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Two Sigma interview questions

Representative quant-interview questions tagged to Two Sigma, concentrated in probability, linear algebra, and other. Preview a free set below, then drill the full bank with hints and worked solutions.

226 Two Sigma questions · 12 free to preview · 2,516 problems total

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FreeLinear algebraEasyTwo SigmaJane StreetCitadel

Three assets AA, BB, CC have correlations ρAB=0.9\rho_{AB}=0.9 and ρBC=0.8\rho_{BC}=0.8. Can ρAC=0.1\rho_{AC}=0.1?

Solution

A correlation matrix must be positive semidefinite (PSD). For three assets with pairwise correlations ρAB=0.9\rho_{AB}=0.9, ρBC=0.8\rho_{BC}=0.8, and a candidate ρAC=0.1\rho_{AC}=0.1, we check whether the 3×33\times3 correlation matrix

R=(10.90.10.910.80.10.81)R = \begin{pmatrix} 1 & 0.9 & 0.1 \\ 0.9 & 1 & 0.8 \\ 0.1 & 0.8 & 1 \end{pmatrix}

is PSD. A necessary and sufficient condition for a 3×33\times3 symmetric matrix with unit diagonal is that all principal minors are nonnegative. The 1×11\times1 and 2×22\times2 minors are clearly nonnegative, so the key condition is det(R)0\det(R) \ge 0.

The determinant of a 3×33\times3 correlation matrix is

det(R)=1+2ρABρBCρACρAB2ρBC2ρAC2.\det(R) = 1 + 2\rho_{AB}\rho_{BC}\rho_{AC} - \rho_{AB}^2 - \rho_{BC}^2 - \rho_{AC}^2.

Substituting the given values:

det(R)=1+2(0.9)(0.8)(0.1)0.920.820.12=1+0.1440.810.640.01=1.1441.46=0.316<0.\begin{aligned} \det(R) &= 1 + 2(0.9)(0.8)(0.1) - 0.9^2 - 0.8^2 - 0.1^2 \\ &= 1 + 0.144 - 0.81 - 0.64 - 0.01 \\ &= 1.144 - 1.46 = -0.316 < 0. \end{aligned}

Since the determinant is negative, RR is not positive semidefinite. Therefore ρAC=0.1\rho_{AC}=0.1 is not possible.

FreeCombinatoricsMediumTwo SigmaJane StreetSIG

Sandy has 55 pairs of socks in a drawer, each pair a distinct color. On Monday, she randomly picks two socks from the 1010 total; on Tuesday, she picks two from the remaining 88; on Wednesday, she picks two from the remaining 66. What is the probability that Wednesday is the first day she selects a matching pair?

Solution

Let MM, TT, WW be the events that Monday, Tuesday, Wednesday produce a matching pair, respectively. We want

P(McTcW)=P(Mc)P(TcMc)P(WMcTc).P(M^c \cap T^c \cap W) = P(M^c)\,P(T^c \mid M^c)\,P(W \mid M^c \cap T^c).

1. P(Mc)P(M^c). Total ways to choose 22 socks from 1010: (102)=45\binom{10}{2}=45. Matching pairs: 55 (one per color). So P(M)=5/45=1/9P(M)=5/45=1/9, hence

P(Mc)=119=89.P(M^c)=1-\frac19=\frac89.

2. P(TcMc)P(T^c \mid M^c). After a non‑matching Monday, two socks of different colors are removed. The remaining 88 socks consist of 33 colors with 22 socks each and 22 colors with 11 sock each. Total pairs from 88: (82)=28\binom{8}{2}=28. Only the three colors with 22 socks can produce a matching pair, so 33 matching pairs. Thus P(TMc)=3/28P(T\mid M^c)=3/28 and

P(TcMc)=1328=2528.P(T^c \mid M^c)=1-\frac{3}{28}=\frac{25}{28}.

3. P(WMcTc)P(W \mid M^c \cap T^c). After Monday and Tuesday are both non‑matching, the composition of the remaining 66 socks depends on how Tuesday's non‑matching pair was formed. Starting from the Monday state (33 colors with 22 socks, 22 colors with 11 sock), Tuesday's non‑matching pair can be of three types:

  • Case A: Both socks from the 22-sock colors. Number of ways: (32)22=12\binom{3}{2}\cdot2\cdot2 = 12. Resulting state: 11 color with 22 socks, 44 colors with 11 sock. Probability of a match on Wednesday: 1(62)=115\frac{1}{\binom{6}{2}} = \frac{1}{15}.
  • Case B: One sock from a 22-sock color and one from a 11-sock color. Number of ways: 3221=123\cdot2\cdot2\cdot1 = 12. Resulting state: 22 colors with 22 socks, 22 colors with 11 sock, 11 color with 00 socks. Probability of a match on Wednesday: 215\frac{2}{15}.
  • Case C: Both socks from the two 11-sock colors. Number of ways: 11. Resulting state: 33 colors with 22 socks, 22 colors with 00 socks. Probability of a match on Wednesday: 315=15\frac{3}{15} = \frac15.

Total non‑matching pairs on Tuesday: 12+12+1=2512+12+1 = 25, so the conditional probabilities of the cases given TcT^c are 1225,1225,125\frac{12}{25},\frac{12}{25},\frac{1}{25}. Hence

P(WMcTc)=1225115+1225215+125315=12+24+3375=39375=13125.P(W \mid M^c \cap T^c) = \frac{12}{25}\cdot\frac{1}{15} + \frac{12}{25}\cdot\frac{2}{15} + \frac{1}{25}\cdot\frac{3}{15} = \frac{12+24+3}{375} = \frac{39}{375} = \frac{13}{125}.

4. Multiply.

P(McTcW)=89252813125=82513928125=260031500=26315.P(M^c \cap T^c \cap W) = \frac{8}{9} \cdot \frac{25}{28} \cdot \frac{13}{125} = \frac{8\cdot25\cdot13}{9\cdot28\cdot125} = \frac{2600}{31500} = \frac{26}{315}.

Thus the desired probability is 26315\boxed{\frac{26}{315}}.

(One can verify by counting all sequences: total sequences (102)(82)(62)=18900\binom{10}{2}\binom{8}{2}\binom{6}{2}=18900, favorable sequences 52413=15605\cdot24\cdot13=1560, giving the same fraction.)

FreeProbabilityHardTwo SigmaJane StreetSIG

Two players share a fair coin and flip it repeatedly, recording the sequence of heads (HH) and tails (TT) that appears. The first player wins if HTHHTH occurs before HHTHHT; otherwise, the second player wins. What is the probability that the first player wins?

Solution

Idea

Track the game state as the longest suffix of the flip sequence that is a prefix of either target pattern. This yields a small Markov chain whose first-step equations determine the win probability exactly.

States and Transitions

Target patterns: Player 1 wins on HTHHTH; Player 2 wins on HHTHHT.

The transient states are {ε,H,HH,HT}\{\varepsilon,\, H,\, HH,\, HT\}, where ε\varepsilon denotes no useful suffix (start, or after a progress-resetting tail).

StateFlip HHFlip TT
ε\varepsilonHHε\varepsilon
HHHHHHHTHT
HHHHHHHHP2 wins
HTHTP1 winsε\varepsilon

Remark on HHHHHHH \xrightarrow{H} HH: after any run of heads, the longest suffix that prefixes a target is still HHHH (the length-2 prefix of HHTHHT).

System of Equations

Let psp_s denote the probability that Player 1 wins from state ss.

pε=12pH+12pε    pε=pH(1)p_{\varepsilon} = \tfrac{1}{2}p_H + \tfrac{1}{2}p_{\varepsilon} \implies p_{\varepsilon} = p_H \tag{1} pH=12pHH+12pHT(2)p_H = \tfrac{1}{2}p_{HH} + \tfrac{1}{2}p_{HT} \tag{2} pHH=12pHH+120    pHH=0(3)p_{HH} = \tfrac{1}{2}p_{HH} + \tfrac{1}{2}\cdot 0 \implies p_{HH} = 0 \tag{3} pHT=121+12pε(4)p_{HT} = \tfrac{1}{2}\cdot 1 + \tfrac{1}{2}\,p_{\varepsilon} \tag{4}

Equation (3) reflects that HHHH is a trap: every additional HH keeps the game in HHHH, and the inevitable first TT completes HHTHHT, so Player 1 cannot win from HHHH.

Solution

Substituting (3)(3) into (2)(2):

pH=12pHT.p_H = \tfrac{1}{2}p_{HT}.

Combined with (1)(1), we have pε=pH=12pHTp_{\varepsilon} = p_H = \tfrac{1}{2}p_{HT}. Substituting into (4)(4):

pHT=12+1212pHT=12+14pHT.p_{HT} = \frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}p_{HT} = \frac{1}{2} + \frac{1}{4}p_{HT}. 34pHT=12    pHT=23.\frac{3}{4}p_{HT} = \frac{1}{2} \implies p_{HT} = \frac{2}{3}. pε=1223=13.p_{\varepsilon} = \frac{1}{2}\cdot\frac{2}{3} = \boxed{\dfrac{1}{3}}.

The game begins in state ε\varepsilon, so Player 1 wins with probability 13\dfrac{1}{3}.

ProbabilityEasyTwo SigmaJane StreetSIG

14 slips numbered 1141-14 are placed in a random order. A position ii is called a local maximum if the slip there is strictly greater than each of its immediate neighbors. Find the expected number of local maxima. For example, with 6 numbers the arrangement 513246 has local maxima at positions 1, 3, and 6, giving 3 local maxima.

Approach

Write the total number of local maxima as a sum of indicator random variables, one for each position.

ProbabilityEasyTwo SigmaJane StreetCitadel

Michael rides a remote-control skateboard around campus. The front of the Hopkins sign is the origin (0,0)(0,0); rightward is positive xx and into campus (upward) is positive yy. Every second he picks an angle uniformly from [0,2π)[0, 2\pi) and moves 1 foot in that direction from his current position. After 1616 seconds, what is the expected squared distance from the Hopkins sign?

Approach

Write the squared distance as the squared norm of a sum of random unit vectors and expand the square.

ProbabilityEasyTwo SigmaJane StreetCitadel

There are NN employees, each driving a separate car to QuantEssential. The cars are initially well-spaced and travel at distinct speeds assigned uniformly at random. Whenever a faster car catches up to a slower one, it adopts the slower car's speed. After a long time, the cars form KK clusters, each moving at a distinct speed. Find the expected value of KK when N=10N = 10.

Approach

Consider the speeds of the cars in order from front to back and think about which cars become cluster leaders.

ProbabilityEasyTwo SigmaJane StreetCitadel

Marbles are drawn without replacement from a bag containing 50 red and 50 blue marbles until the bag is empty, and the sequence of colors is recorded. A run is a maximal block of consecutive marbles of the same color; for instance, RBBRRRBRRRBBRRRBRR has 5 runs. Find the expected number of runs in the full sequence.

Approach

Express the number of runs as 1 plus the number of times consecutive draws differ.

CombinatoricsEasyTwo SigmaJane StreetSIG

In a single-elimination tournament with 2n2^n strictly ranked teams (higher always beats lower), the bracket is drawn uniformly at random. Find the probability that the top-ranked and second-ranked teams play each other in the final.

Verification. For a tournament with n = 2 (i.e., 4 teams total), what is the probability that the top two teams meet in the final?

Approach

Since the top-ranked team beats everyone, it always reaches the final. What must be true about the bracket placement of the second-ranked team for it to also reach the final?

ProbabilityMediumTwo SigmaJane StreetCitadel

There are nn identical urns, each containing white and black balls. The iith urn (1in1 \le i \le n) holds 11 white ball and 2i12^i - 1 black balls. An urn is chosen uniformly at random, and a ball is drawn from it; the ball is white. After replacing that ball, a second ball is drawn from the same urn. Let p(n)p(n) denote the probability that the second ball is also white. Compute limnp(n)\displaystyle \lim_{n \to \infty} p(n).

Approach

Express the desired conditional probability using the law of total probability over the choice of urn.

Linear algebraEasyTwo SigmaCubistDE Shaw

Consider a linear regression on a dataset yielding coefficients β^OLS\hat{\beta}_{\text{OLS}}. With data matrix XX and IID normal errors of variance σ2\sigma^2, we have Var(β^OLS)=σ2(XTX)1\operatorname{Var}(\hat{\beta}_{\text{OLS}}) = \sigma^2 (X^T X)^{-1}. If the regression is rerun on a dataset where every point's values are doubled, producing coefficients β^OLS\hat{\beta}_{\text{OLS}}', find the constant cc such that Var(β^OLS)=cVar(β^OLS)\operatorname{Var}(\hat{\beta}_{\text{OLS}}') = c \operatorname{Var}(\hat{\beta}_{\text{OLS}}). If no such constant exists, output 1-1.

Approach

Consider how the data matrix $X$ and the response vector $y$ change when every point's values are doubled.

Linear algebraEasyTwo SigmaCitadelDE Shaw

Let xn=[1,2,,n]TRnx_n = [1, 2, \dots, n]^T \in \mathbb{R}^n and define An=xnxnTA_n = x_n x_n^T. For any fixed nn, AnA_n has exactly one non-zero eigenvalue, denoted λn\lambda_n. Determine the constant kk such that λn/nk\lambda_n / n^k converges to a finite, non-zero limit.

Approach

Recognize that $A_n$ is a rank‑1 matrix, so its only non‑zero eigenvalue equals its trace.

Other firms & topics

Frequently asked questions

How many Two Sigma interview questions are on QuantGrind?

226 questions are tagged to Two Sigma across our 2,516-problem set, concentrated in probability, linear algebra, and other. 12 are free to preview on this page; the full set unlocks with a membership, each with hints, the accepted answer, and a worked solution.

How hard is the Two Sigma quant interview?

Expect a phone screen heavy on mental math and probability, then an onsite or final round mixing brainteasers, market-making games, and topic depth. The bar is less about exotic tricks and more about speed, accuracy, and explaining your thinking clearly under time pressure.

How should I prepare for the Two Sigma quant interview?

Practice timed, out loud, and without a calculator. Rebuild each answer from first principles rather than recognizing it, and review the ones you got slowly — interview signal comes from how fast and cleanly you reason, not whether you've seen the exact prompt before.

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