Can You Solve All 5 Levels of a Quant Interview?
From a two-dice warm-up to a stochastic-process problem that trips up strong candidates, here's how the difficulty actually escalates in a real interview.
Introduction
A single quant interview can move through several completely different styles of thinking in the space of thirty minutes.
This article walks through five questions, arranged by difficulty, that reflect exactly that range. We have created a genuinely useful way to test several kinds of reasoning that show up again and again in trading and research interviews: conditioning correctly, updating beliefs on new evidence, thinking algorithmically, reasoning about information asymmetry, and modelling a random process with states.
Each level includes the question, how to approach it, a full solution, and an example of how a strong candidate might talk through it out loud. Let's start with Level 1.
Level 1: Easy
Concept: Conditional probability
The Question
You roll two fair six-sided dice. Given that at least one die shows a 4, what is the probability that the sum is 9?
How to Think About It
The word "given" is doing all the work here. This isn't a question about the probability of two independent events; it's a question about a shrunken sample space. Once you know at least one die is a 4, you should stop thinking about all 36 outcomes and instead think only about the outcomes consistent with that information.
Solution
Out of 36 equally likely outcomes, the outcomes where at least one die shows a 4 are: (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4). That's 11 outcomes in total, not 12, because (4,4) would otherwise be double counted.
Within those 11 outcomes, the ones that sum to 9 are (4,5) and (5,4). That's 2 outcomes.
P(sum = 9 | at least one die is 4) = 2/11.
How I Would Answer This in an Interview
"I'll condition on the event directly rather than using the formula. If at least one die shows a 4, there are 11 outcomes out of the usual 36, since (4,4) only counts once. Of those 11, only (4,5) and (5,4) sum to 9. So the answer is 2/11."
What the Interviewer Is Testing
This question checks whether you actually shrink the sample space when you condition, rather than multiplying probabilities as if the events were independent. It's a small trap, but it's a common one, and getting it wrong early in an interview creates doubt about your fundamentals before the harder questions even start.
Level 2: Easy
Concept: Statistics and Bayesian inference
The Question
A coin is equally likely to be fair (P(H) = 1/2) or biased (P(H) = 3/4). You pick one coin at random and observe 3 heads in 3 flips. What is the probability that you selected the biased coin?
How to Think About It
You have a prior belief about which coin you're holding, and then you get evidence. The right move is to compare how likely that evidence is under each hypothesis, and let the more consistent hypothesis pick up a larger share of the posterior probability.
Solution
The prior on each coin is 1/2. The likelihood of 3 heads in 3 flips under each coin is:
P(3H | fair) = (1/2)^3 = 1/8
P(3H | biased) = (3/4)^3 = 27/64
By Bayes' rule:
P(biased | 3H) = [1/2 · 27/64] / [1/2 · 27/64 + 1/2 · 1/8] = (27/64) / (27/64 + 8/64) = 27/35
P(biased | 3 heads) = 27/35 ≈ 0.771.
How I Would Answer This in an Interview
"I want the posterior probability that the coin is biased. Both coins started equally likely, so the prior cancels out and I just need to compare the likelihoods. Three heads under the fair coin is 1/8, under the biased coin it's 27/64. Normalising those two numbers so they sum to 1 gives 27/35 for the biased coin."
What the Interviewer Is Testing
This is a direct test of Bayesian updating, and specifically whether you can set it up as a ratio of likelihoods rather than reaching for a formula you don't fully understand. Interviewers care less about whether you remember Bayes' rule and more about whether you can explain why the biased coin picks up more posterior weight.
Level 3: Medium
Concept: Programming and computational thinking
The Question
You are given an array of daily P&Ls: pnl = [3, -5, 2, 4, -1, 6, -7]. You want the maximum total P&L over any contiguous block of days. What is the maximum P&L, and what O(n) algorithm computes it for an arbitrary array?
How to Think About It
The brute-force instinct is to check every possible starting and ending day, which is O(n^2). The faster approach comes from asking a simpler question at each day: if I were forced to end my trading window today, what's the best total I could have achieved? That local question turns out to be enough to solve the global one.
Solution
Walk through the array once, keeping a running "best P&L ending here." At each day, that running value is either the day's P&L on its own, or the previous running value plus today's P&L, whichever is larger. Separately, track the best value seen so far across the whole walk.
Running through pnl = [3, -5, 2, 4, -1, 6, -7]:
Day 1 (3): running = 3, best = 3
Day 2 (-5): running = max(-5, 3-5) = -2, best = 3
Day 3 (2): running = max(2, -2+2) = 2, best = 3
Day 4 (4): running = max(4, 2+4) = 6, best = 6
Day 5 (-1): running = max(-1, 6-1) = 5, best = 6
Day 6 (6): running = max(6, 5+6) = 11, best = 11
Day 7 (-7): running = max(-7, 11-7) = 4, best = 11
The maximum P&L is 11, coming from the block [2, 4, -1, 6] on days 3 through 6.
This running-max approach is Kadane's algorithm. It works because a negative running total can never help a future block, so the moment your running sum dips below the value of the current day alone, you're better off restarting from that day. That single observation is what takes the problem from O(n^2) to O(n).
How I Would Answer This in an Interview
"I'll track a running total that resets whenever carrying the previous days forward stops being worth it. Each day I compare the day's value on its own against the running total plus that value, and keep the bigger of the two, while separately remembering the best running total I've seen. Walking through the array that way, the best block is days 3 through 6, giving 11."
What the Interviewer Is Testing
This question isn't really about memorising Kadane's algorithm by name. It's testing whether you can turn a global optimisation into a simple local decision made once per element, which is the same instinct behind a lot of dynamic programming problems you'll see in a technical round.
Level 4: Hard
Concept: Trading and Bayesian reasoning
The Question
An asset's true value is equally likely to be 100 or 110. A trader arrives. With probability 0.5 they are informed and know the true value; with probability 0.5 they are uninformed and buy or sell with equal probability regardless of value. An informed trader buys if the value is 110 and sells if the value is 100. You observe that the trader buys. What is the asset's fair value immediately after observing the buy?
How to Think About It
This is the standard adverse-selection problem that sits underneath market making. A buy order isn't just a random event; it's more likely to come from someone who knows the price is going up than from someone flipping a coin. Seeing a buy should shift your estimate of the value upward, and the size of that shift depends entirely on how much more likely a buy is when the trader is informed.
Solution
Condition on the true value and work out how likely a buy is in each case.
If the value is 110: an informed trader (probability 0.5) always buys, and an uninformed trader (probability 0.5) buys half the time. P(buy | value = 110) = 0.5(1) + 0.5(0.5) = 0.75
If the value is 100: an informed trader always sells, and an uninformed trader still buys half the time. P(buy | value = 100) = 0.5(0) + 0.5(0.5) = 0.25
Using Bayes' rule with equal priors on 100 and 110:
P(value = 110 | buy) = [0.5 · 0.75] / [0.5 · 0.75 + 0.5 · 0.25] = 0.375 / 0.5 = 0.75
So P(value = 110 | buy) = 0.75 and P(value = 100 | buy) = 0.25. The fair value is the expectation under this updated belief:
Fair value = 0.75(110) + 0.25(100) = 82.5 + 25 = 107.5
How I Would Answer This in an Interview
"As a market maker, I care about how much a buy order tells me. If the true value is 110, a buy happens 75% of the time, since the informed trader always buys and the uninformed trader buys half the time. If the value is 100, a buy only happens 25% of the time. So a buy is three times more likely under 110 than under 100, which means my updated belief puts 75% weight on 110. My fair value is 0.75 times 110 plus 0.25 times 100, which is 107.5. I'd move my quotes up from the prior midpoint of 105 to reflect that."
What the Interviewer Is Testing
This question is really about extracting information from order flow rather than treating a trade as noise. It's the same logic that underlies bid-ask spreads: market makers widen their quotes precisely because a trade against them is evidence, not just a random transaction, and this problem asks you to quantify exactly how much evidence one buy order carries.
Level 5: Very Hard
Concept: Probability and stochastic processes
The Question
A fair coin is flipped repeatedly. You stop as soon as either HHT or THH appears as three consecutive flips. What is the probability that HHT appears first?
How to Think About It
The trap here is treating HHT and THH as symmetric just because they use the same letters. They aren't. What matters is how each pattern can "build on itself." HHT can never partially overlap with a failed attempt at HHT in a way that helps it, but as you'll see, it's actually the pattern HH that behaves differently from what intuition suggests. The clean way to solve this is to track the last two flips as a state, since that's all the information you need to know how close you are to either pattern.
Solution
After the first two flips, you land in one of four equally likely states: HH, HT, TH, or TT, each with probability 1/4, since neither pattern can complete in only two flips. From each state, let p(state) be the probability that HHT appears first.
State HH: flip H, you stay in HH. Flip T, you complete HHT immediately and win. So p(HH) = 1/2 · p(HH) + 1/2 · 1, which gives p(HH) = 1. Once you have HH, THH can never appear first, because any T you flip next turns into HHT on the spot rather than building toward THH.
State TH: flip H, you complete THH and lose. Flip T, your last two flips become HT. So p(TH) = 1/2 · 0 + 1/2 · p(HT).
State HT: flip H, last two flips become TH. Flip T, last two flips become TT. So p(HT) = 1/2 · p(TH) + 1/2 · p(TT).
State TT: flip H, last two flips become TH. Flip T, you stay in TT. So p(TT) = 1/2 · p(TH) + 1/2 · p(TT), which gives p(TT) = p(TH).
Substituting p(TT) = p(TH) into the HT equation gives p(HT) = p(TH). Substituting that into the TH equation gives p(TH) = 1/2 · p(TH), which forces p(TH) = 0. It follows that p(HT) = 0 and p(TT) = 0 as well.
Averaging over the four equally likely starting states:
P(HHT first) = 1/4(1) + 1/4(0) + 1/4(0) + 1/4(0) = 1/4
So HHT wins with probability 1/4, and THH wins with probability 3/4, despite the two patterns looking superficially similar. This can be independently checked using Conway's leading number algorithm for pattern races, which agrees with the 1:3 odds derived above.
How I Would Answer This in an Interview
"I'll track my state as the last two flips I've seen, since that's all I need to know how close I am to either pattern. The interesting case is landing in HH: from there, THH is dead, because the very next T gives me HHT directly rather than resetting toward THH. That asymmetry is the whole problem. Setting up the four states HH, HT, TH, and TT and solving the resulting equations, everything collapses to zero except the HH state, which is a guaranteed win for HHT. Averaging over the four equally likely starting states after two flips gives 1/4."
What the Interviewer Is Testing
This question tests whether you can build your own state space for a random process rather than reaching for a memorised formula. It also rewards noticing structural asymmetry, in this case that reaching HH is an absorbing near-certainty for HHT, which is exactly the kind of observation that separates a candidate who understands Markov chains from one who has only seen textbook examples of them.
Closing Remarks
These five questions move from a simple conditional probability exercise to a stochastic process problem that requires building your own model from scratch. That range is a fair reflection of what a real quant interview process can look like, sometimes across a single conversation.
The common thread isn't a specific formula. It's the habit of slowing down enough to identify what kind of problem you're actually facing, whether that's a shrunken sample space, a belief that needs updating, a local decision that solves a global problem, information hidden in someone else's action, or a process that needs its own states defined before you can reason about it.
These five questions only cover a small part of the interview landscape. If you want more practice with this style of thinking, our probability interview questions and brainteasers go much deeper on Levels 1 and 2. For the trading intuition behind Level 4, our market-making simulations let you practice reasoning about order flow in real time. And if you want a structured path through the underlying material, our probability for quant finance course and online assessments are built specifically for candidates preparing for these interviews.
