Real questions from quantitative trading interviews, each with the full reasoning rather than just an answer. Every numerical result on this page has been verified by computation.
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Answer: 1/6
There are 36 equally likely outcomes. The sum is 7 for (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) — six of them. So 6/36 = 1/6. Worth memorising: 7 is the most likely sum, and each sum s from 2 to 7 has s−1 combinations.
Answer: 2
Let E be the expectation. With probability 1/2 you finish in one flip; otherwise you have used a flip and are back where you started: E = 1 + (1/2)E, so E = 2. In general, for probability p per trial the expected wait is 1/p.
Answer: 671/1296 ≈ 51.8%
Work with the complement. No six on a single roll has probability 5/6, so no six in four rolls is (5/6)⁴ = 625/1296. Therefore at least one six is 1 − 625/1296 = 671/1296 ≈ 51.8%. This is the classic de Méré problem — the answer is just above a coin flip.
Answer: 1/221
P = (4/52) × (3/51) = 12/2652 = 1/221 ≈ 0.45%. Note the second factor is 3/51, not 4/52 — the deck has changed.
Answer: 1/3
The equally likely cases are BB, BG, GB, GG. The information rules out GG, leaving three cases, of which one is BB. So 1/3. The common wrong answer is 1/2, which comes from imagining a specific child rather than conditioning on the stated event. If instead you were told "the elder is a boy", the answer really would be 1/2.
Answer: 23
P(all distinct) = 365/365 × 364/365 × … × (365−n+1)/365. At n = 22 this is about 0.524, and at n = 23 about 0.493. So 23 people give a shared birthday probability of roughly 50.7%. It is counter-intuitive because there are C(23,2) = 253 pairs, not 23.
Answer: 3/8
There are 8 outcomes; exactly two heads occurs as HHT, HTH, THH. So 3/8. Generally C(n,k)/2ⁿ.
Answer: Yes — switching wins 2/3 of the time
Your first pick is right with probability 1/3, and that does not change when the host — who was always going to open a goat door — reveals one. So the remaining door carries the other 2/3. The key is that the host's choice is not random: he never opens the car door. If he opened a door at random and it happened to show a goat, switching would gain nothing.
Answer: 14.7
This is the coupon collector problem. After collecting k distinct faces, the wait for a new one is geometric with p = (6−k)/6, so expected 6/(6−k). Total = 6(1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1) = 6 × 2.45 = 14.7.
Answer: 1/4
With break points x and y on [0,1], a triangle needs every piece below 1/2. Plotting the constraints in the unit square leaves two triangular regions each of area 1/8, totalling 1/4. Intuition: the longest piece must be under half the stick, which usually fails.
Answer: 10.6
The 4 aces split the other 48 cards into 5 gaps, and by symmetry each gap has expected size 48/5 = 9.6. The cards before the first ace are one such gap, so you draw 9.6 + 1 = 10.6 cards on average. Symmetry arguments like this replace a great deal of algebra.
Answer: 63/256 ≈ 24.6%
C(10,5)/2¹⁰ = 252/1024 = 63/256 ≈ 24.6%. Note that the single most likely outcome is still under 25% — 'about half' is likely, but exactly half is not.
Answer: $3.50
E = (1+2+3+4+5+6)/6 = 3.5. Any price below 3.5 is +EV for you.
Answer: $4.25
Reroll only when the first roll is below the reroll's expectation of 3.5 — so reroll on 1, 2 or 3. Then E = (1/2)(3.5) + (1/6)(4+5+6) = 1.75 + 2.5 = 4.25. The general principle: continue when the current value is below the continuation value.
Answer: ≈ $4.67
With one reroll left the game is worth 4.25, so on the first roll you should reroll anything below 4.25 — that is 1 through 4. E = (4/6)(4.25) + (1/6)(5) + (1/6)(6) = 2.833 + 1.833 = 4.67.
Answer: 7
Each rank 1–13 is equally likely, so E = (1+13)/2 = 7. Suits are irrelevant, which candidates often overcomplicate.
Answer: +$0.50 per flip
E(payout) = 0.5 × 3 = 1.5. Subtract the $1 cost for an edge of +$0.50, a 50% return per flip. Note this says nothing about how much to bet — that is a Kelly question.
Answer: 1.2
By linearity of expectation, each draw is red with probability 3/5, so E = 2 × 3/5 = 1.2. Linearity holds even though the draws are dependent — which is exactly why it is such a powerful shortcut.
Answer: 20%
Fractional odds of b:1 imply 1/(b+1) = 1/5 = 20%. If you believe the true probability is above 20%, the price is generous.
Answer: 6.67%
f* = (bp − q)/b with b = 3, p = 0.3, q = 0.7: (0.9 − 0.7)/3 = 0.2/3 = 6.67%. A modest edge justifies a modest stake — the most common sizing error is betting far more than this.
Answer: Higher, and skewed to buy
Being lifted instantly is information: the buyer thinks it is worth more than 44. A sensible response is to move the whole market up and skew — perhaps 43 / 47 — so you are more likely to buy than sell while you reassess. Interviewers watch specifically for whether you update. Requoting 40 / 44 after being hit signals you will not learn from the market.
Answer: Below fair value on both sides
You want to sell, not buy more. Rather than quoting symmetrically around 100, shift down — say 97 / 101 — so a seller finds you unattractive and a buyer finds you cheap. This is inventory management: your quote reflects both your view and your position, and the position often matters more.
Answer: Yes — buy from A at 52, sell to B at 53, for 1 point risk-free
A will sell to you at 52 and B will buy from you at 53. You lock 1 point per unit with no exposure to the underlying value. This is a crossed market: it exists whenever one trader's ask is below another's bid, and spotting it instantly is the reflex being tested.
Answer: Because a market maker who never trades earns nothing
A wide market cannot lose money to a pick-off, but it also attracts no flow. The business only works by trading often at a small edge. Interviewers read an excessively wide quote as either not understanding the business or being unwilling to commit to a view. The skill is quoting as tight as your genuine uncertainty allows — no tighter.
Answer: Because ruin is permanent
Kelly gives f* = (1 × 0.6 − 0.4)/1 = 20%. Betting more than Kelly raises volatility while lowering long-run growth, and betting everything guarantees eventual ruin — one loss ends the sequence. Growth compounds; zero does not.
Answer: Light rope 1 at both ends and rope 2 at one end; when rope 1 finishes, light rope 2's other end
Rope 1 lit at both ends burns out in 30 minutes regardless of unevenness. At that moment rope 2 has 30 minutes of burn left; lighting its second end halves that to 15. Total 30 + 15 = 45 minutes. The trick is that burning from both ends always halves the remaining time, even when the rate varies.
Answer: 7
Race 5 groups of 5 (5 races), then race the 5 winners (race 6). The overall winner is first. Only 5 horses can still place: 2nd and 3rd from the winner's group, 1st and 2nd from the runner-up's group, and 1st from the third group. One final race among those 5 settles 2nd and 3rd. Total 7. The insight is eliminating horses that cannot possibly place before racing.
Answer: 10
Number the bottles 0–999 in binary — 10 bits suffice since 2¹⁰ = 1024. Assign each prisoner one bit position and have them drink from every bottle with a 1 in that position. The pattern of deaths reads off the bottle number in binary. So 10 prisoners.
Answer: 17 minutes
Send 1 and 2 across (2), return 1 (1), send 5 and 10 (10), return 2 (2), send 1 and 2 (2). Total 2+1+10+2+2 = 17. The key insight is pairing the two slowest together so 10 and 5 cross once rather than 10 crossing twice. The obvious greedy solution gives 19.
Answer: Fill 5, pour into 3, empty 3, transfer the 2, refill 5, top up the 3
Fill the 5 and pour into the 3, leaving 2 in the 5-litre jug. Empty the 3 and move those 2 litres into it. Refill the 5, then top up the 3-litre jug — which needs only 1 more litre. That leaves exactly 4 litres in the 5-litre jug.
Answer: A circle cannot fall through its own hole
Every diameter of a circle is equal, so the cover cannot be turned to drop in. A square cover falls through diagonally, since the diagonal exceeds the side. Circles also roll and need no alignment when replaced. The question tests whether you reason from properties rather than recall an answer — the falling-through argument is the one that matters.
Answer: 1/2
The process ends when someone takes either seat 1 or seat 100, and by symmetry each is equally likely to be taken first. So the answer is 1/2, independent of the number of passengers. Attempting this by direct enumeration is very hard; the symmetry argument makes it a one-liner.
Answer: 2,598,960
C(52,5) = 52!/(5!·47!) = 2,598,960. Worth knowing by heart — it is the denominator of every poker probability.
Answer: ≈ 0.197%
Flushes of any kind: 4 × C(13,5) = 4 × 1287 = 5148. Remove 40 straight flushes to get 5108. Divide by 2,598,960 for 0.197%, about 1 in 508.
Answer: 60
6 letters with A repeated 3 times and N twice: 6!/(3!·2!) = 720/12 = 60.
Answer: 45
C(10,2) = 45. Equivalently 10 × 9/2, halving because a handshake is shared.
Answer: 1024
Each element is either in or out, giving 2¹⁰ = 1024, including the empty set and the full set.
Answer: Roughly 100 — the method matters, not the number
Chicago has about 3 million people, say 1 million households. Perhaps 1 in 20 owns a piano, giving 50,000 pianos. Tuned roughly once a year, that is 50,000 tunings. A tuner doing 4 a day for 250 days handles 1,000 — so about 50 tuners, plus institutional pianos, call it 50–100. State assumptions aloud and keep numbers round. Interviewers score the decomposition, not the answer.
Answer: Roughly 500,000
A bus is about 12m × 2.5m × 2m ≈ 60 m³, or 60 million cm³. A golf ball is about 4.3cm across, occupying roughly 80 cm³ in its bounding cube. That gives 750,000, and sphere packing wastes about a third of the space, so around 500,000. Round aggressively; the packing correction is the detail that impresses.
Answer: Around 400 tonnes
Maximum take-off weight is close to 400 tonnes (about 875,000 lb). A sanity check: roughly 180 tonnes empty, 140 tonnes of fuel, and 400 passengers with luggage at 100kg each adds 40 tonnes.
Answer: Around 60,000 litres
About 2 litres a day × 365 days ≈ 730 litres a year. Over 80 years that is roughly 60,000 litres — near enough to a small swimming pool, which is a useful way to check the order of magnitude.
Answer: 391
Use the difference of squares: both numbers sit 3 from 20, so 17 × 23 = 20² − 3² = 400 − 9 = 391. This works whenever two numbers are equally spaced around a round number.
Answer: 87.5%
Eighths are worth memorising: 1/8 = 12.5%, so 7/8 = 100 − 12.5 = 87.5%. Knowing every eighth, sixth and twelfth by heart removes most fraction work in a timed test.
Answer: 1200
25 = 100/4, so 48 × 25 = 4800/4 = 1200. Multiplying by 25 is always divide by 4 then ×100; by 5 is halve then ×10.
Answer: 36
10% is 24, and 5% is half of that, 12. Sum for 36. Splitting percentages into 10% and 5% chunks is faster than any formula.
Answer: ≈ 44.7
44² = 1936 and 45² = 2025, so the answer sits just under 45. Interpolating gives about 44.7 (the true value is 44.72). Knowing squares to 25 and the round hundreds makes this instant.
Answer: 0.0625
Halve repeatedly: 1/2 = 0.5, 1/4 = 0.25, 1/8 = 0.125, 1/16 = 0.0625. These powers of two appear constantly in option pricing and tick sizes.
Answer: Bet sizing — a positive edge still ruins you if you overbet
Each flip has EV +$0.50, but that says nothing about survival. Staking your whole bankroll means a single tail ends the sequence permanently. Kelly here is f* = (2 × 0.5 − 0.5)/2 = 25% of bankroll. Edge tells you whether to bet; Kelly tells you how much. Confusing the two is the most expensive mistake in trading.
Answer: The two-way trader
Identical P&L conceals very different risk. The one-way trader was exposed to a single outcome and was paid for luck as much as skill; the two-way trader earned the spread repeatedly with less exposure. Firms hire for repeatable process, and in a group exercise this distinction is usually what separates candidates.
Answer: Zero at fair odds, and it guarantees eventual ruin with a finite bankroll
Doubling does not change the odds of any individual bet, so the EV of the sequence stays zero at fair odds and negative with any house edge. The strategy converts many small wins into one catastrophic loss: with a finite bankroll, the losing streak that exceeds your capital is certain to arrive eventually. The apparent reliability is exactly what makes it dangerous.
Reading solutions is not the same as producing them under a clock. The drills put the same skills under time pressure, which is where interviews are actually decided.
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