DRW's Quant/QR intern OA keeps its signature "hardcore" style: large question count, tight time, low error tolerance. Beyond conventional programming, math reasoning, probability and statistics, and classic brainteasers remain the focus. Many candidates stumble the first time by mis-allocating time or getting stuck on a single brainteaser and losing points on details. This breakdown is organized by question type so you can hit every point in 45 minutes instead of answering blindly.
DRW's business centers on three areas: Liquidity Providing, Risk Taking, and Latency Sensitive Trading, and the OA leans toward "fast quantitative judgment." The whole OA is 6 math questions in 45 minutes at a brisk pace. Here is the breakdown.
1. Expected Payoff of a Biased Coin
Heads probability 0.8, tails 0.2. If you pick "heads," each flip lands heads with probability 0.8. Over 100 flips the expected number of heads is 100 × 0.8 = 80, and each heads pays $80.
The key is linearity of expectation: total expectation = per-trial expectation × count. Don't let "consecutive/independent" wording lure you into conditional probability. Expected payoff = 80 × $80 = $6400 (per the problem's payout convention).
2. Projection L2 Norm onto a Matrix Null Space
First find the null space of matrix A, then project vector y onto that space, and finally take the L2 norm of the projection.
Approach: do an SVD of A or solve Ax = 0 to get an orthonormal basis {u_i} of the null space; the projection is Σ (yᵀuᵢ) uᵢ, and its norm is sqrt(Σ (yᵀuᵢ)²). In code, scipy.linalg.null_space(A) gives the orthonormal basis directly, saving hand computation.
import numpy as np
from scipy.linalg import null_space
def proj_norm(A, y):
N = null_space(A) # columns are the null-space orthonormal basis
coeffs = N.T @ y # projection coefficients along each basis vector
proj = N @ coeffs
return np.linalg.norm(proj)
3. Quantization-Error Minimization (Numerical Solve)
Find a constant c that minimizes the quantization error. Error definition: for x ≥ 0 compute (x − c)², for x < 0 compute (x + c)², with x following a normal distribution of mean 0 and variance 3. Just write the objective and hand it to a numerical optimizer:
import numpy as np
from scipy.optimize import minimize
from scipy.stats import norm
def quantization_error(c):
mean, var = 0, 3
std = var ** 0.5
pos = lambda x: (x - c) ** 2 * norm.pdf(x, loc=mean, scale=std)
neg = lambda x: (x + c) ** 2 * norm.pdf(x, loc=mean, scale=std)
ip = norm.expect(pos, loc=mean, scale=std, lb=0, ub=np.inf)
ineg = norm.expect(neg, loc=mean, scale=std, lb=-np.inf, ub=0)
return ip + ineg
c_opt = round(minimize(quantization_error, x0=0.5).x[0], 3)
Don't force the integral by hand for these — using scipy.optimize + norm.expect is steadier under the clock.
4. Expected Tosses to Collect All 7 Faces
The probability of tossing a face you have already seen changes as the collected set grows. Let E be the total expected tosses to collect all 7 — this is the classic Coupon Collector:
E = 7 × (1/7 + 1/6 + 1/5 + ... + 1/1) = 7 × H₇
where H₇ is the 7th harmonic number. The problem's "after seeing 6 distinct faces, the probability of tossing a new face is 6/7" actually corresponds to the per-step expectation of the 7th stage = 7/(7−6) = 7. Sum the geometric-distribution expectation of each stage.
5. Coin-Walk Staircase Reach Probability DP
Each toss moves you 1 or 2 steps forward depending on heads/tails. Find Pₙ = "the probability of landing exactly on step n," then P₄ and P₁₀, and finally 1000 × (P₄ + P₁₀).
Recurrence: P[n] = 0.5 * P[n-1] + 0.5 * P[n-2], with boundaries P[0] = 1 and P[1] = 0.5 (step 1 is reachable only by a single 1-step move from 0).
def reach_prob(n, p=0.5):
P = [0.0] * (n + 1)
P[0] = 1.0
if n >= 1:
P[1] = p
for i in range(2, n + 1):
P[i] = p * P[i - 1] + (1 - p) * P[i - 2]
return P[n]
ans = round(1000 * (reach_prob(4) + reach_prob(10)), 3)
"Land exactly on step n" means you cannot jump over it, so each step's probability transitions from the previous two — don't confuse it with "reach ≥ n."
6. Time-Allocation Mindset
45 minutes for 6 questions = 7.5 minutes average, but difficulty is uneven. Suggested:
| Phase | Action |
|---|---|
| First 3 min | Read all questions, tag each "mental math" / "needs code" / "skip first" |
| Mid | Harvest the expectation/probability mental-math ones first (coin, Coupon Collector) |
| Late | Hand matrix projection and quantization error to scipy numerical solvers |
| End | Leave time for the DP question to verify boundaries; don't grind on a single brainteaser |
7. Summary
The core of the DRW OA is not whether you know it, but whether you can hit every point under the clock. Stock three toolkits — probability/expectation (linearity, geometric distribution, Coupon Collector), linear algebra (null space, projection, norm), and numerical optimization (scipy) — then drill time allocation and skip strategy, and even a hardcore OA becomes a steady score.
FAQ
Q1: What does the DRW intern OA test?
6 questions in 45 minutes, dominated by math reasoning, probability and statistics, and brainteasers, plus a little coding. Large count, tight time, and low error tolerance are its hallmarks.
Q2: Do I need to force integrals and matrix decompositions by hand?
No. Most questions are faster and steadier with numerical solves via scipy.optimize / scipy.stats / scipy.linalg, leaving your time for thinking rather than hand computation.
Q3: How do I prepare for these probability brainteasers?
Get fluent with linearity of expectation, geometric distribution, Coupon Collector, and Markov reach probabilities so you can recognize the pattern and apply the formula within 30 seconds.
Q4: I always run out of time — what do I do?
Read all questions first and classify them, harvest the mental-math ones early, hand the hard ones to numerical solvers, and reserve time to verify DP boundaries. For timed DRW Quant OA practice and question-type prediction, reach out for the high-frequency questions and review materials for your target role.
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