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.

The Problems

The Unfair Coin

Problem

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.

The Monty Hall Problem — Extended

Problem

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.

Expected Flips: HH vs. HT

Problem

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.

Gambler's Ruin

Problem

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.

The Birthday Problem

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.

Russian Roulette: Spin or Don't Spin?

Problem

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.

The Two-Envelope Paradox

Problem

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.

The Secretary Problem (Optimal Stopping)

Problem

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.

Conditional Dice Probability

Problem

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.

The Random Walk and Recurrence

Problem

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.

Practice these 10 problems until the logic feels instinctive — then work backward to explain each solution clearly under real interview pressure. Ready to go deeper? Myntbit's quant prep courses cover expected value derivations, Markov chains, and interview-style probability drills used by Jane Street, Optiver, and Citadel candidates.