← Jane Street Interview Insights
I jumped straight to the independence formula and got a number, then the interviewer asked me to slow down and think about what the actual bounds were.
Use the inclusion-exclusion principle to express the probability of at least one pizza day as P(Sat) + P(Sun) - P(Sat and Sun). Recognize that the only unknown is the joint probability, which is constrained by the marginal probabilities. Then find the minimum and maximum possible values of the joint probability and construct dependence structures that achieve them.
Pro tip: Emphasize that the range is determined by the Fréchet–Hoeffding bounds, and explicitly describe the dependence structures (e.g., mutually exclusive events for the minimum, and one event implying the other for the maximum). This shows you understand both the math and the practical implications of dependence.
Let A be the event of eating pizza on Saturday, B on Sunday. We want P(A ∪ B). Use inclusion-exclusion: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.4 + 0.3 - P(A ∩ B) = 0.7 - P(A ∩ B).
The joint probability P(A ∩ B) must satisfy max(0, P(A)+P(B)-1) ≤ P(A ∩ B) ≤ min(P(A), P(B)). Here, max(0, 0.7-1) = 0, and min(0.4, 0.3) = 0.3. So 0 ≤ P(A ∩ B) ≤ 0.3.
Substitute the bounds into the inclusion-exclusion formula: minimum P(A ∪ B) = 0.7 - 0.3 = 0.4; maximum P(A ∪ B) = 0.7 - 0 = 0.7. Thus the range is [0.4, 0.7].
For the minimum (0.4), make A and B mutually exclusive (P(A ∩ B)=0). For the maximum (0.7), make B a subset of A (P(A ∩ B)=0.3), so that Sunday pizza always implies Saturday pizza.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The constraint 3g + s = 900 pins s once you pick g, so you're really maximizing a product of two binomial coefficients.
Model the total score as a sum of independent random variables: gold heads ~ Binomial(250, 0.5) and silver heads ~ Binomial(250, 0.5). The total score S = 3G + H. Given S = 900, find the conditional distribution of G (or H) and identify the mode. Use the fact that G and H are independent and compute P(G=g, H=h | S=900) ∝ P(G=g)P(H=h) with 3g + h = 900.
Pro tip: Leverage the symmetry and the fact that the conditional distribution is approximately normal; the mode occurs where the quadratic exponent is maximized, which can be found by solving a simple equation. Also, note that the most likely combination is not necessarily the one that maximizes the individual probabilities without conditioning.
Let G be the number of gold heads and H be the number of silver heads. Then G ~ Binomial(250, 0.5) and H ~ Binomial(250, 0.5), independent. The total score is S = 3G + H.
We want to find (g, h) that maximizes P(G=g, H=h | S=900). Since S=900 implies h = 900 - 3g, we need to maximize P(G=g) * P(H=900-3g) over integer g such that 0 ≤ g ≤ 250 and 0 ≤ 900-3g ≤ 250.
Use normal approximations: G ≈ N(125, 62.5) and H ≈ N(125, 62.5). Then the log-likelihood is proportional to -(g-125)^2/(2*62.5) - (h-125)^2/(2*62.5) with h=900-3g. Maximize this quadratic.
Differentiate the quadratic with respect to g and set to zero: -(g-125)/62.5 - 3*(h-125)/62.5 * (-3?) Wait, careful: h = 900 - 3g, so dh/dg = -3. The derivative of the log-likelihood is -(g-125)/62.5 + 3*(h-125)/62.5 = 0. Solve for g, then h = 900 - 3g.
The solution may not be integer; check nearby integers. Also ensure g and h are within [0,250]. The most likely combination is the one that maximizes the product of binomial probabilities.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the ant's movement as a Markov chain on the cube's vertices, grouping states by distance from the start (0, 1, 2, 3). Set up equations for the expected hitting time from each distance and solve them to find the expected steps from the start to the opposite vertex.
Pro tip: Emphasize that the cube's symmetry allows you to collapse the state space to just four states, which is a powerful technique for simplifying complex random walks. Also, verify your answer by checking that the expected time from the opposite vertex is zero and that the equations are consistent.
Label the starting vertex as state 0, vertices adjacent to start as state 1, vertices adjacent to the opposite as state 2, and the opposite vertex as state 3. Note that the ant moves to a uniformly random adjacent vertex each second.
From state 0, the ant always moves to state 1. From state 1, it moves to state 0 with probability 1/3 and to state 2 with probability 2/3. From state 2, it moves to state 1 with probability 2/3 and to state 3 with probability 1/3. State 3 is absorbing.
Let E_i be the expected number of steps to reach state 3 from state i. Write equations: E_3 = 0, E_2 = 1 + (2/3)E_1 + (1/3)E_3, E_1 = 1 + (1/3)E_0 + (2/3)E_2, E_0 = 1 + E_1.
Substitute E_3 = 0 into E_2, then solve the linear system for E_0. The solution yields E_0 = 10 steps.
Check that the solution satisfies all equations and makes intuitive sense: the ant must pass through intermediate states, and the expected time is finite. The answer is 10 steps.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use linearity of expectation by defining indicator variables for each potential loop closure. Recognize that the process is equivalent to a random pairing of the 2N ends, and compute the probability that a given set of ends forms a loop, then sum over all possible loops.
Pro tip: Start with small N (e.g., N=1,2,3) to identify the pattern and verify your formula. This demonstrates systematic thinking and helps catch mistakes early.
Clarify that we randomly pair all 2N ends, and a loop is formed when a sequence of ties connects ends of the same lace or previously connected components.
For each possible loop, define an indicator variable that is 1 if that loop is formed. The total number of loops is the sum of these indicators.
Calculate the probability that a given set of ends forms a loop. For a loop involving k laces, the probability is 1/(2N-1) * 1/(2N-3) * ... * 1/(2k-1) times the number of ways to order the laces.
Sum the probabilities over all possible loops. Use combinatorial counting to account for the number of loops of each size and simplify the sum.
Evaluate the sum to obtain the expected number of loops, which is the Nth harmonic number H_N = 1 + 1/2 + ... + 1/N.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize that this is a conservation problem: the amount of red in the white bucket equals the amount of white in the red bucket after the two transfers. Therefore, both buckets experience the same magnitude of color change, just in opposite directions. Explain this using the principle of conservation of volume and concentration, avoiding unnecessary calculations.
Pro tip: Emphasize that the answer is independent of the volumes and mixing ratios, as long as the final volumes are equal. This demonstrates strong quantitative reasoning and the ability to generalize.
Identify the initial state: two equal-volume buckets, one pure white, one pure red. Note the sequence: transfer a quarter bucket of red to white, mix, then transfer the same volume of mixture back to red.
Let the initial volume of each bucket be V. After the first transfer, the white bucket has V + V/4 = 5V/4 of mixture, and the red bucket has 3V/4 of pure red.
The second transfer moves V/4 of the mixture from the white bucket back to the red bucket. This mixture contains both red and white in some proportion.
Since the final volumes are equal (V each), the amount of red in the white bucket must equal the amount of white in the red bucket. Thus, the color change is symmetric.
Both buckets undergo the same magnitude of color change: the white bucket gains red, the red bucket gains white, and the amounts are equal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.