   Answer: 5.1229

   Key idea. Put–call parity, which holds without any model: C − P = S − K·e^−rT.

   1. S = K = 100, r = 0.05, T = 1, C = 10.
   2. K·e^−0.05 ≈ 100 × 0.9512 = 95.12.
   3. P = C − S + Ke^−rT = 10 − 100 + 95.12 = 5.12.

   Check. With positive rates the ATM call is worth more than the ATM put, and it is here.

2. (b) Roughly how much does the call's value change if implied volatility falls by 1 point?
   Answer and worked solution

   Answer: -0.4

   Key idea. Vega: how much the price moves per unit of volatility. ATM, vega ≈ 0.4·S·√T per 1.00 of vol.

   1. 0.4 × 100 × 1 = 40 per 1.00 (100 points) of volatility.
   2. 1 point is 0.01: 40 × 0.01 = 0.40.
   3. Vol falls, so the call's value changes by about −$0.40.

   Check. Sanity check: 1 vol point on a $10 option is a few percent of its value.

   Trap. Some prep material says $4, which is off by a factor of ten. Say the units out loud.

Interviewers may push further: What is the ATM straddle worth at 30% vol? (≈ 0.8 × 0.3 × 100 = $24.)

Source: Quantt: quant trader interview questions · Try it in the app →

Any trading role · Technical · Market making · practice question on a reported topic

The market is 10 bid, 11 offered. You think fair value is 10.4.

1. (a) What is your expected profit per share if you sell at 10?
   Answer and worked solution

   Answer: -0.4

   Key idea. Compare the trade price with your fair value, not with the other side of the market.

   1. You sell at 10 something worth 10.4.
   2. Profit per share = 10 − 10.4 = −0.4.
2. (b) If you buy at 11?
   Answer and worked solution

   Answer: -0.6

   Key idea. Same comparison for the buy side.

   1. You buy at 11 something worth 10.4.
   2. Profit per share = 10.4 − 11 = −0.6.

   Check. Both trades lose. Don't hit or lift: quote your own market around 10.4 (e.g. 10.2 @ 10.6) and let others trade with you.

Interviewers may push further: When would you cross the spread anyway? (Urgent hedging, strong short-term signal.)

Source: Quantt: quant trader interview questions · Try it in the app →

Any · Interview · Conditional probability · practice question on a reported topic

There are three doors: a car behind one, goats behind the others. You pick a door. The host, who knows where the car is, opens another door to show a goat and offers you the chance to switch.

1. (a) What is your chance of winning if you switch?
   Answer and worked solution

   Answer: 2/3 ≈ 0.66667

   Key idea. The host's reveal doesn't change the chance your first pick was right; it concentrates the rest on the other door.

   1. Your first pick is right with probability 1/3.
   2. The host can always open a goat door, so nothing about your door changes.
   3. Switching wins exactly when your first pick was wrong: 2/3.

   Check. Enumerate: car behind 1, 2 or 3, you pick 1. Switching wins in 2 of the 3 cases.

   Trap. Answering ½ because "two doors are left". The host's choice wasn't random.

2. (b) Same game with 100 doors: the host opens 98 goat doors. Chance of winning if you switch?
   Answer and worked solution

   Answer: 0.99

   Key idea. Same logic with 100 doors.

   1. Your first pick is right 1% of the time.
   2. The single other closed door carries the rest: 99/100.

   Check. The 100-door version makes the intuition obvious.

Interviewers may push further: What if the host opened a door at random and it happened to show a goat? (Then switching wins 1/2.)

Source: Wikipedia: Monty Hall problem · Try it in the app →

Any · Interview · Probability · practice question on a reported topic

Ignore leap years and assume birthdays are uniform over 365 days.

1. (a) What is the smallest group in which the chance of a shared birthday exceeds 1/2?
   Answer and worked solution

   Answer: 23

   Key idea. Complement: everyone has a different birthday.

   1. P(all different, n people) = Π_k=0ⁿ⁻¹ (365 − k)/365.
   2. Approximation: there are C(n, 2) pairs, each matching with probability 1/365: P(no match) ≈ e^{−n(n−1)/730}.
   3. This drops below ½ when n(n − 1)/730 > ln 2 ≈ 0.693, i.e. n(n − 1) > 506.
   4. n = 23 (23 × 22 = 506).

   Check. It feels too small because there are 253 pairs among 23 people, not 23.

2. (b) What is the probability for a group of 23?
   Answer and worked solution

   Answer: 0.5073

   Key idea. Evaluate the product exactly.

   1. 1 − Π_k=0²² (365 − k)/365.
   2. ≈ 0.507.

   Check. The approximation 1 − e^−253/365 ≈ 0.500 is very close.

Source: Wikipedia: Birthday problem · Try it in the app →

Any · Interview · Optimal stopping · practice question on a reported topic

n candidates are interviewed in random order. After each one you must hire them on the spot or reject them for good, and you can only rank each against those you've already seen. You want the single best.

1. (a) With n = 3, what is your best chance of hiring the best candidate?
   Answer and worked solution

   Answer: 0.5

   Key idea. Try "reject the first, then take the first one better than everyone so far".

   1. Rank candidates 1 (worst) to 3 (best). List the 6 orders.
   2. The strategy hires the best candidate in 1-3-2, 2-1-3 and 2-3-1, and fails in 1-2-3, 3-1-2 and 3-2-1.
   3. It wins in 3 of 6 orders: ½.

   Check. Hiring the first candidate wins only 1/3.

2. (b) For large n, the best strategy (reject the first n/e, then take the next record) wins with what probability?
   Answer and worked solution

   Answer: 0.36788

   Key idea. Reject the first r, then take the next record. Optimise r.

   1. P(win) ≈ (r/n) ln(n/r) for large n.
   2. Let x = r/n: maximise −x ln x. Derivative −ln x − 1 = 0 gives x = 1/e.
   3. Value at x = 1/e: (1/e) × 1 = 1/e ≈ 0.368.

   Check. Remarkably, the chance doesn't go to 0 as n grows.

Source: Wikipedia: Secretary problem · Try it in the app →

Any · Interview · Probability · practice question on a reported topic

Parallel lines are drawn on the floor one needle-length apart. A needle is dropped at random.

1. What is the probability it crosses a line?
   Answer and worked solution

   Answer: 0.63662

   Key idea. Parametrise a drop by the centre's distance to the nearest line and the needle's angle.

   1. Length = spacing = 1. Distance x uniform on [0, ½]; angle θ uniform on [0, π/2].
   2. It crosses when x ≤ ½ sin θ.
   3. P = (2/π) ∫₀^π/2 sin θ dθ = 2/π ≈ 0.637.

   Check. Drop many needles, and the crossing rate estimates π.

Source: Wikipedia: Buffon's needle problem · Try it in the app →

Any · Interview · Combinatorics · practice question on a reported topic

Four people check their hats, and the hats are handed back at random.

1. (a) What is the probability nobody gets their own hat back?
   Answer and worked solution

   Answer: 0.375

   Key idea. Inclusion–exclusion on "person i gets their own hat".

   1. P(no one) = 1 − 1/1! + 1/2! − 1/3! + 1/4!.
   2. = 1 − 1 + 0.5 − 0.1667 + 0.0417 = 9/24.
   3. = 3/8.

   Check. For large n this tends to 1/e ≈ 0.368; already close at n = 4.

2. (b) What is the expected number who get their own hat?
   Answer and worked solution

   Answer: 1

   Key idea. Linearity: each person gets their own hat with probability 1/4.

   1. E = 4 × 1/4 = 1.

   Check. True for any n, despite the matches not being independent.

Source: Wikipedia: Derangement · Try it in the app →

Any · Interview · Combinatorics · practice question on a reported topic

Candidate A gets 6 votes and B gets 4. The votes are counted in random order.

1. What is the probability A is strictly ahead throughout the count?
   Answer and worked solution

   Answer: 0.2

   Key idea. Ballot theorem, proved by reflection.

   1. A must take the first vote.
   2. Bad orders starting with A (B catches up at some point) match one to one with orders starting with B: reflect the votes before the first tie.
   3. P = P(A first) − P(B first) = 6/10 − 4/10.
   4. = 1/5.

   Check. In general: (a − b)/(a + b).

Source: Wikipedia: Bertrand's ballot theorem · Try it in the app →

Any · Interview · Probability · practice question on a reported topic

100 numbered prisoners. 100 boxes each hold one prisoner's number, shuffled. Each prisoner may open 50 boxes, alone and without communicating afterwards. All are freed only if every prisoner finds their own number. They may agree a strategy beforehand.

1. With the best strategy, what is the chance they all succeed?
   Answer and worked solution

   Answer: 0.31183

   Key idea. Follow the cycle: open your own number's box, then the box named inside, and so on.

   1. You find your number exactly when its cycle has length ≤ 50.
   2. Everyone succeeds when the shuffle has no cycle longer than 50.
   3. P(some cycle of length k > 50) = 1/k, and at most one such cycle exists.
   4. P(success) = 1 − (1/51 + … + 1/100) ≈ 1 − ln 2 ≈ 0.31.

   Check. Random opening would give (½)¹⁰⁰. The strategy makes everyone's successes depend on the same event.

Source: Wikipedia: 100 prisoners problem · Try it in the app →

Any · Interview · Expected value · practice question on a reported topic

You turn over cards from a shuffled 52-card deck until the first ace.

1. What is the expected number of cards turned over, including the ace?
   Answer and worked solution

   Answer: 10.6

   Key idea. Symmetry: the 4 aces split the 48 other cards into 5 gaps of equal expected size.

   1. Each gap holds 48/5 = 9.6 cards on average.
   2. You turn over the first gap, then the ace.
   3. 9.6 + 1 = 10.6 (53/5).

   Check. General formula: (N + 1)/(K + 1) for the first of K special cards among N.

Source: Wikipedia: Negative hypergeometric distribution · Try it in the app →

Any · Interview · Probability · practice question on a reported topic

Three points are chosen at random on a circle.

1. What is the probability that the triangle they form contains the centre?
   Answer and worked solution

   Answer: 0.25

   Key idea. The triangle misses the centre exactly when all three points lie in a semicircle.

   1. For each point, P(the other two lie in the semicircle clockwise from it) = (½)² = ¼.
   2. These three events can't overlap, so P(semicircle) = 3/4.
   3. P(contains centre) = 1/4.

   Check. For n points: P(semicircle) = n/2ⁿ⁻¹.

Interviewers may push further: Four points: chance the quadrilateral contains the centre? (1/2.)

Source: Wikipedia: Wendel's theorem · Try it in the app →

Any · Interview · Markov chains · practice question on a reported topic

An urn starts with one red and one blue ball. Repeatedly draw a ball at random and put it back together with another ball of the same colour.

1. In the first 10 draws, what is the probability exactly 5 are red?
   Answer and worked solution

   Answer: 1/11 ≈ 0.090909

   Key idea. Every sequence with the same number of reds has the same probability.

   1. P(a given order with k reds in n draws) = k!(n − k)!/(n + 1)!.
   2. Multiply by C(n, k) orders: 1/(n + 1).
   3. For n = 10, k = 5: 1/11.

   Check. Every count from 0 to 10 is equally likely: the long-run red fraction is uniform on [0, 1].

Source: Wikipedia: Pólya urn model · Try it in the app →

Any · Interview · Probability · practice question on a reported topic

Which is most likely: at least one six with 6 dice, at least two sixes with 12 dice, or at least three sixes with 18 dice?

1. (a) P(at least one six in 6 dice)?
   Answer and worked solution

   Answer: 0.6651

   Key idea. Complement: no sixes.

   1. (5/6)⁶ ≈ 0.335.
   2. P(at least one) ≈ 0.665.
2. (b) P(at least two sixes in 12 dice)?
   Answer and worked solution

   Answer: 0.61867

   Key idea. Complement: zero or one six.

   1. P(0) = (5/6)¹² ≈ 0.112; P(1) = 12(1/6)(5/6)¹¹ ≈ 0.269.
   2. P(at least two) ≈ 1 − 0.381 = 0.619.

   Check. So the 6-dice bet is the best: the distributions spread out more around the same expected count.

Source: Wikipedia: Newton–Pepys problem · Try it in the app →

Any · Interview · Bayes · practice question on a reported topic

1% of people have a condition. A test catches 99% of those who have it, but also flags 5% of those who don't. Someone tests positive.

1. What is the probability they have the condition?
   Answer and worked solution

   Answer: 1/6 ≈ 0.16667

   Key idea. Count people instead of multiplying probabilities.

   1. In 10,000 people: 100 have it, 99 test positive.
   2. 9,900 don't; 5% of them, 495, test positive anyway.
   3. P(has it | positive) = 99/(99 + 495) = 1/6 ≈ 0.17.

   Check. False positives swamp the true ones because the condition is rare.

   Trap. Answering 99%. That's P(positive | has it), not the question asked.

Source: Wikipedia: Base rate fallacy · Try it in the app →

Any · Interview · Mental maths · practice question on a reported topic

Count the zeros at the end of 100! (100 factorial).

1. How many?
   Answer and worked solution

   Answer: 24

   Key idea. Each trailing zero needs a factor 10 = 2 × 5; count the 5s (they're scarcer).

   1. Multiples of 5 up to 100: 20.
   2. Multiples of 25 add one more each: 4.
   3. Total: 24.

   Check. Legendre's formula: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0.

Source: Wikipedia: Trailing zero · Try it in the app →

Any · Interview · Expected value · practice question on a reported topic

Add independent uniform random numbers from [0, 1] until the total exceeds 1.

1. What is the expected number of numbers you add?
   Answer and worked solution

   Answer: 2.7183

   Key idea. Use E[N] = Σ_n≥0 P(N > n).

   1. N > n means the first n numbers sum to at most 1.
   2. That's the volume of a simplex: 1/n!.
   3. E[N] = Σ 1/n! = e ≈ 2.718.

   Check. At least 2 numbers are always needed, so the answer is above 2.

Source: Wikipedia: e (mathematical constant) · Try it in the app →

Any · Interview · Logic · practice question on a reported topic

You have two identical eggs and a 100-storey building. An egg breaks if dropped from floor f or above, for some unknown f, and survives otherwise. A broken egg can't be reused.

1. What is the fewest drops that guarantees finding f?
   Answer and worked solution

   Answer: 14

   Key idea. With d drops, the first drop should be from floor d, then d − 1 floors higher each time.

   1. If the first egg breaks at floor d, test floors 1 to d − 1 one by one: d drops in total.
   2. Each survival uses a drop, so the next jump is one floor smaller.
   3. d drops cover d + (d − 1) + … + 1 = d(d + 1)/2 floors.
   4. d(d + 1)/2 ≥ 100 first holds at d = 14 (105 floors).

   Check. Equal jumps of 10 floors need up to 19 drops; shrinking jumps balance the worst case.

Source: Wikipedia: egg dropping puzzle · Try it in the app →

Any · Interview · Decision making · practice question on a reported topic

Two envelopes hold money, one twice the other. You open one and see $100. "The other has $50 or $200 with equal chance, worth $125 on average, so always switch." What's wrong with this argument?

Source: Wikipedia: Two envelopes problem · Try it in the app →