How Wincent's process runs, as candidates and the firm describe it, and the questions candidates report, each with a worked solution. Free, no sign-up.
Official (Quant Research/Trading internship, Bratislava): four stages. An online maths test taken at home (probability, logic and maths, low-to-medium difficulty), a 45-minute quant interview (medium problems), a 90-minute quant interview (your background plus medium-hard problems), then an on-site round in Bratislava built around betting games.
Online test, as candidates describe it: about 100 minutes for 12 numeric-answer problems in two sections (6 easier, then 6 medium/hard), proctored, pen and paper only. Several mention needing roughly 7 of 10–12 to pass. Most also get a separate coding test (up to 4 hours, auto-graded; one London 2026 candidate had graph problems).
The first technical interview is reported as 30–45 minutes with about 3 probability or combinatorics questions, some Olympiad-style. Two candidates say it ended early once the first question went badly, so say your reasoning out loud from the first second and ask for clarification if a question is read quickly.
Reported interview themes: probability, Bayes, expected value and variance, combinatorics, Markov chains, and mental maths under time pressure. Several candidates say the questions feel like variants of the "Green Book" (Zhou, A Practical Guide to Quantitative Finance Interviews).
Wincent is a high-frequency crypto market maker. Interns describe morning betting games and an on-site game where "the stakes kept building up", so practise bet sizing and staying calm as stakes grow.
Try each one before opening the solution. Interviewers change the numbers, so learn the method.
Quant researcher · Interview · Probability · reported by a candidate
Table tennis: the score is 10–10, so you need to win by two points. You win each point with probability 40%, independently.
(a) What is the probability that you win the game?
Answer and worked solution
Answer: 4/13 ≈ 0.30769
Key idea. At 10–10 you need two points in a row; a split returns you to 10–10. So look at pairs of points.
Win–win: 0.4² = 0.16, you win. Lose–lose: 0.6² = 0.36, you lose. Split: 0.48, back to deuce.
Ignore the splits: P(win) = 0.16/(0.16 + 0.36) = 16/52.
= 4/13 ≈ 0.308.
Check. Below 0.4: the deuce rule amplifies the stronger player's edge, and here that's your opponent.
(b) Follow-up (ours): how many more points are played, on average?
Answer and worked solution
Answer: 50/13 ≈ 3.8462
Key idea. Each pair of points ends the game with a fixed probability, so the number of pairs is geometric.
A pair ends the game with probability 0.16 + 0.36 = 0.52.
Expected pairs = 1/0.52.
Expected points = 2/0.52 ≈ 3.85.
Check. At least 2 points are always needed, and long rallies of splits are fairly rare.
Interviewers may push further: What if you win each point on your own serve with probability 0.5 and on theirs with 0.3, alternating serves every point? At what point-win probability is your chance of winning from deuce exactly 1/3?
Quant trading intern · Interview · Expected value · reported by a candidate
Threes: roll five dice. Threes count as zero; other faces count at face value; you want the lowest total. After each roll you must keep at least one die (kept dice add to your score) and re-roll the rest. Play optimally.
A candidate reports being asked "a very tricky variant" of the game of Threes from Matthew Conroy's A Collection of Dice Problems (problem 56). The variant itself isn't public, so practise the original until the backward-induction method is automatic.
(a) With one die left to roll, what is your expected score from it?
Answer and worked solution
Answer: 3
Key idea. With one die you just take whatever comes up.
Face values: 1, 2, 0 (a three), 4, 5, 6.
Average: 18/6 = 3.
(b) With two dice left, what is your expected score under optimal play?
Answer and worked solution
Answer: 79/18 ≈ 4.3889
Key idea. You must keep the lower die. Keep the other too only if it beats re-rolling, which is worth 3.
Keep the higher die only if it shows 0 (a three), 1 or 2; otherwise re-roll it for an expected 3.
Your score is min + min(other, 3), with threes counted as 0.
Averaging over all 36 rolls gives 158/36.
= 79/18 ≈ 4.39.
Check. Always keeping both would average 6; the re-roll option saves a lot.
(c) Follow-up (from Conroy's solution): the expected score of the full five-dice game under optimal play?
Answer and worked solution
Answer: 6.254
Key idea. Backward induction: the value with n dice uses the values with fewer dice.
E₁ = 3, E₂ = 79/18 ≈ 4.389.
With n dice: keep the lowest, then keep the next-lowest dice only while their total beats the value of re-rolling the rest.
This gives E₃ ≈ 5.234 and E₄ ≈ 5.834.
And E₅ ≈ 6.254.
Check. Each extra die adds less and less, because you only keep cheap dice.
Trap. Keeping every 2. With three dice, a 2 is worth keeping next to another 2 but not next to a 4.
Interviewers may push further: Why must the optimal strategy keep the lowest dice first? How does the answer change if you may keep zero dice on a roll (but still at most five rolls)?
Quant trader · Interview · Optimal stopping · reported by a candidate
A hat holds 100 one-dollar bills, 10 ten-dollar bills and one hundred-dollar bill. Bills are drawn one at a time at random, without replacement, and you see each one. At any moment (including before the first draw) you may say "stop": the next bill drawn is yours.
With the best strategy, what is the game worth (in dollars)?
Answer and worked solution
Answer: 100/37 ≈ 2.7027
Key idea. The amount you'd win by stopping now is the average of the bills left. That average is a martingale.
If you stop, the next bill is a random one of those left, so you expect the current average.
After a draw, the expected new average equals the current average (whatever comes out, the rest reweights exactly).
So no stopping rule can change your expected winnings: every strategy is worth the same as stopping immediately.
Check. Try "wait for the $100 to come out, then stop": sometimes the average rises, sometimes it falls, and it balances exactly.
Trap. Waiting for small bills to clear "to improve the odds". It improves some paths and ruins others equally.
Interviewers may push further: Does the answer change if you must stop before the hat is empty? (No.) What if you can see the bills but the payout is the bill drawn two after you stop?
Quant (online test topic) · Online test · Probability · practice question on a reported topic
A cell, each minute, dies with probability 1/4, stays a single cell with probability 1/4, or splits into two with probability 1/2. Every cell behaves independently. You start with one cell.
Practice on a topic a prep site lists for Wincent's online test. The wording is ours.
What is the probability that the population eventually dies out?
Answer and worked solution
Answer: 0.5
Key idea. Condition on the first minute: the line dies out only if every child's line dies out, independently.
Quant trading intern · Online test · Probability · reported by a candidate
500 ants are dropped at uniformly random points on a 1-metre stick, each facing left or right at random, walking at 1 cm/s. When two ants meet, both turn around. Ants fall off at the ends.
Candidates mention "500 ants falling off a 1 m stick" and "the ant collision problem from the Green Book" in Wincent's online test. The exact wording isn't public.
What is the expected time, in seconds, until the last ant falls off?
Answer and worked solution
Answer: 99.8
Key idea. Bouncing ants are indistinguishable from ants that pass through each other, which just swaps their labels.
So each "ghost" ant walks straight off the end it faces.
Its time is its distance to that end ÷ 1 cm/s: uniform on 0–100 s.
The last ant falls at the maximum of 500 independent uniforms on [0, 100].
E[max of n uniforms on [0, 1]] = n/(n + 1): 100 × 500/501 ≈ 99.8 s.
Check. With many ants one is almost certainly near an end facing inward, so the answer is close to 100 s.
Trap. Trying to track collisions. They don't change when ants fall off.
Interviewers may push further: What is the expected number of collisions? Which ant falls off last, and does it depend on the directions?
Quant (interview topic) · Interview · Bayes · practice question on a reported topic
20% of traders are skilled. A skilled trader makes money on a given day with probability 70%, an unskilled one with probability 50%. A trader made money on each of their first 3 days.
Our question on a theme candidates report from Wincent interviews (Bayesian updating).
What is the probability that they are skilled?
Answer and worked solution
Quant trading (on-site betting games) · On-site · Bet sizing · practice question on a reported topic
You may bet any fraction of your bankroll, again and again, at even money on a coin that lands heads 60% of the time.
Wincent's on-site round is built around betting games. This is our warm-up for that format, not a reported game.
(a) What fraction of your bankroll should you bet each time to maximise long-run growth?
Answer and worked solution
Answer: 0.2
Key idea. Kelly maximises the expected log of your bankroll. For a bet paying b to 1 with win probability p: f* = p − q/b.
Even money: b = 1.
f* = 0.6 − 0.4 = 0.2 (20% of your bankroll).
Check. Equivalently: f* = edge/odds = 0.2/1.
(b) What is the expected log-growth per bet at that fraction?
Answer and worked solution
Answer: 0.020136
Key idea. Each bet multiplies your bankroll by 1 + f or 1 − f.
Expected log-growth = 0.6 ln(1.2) + 0.4 ln(0.8).
= 0.6 × 0.1823 − 0.4 × 0.2231 = 0.1094 − 0.0893.
≈ 0.0201 per bet, about 2%.
Check. Bet 40% (twice Kelly) and growth becomes slightly negative (−0.0024): a favourable bet that still shrinks your bankroll.
Interviewers may push further: The stakes double every round but the coin is the same. Does your fraction change? You are only 80% sure the coin is 60/40 (otherwise it is fair). What fraction now?