The quant interviews at top HFT shops are deliberately designed to separate candidates who've memorized formulas from those who can build rigorous probabilistic frameworks from scratch under pressure. The problems below aren't trivia, each one isolates a specific reasoning skill: symmetry arguments, conditional probability, recursive expected value, optimal stopping, and recurrence in stochastic processes.
Work through each solution carefully. The goal isn't to memorize answers, it's to internalize the reasoning pattern so you can construct it cleanly on the spot under real interview pressure.
Problem 01
The Unfair Coin
You have a biased coin with P(H) = p (unknown). How do you simulate a fair 50/50 outcome using only this coin?
Solution
Flip the coin twice. Define: HT → "Heads," TH → "Tails," HH or TT → repeat.
Since P(HT) = p(1−p) = P(TH), the procedure is perfectly fair regardless of p.
Expected flips per outcome = 2/[2p(1−p)], minimized at p = 0.5.
Why interviewers ask it. Tests whether you understand symmetry as a tool for constructing fairness without knowing the underlying bias.
Problem 02
The Monty Hall Problem, Extended
You're on a game show with 100 doors. One hides a car; 99 hide goats. You pick door #1. Monty opens 98 doors, all goats. Should you switch?
Solution
Your initial pick has P(car) = 1/100. That probability is fixed.
The remaining door absorbs all the probability mass that wasn't behind your door: P(car | switch) = 99/100.
General formula with n doors: P(win | switch) = (n−1)/n.
P(win | switch) = (n−1)/n
The trap. Many candidates argue "two doors left, 50/50." This ignores the information content of Monty's reveal.
Problem 03
Expected Flips: HH vs. HT
On average, how many fair coin flips does it take to first see HH? How about HT?
Solution
For HH: Let E₁ = expected flips from start, E₂ = expected flips given last flip was H.
E₁ = 1 + (1/2)E₂ + (1/2)E₁ and E₂ = 1 + (1/2)(0) + (1/2)E₁. Solving: E₁ = 6 flips.
For HT: E₁ = 1 + (1/2)E₂ + (1/2)E₁ and E₂ = 1 + (1/2)(0) + (1/2)E₂ → E₂ = 2, so E₁ = 4 flips.
E[HH] = 6 | E[HT] = 4
Why interviewers ask it. HH takes longer because hitting H after a T resets the entire streak. HT is self-correcting, a T after the initial H is progress, not a reset.
Problem 04
Gambler's Ruin
A gambler starts with $k and plays a fair game (p = 0.5) until reaching $n (wins) or $0 (ruined). What's the probability of ruin?
Solution
For a fair game: P(ruin) = 1 − k/n. P(reaching $n) = k/n.
For a biased game (p ≠ 0.5), let q = 1 − p:
P(ruin) = [1 − (p/q)^k] / [1 − (p/q)^n]
P(ruin | fair) = 1 − k/n
Why interviewers ask it. If your edge p is only slightly above 0.5, starting capital k relative to your drawdown limit n determines survival probability over any finite trading horizon.
Problem 05
The Birthday Problem
How many people do you need in a room before there's greater than 50% probability that two share the same birthday?
Solution
P(no match with n people) = (365/365)(364/365)(363/365)···((365−n+1)/365)
The threshold: n = 23 people gives P(match) ≈ 50.7%.
With 57 people, P > 99%.
n = 23 → P(match) ≈ 50.7%
The trap. The answer isn't 183 (half of 365). With 23 people, there are C(23,2) = 253 distinct pairs, each with a small but nonzero collision chance. The growth is combinatorial.
Problem 06
Russian Roulette: Spin or Don't Spin?
A 6-chamber revolver has 2 consecutive bullets loaded. You pull the trigger once and survive (blank). Should you spin again or pull immediately?
Solution
After surviving, 4 of the remaining 5 chambers are safe (the next shot from chambers 3, 4, 5, 6 is blank; only after chamber 1 is fatal).
P(safe | pull now) = 4/5.
If you spin: P(safe) = 4/6 = 2/3.
P(safe | pull) = 4/5 > P(safe | spin) = 2/3
Why interviewers ask it. Since 4/5 > 2/3: Don't spin. Pulling immediately is strictly safer.
Problem 07
The Two-Envelope Paradox
Two envelopes contain X and 2X dollars. You open one and find $100. The other holds either $50 or $200. Should you switch?
Solution
The naïve E(switch) = 0.5($50) + 0.5($200) = $125 > $100 argument is wrong.
X and 2X are fixed, they cannot both vary simultaneously relative to your observation.
Without a prior distribution on X, you cannot compute a valid expected value for switching.
The trap. You cannot determine a rational switching strategy without prior knowledge of X's distribution.
Why interviewers ask it. Tests whether candidates can identify improper conditioning, a critical skill when building pricing models with unknown latent parameters.
Problem 08
The Secretary Problem (Optimal Stopping)
You interview n candidates sequentially and must hire or reject on the spot. You want to maximize the probability of hiring the best candidate. What's your strategy?
Solution
Observe the first ⌊n/e⌋ ≈ 0.368n candidates without hiring (the "exploration phase").
Then hire the first candidate who outperforms everyone seen during exploration.
This strategy yields P(best candidate) → 1/e ≈ 36.8% as n → ∞.
Observe first ⌊n/e⌋ candidates, then hire next best → P ≈ 36.8%
Why interviewers ask it. Structurally identical to optimal order execution, "observe" incoming prices for a calibration window, then execute when you see the best price observed so far.
Problem 09
Conditional Dice Probability
You roll two fair dice. Given that the sum is at least 7, what's the probability the sum is exactly 8?
Solution
Outcomes with sum ≥ 7: 6+5+4+3+2+1 = 21 out of 36 total.
Outcomes summing to exactly 8: (2,6), (6,2), (3,5), (5,3), (4,4) → 5 outcomes.
P(sum = 8 | sum ≥ 7) = 5/21 ≈ 23.8%
P(sum=8 | sum≥7) = 5/21 ≈ 23.8%
The trap. Candidates who compare sums of 7 (6 pairs) and 8 (5 pairs) directly miss the full conditional sample space.
Problem 10
The Random Walk and Recurrence
A particle starts at 0 on a 1D number line, moves +1 or −1 with equal probability each step. What's the probability it eventually returns to 0?
Solution
1D: Return probability = 1 (recurrent). The particle always returns, though the expected return time is infinite.
2D: Also recurrent (P = 1). Proven by Pólya in 1921.
3D and above: Transient. P(never return to origin) ≈ 34.1% in 3D, the particle escapes to infinity with positive probability.
1D & 2D: P(return) = 1 | 3D+: P(return) < 1
Why interviewers ask it. 1D and 2D price dynamics support mean-reversion strategies. Higher-dimensional factor spaces exhibit drift, which is why pure mean-reversion fails in multi-factor models.
Key takeaways
Succeeding in HFT probability interviews requires three skills:
Construct the sample space precisely
Before computing any probability, enumerate what the full outcome space is. Many mistakes come from treating overlapping or incomplete sample spaces as valid.
Identify the correct conditioning event
Conditional probability is where most candidates fail. Be explicit about what you're conditioning on and why the conditional sample space changes.
Communicate your reasoning step-by-step
Interviewers at HFT firms care as much about your reasoning process as the answer itself. Never go silent, narrate the logic as you build it.