Old Mission interview questions

Junior Quant Trader · Interview · Expected value · reported by a candidate

You roll two fair dice.

The report doesn't give the dice; we use two fair six-sided dice.

1. What is the expected absolute difference between them?
   Answer and worked solution

   Answer: 35/18 ≈ 1.9444

   Key idea. List the differences and how often they occur.

   1. Difference d ≥ 1 occurs in 2(6 − d) of 36 rolls.
   2. Σ d × 2(6 − d) = 2(5 + 8 + 9 + 8 + 5) = 70.
   3. E = 70/36 = 35/18 ≈ 1.94.

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Quant Trading intern · Interview · Markov chains · reported by a candidate

You and I play chess repeatedly. Half of our games are draws; of the decisive ones, I win 2/3. The match ends when someone wins three games in a row.

We read "three consecutive games" strictly: a draw breaks a winning streak.

1. What is the probability that I win the match?
   Answer and worked solution

   Answer: 0.86869

   Key idea. A Markov chain on the current streak: who holds it and its length.

   1. Per game: I win 1/3, you win 1/6, draw 1/2 (which resets streaks).
   2. Equations: P₀ = (1/3)M₁ + (1/6)Y₁ + (1/2)P₀; M₁ = (1/3)M₂ + (1/6)Y₁ + (1/2)P₀; M₂ = 1/3 + (1/6)Y₁ + (1/2)P₀; Y₁ = (1/3)M₁ + (1/6)Y₂ + (1/2)P₀; Y₂ = (1/3)M₁ + (1/2)P₀.
   3. Solving gives P₀ = 86/99 ≈ 0.869.

   Check. I'm twice as likely to win a game, and needing three in a row amplifies that edge (roughly like 2³ : 1).

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Quant Trading intern · Interview · Probability · reported by a candidate

Four points are chosen independently and uniformly on the surface of a sphere.

1. What is the probability that the centre of the sphere lies inside the tetrahedron they form?
   Answer and worked solution

   Answer: 0.125

   Key idea. Pair each random point with its antipode: of the 16 tetrahedra you can form by picking one point from each pair, exactly 2 contain the centre.

   1. Choosing a random point on the sphere is the same as choosing a random antipodal pair and then one point of it with probability ½.
   2. For four pairs in general position, exactly one tetrahedron and its mirror image (all points swapped to their antipodes) contain the centre: 2 of the 16 choices.
   3. P = 2/16 = 1/8.

   Check. The 2D version gives 1/4 (three points on a circle); each extra dimension halves it here.

Interviewers may push further: The 2D version: three points on a circle and the triangle containing the centre?

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Trading intern · Interview · Expected value · reported by a candidate

Quick-fire questions from one trading-intern interview.

For (b) we assume one reroll, costing $1, and you keep the second roll. For (c), positive whole numbers where order matters.

1. (a) X is uniform on [0, 10]. What is its variance?
   Answer and worked solution

   Answer: 25/3 ≈ 8.3333

   Key idea. A uniform on [a, b] has variance (b − a)²/12.

   1. (10 − 0)²/12 = 100/12 ≈ 8.33.

   Check. The mean is 5; the standard deviation ≈ 2.89.

2. (b) A die pays $1 per dot. After seeing it you may reroll once for $1, keeping the second roll. What is the game worth?
   Answer and worked solution

   Answer: 23/6 ≈ 3.8333

   Key idea. A reroll costs $1 and averages 3.5, so it's worth 2.5. Keep anything above that.

   1. Keep 3, 4, 5, 6: (3 + 4 + 5 + 6)/6 = 3.
   2. Reroll 1 and 2: (2/6) × 2.5 = 5/6.
   3. Value = 3 + 5/6 = 23/6 ≈ 3.83.

   Trap. Rerolling a 2 or keeping a 2? Keep 3, reroll 2: 2 < 2.5 < 3.

3. (c) In how many ways can three positive whole numbers (in order) add up to 10?
   Answer and worked solution

   Answer: 36

   Key idea. Stars and bars.

   1. 10 stars in a row have 9 gaps between them.
   2. Choosing 2 gaps cuts them into 3 positive parts: C(9, 2) = 36.

   Check. For a total of n: C(n − 1, 2). For 11: 45.

Source: Wall Street Oasis: Old Mission interview reports · Try it in the app →

Quantitative Research intern · Interview · Probability · reported by a candidate

Three questions from one interview.

1. (a) What is the probability that a 5-card poker hand is four of a kind?
   Answer and worked solution

   Answer: 0.0002401

   Key idea. Count favourable hands over all hands.

   1. Choose the rank of the four (13 ways) and the fifth card (48 ways): 624 hands.
   2. All hands: C(52, 5) = 2,598,960.
   3. P = 624/2,598,960 = 1/4165 ≈ 0.00024.
2. (b) 56 × 67, quickly.
   Answer and worked solution

   Answer: 3752

   Key idea. Round one factor and correct.

   1. 56 × 70 = 3920; subtract 56 × 3 = 168.
   2. 3752.
3. (c) You want to draw a black card and then a red card. With a normal 52-card deck, what is the probability?
   Answer and worked solution

   Answer: 13/51 ≈ 0.2549

   Key idea. Multiply the two conditional probabilities.

   1. Black first: 26/52 = ½. Then red: 26/51.
   2. P = 13/51 ≈ 0.255.

   Check. With 104 cards: (52/104)(52/103) = 26/103 ≈ 0.252. The smaller deck is slightly better: drawing a black makes red relatively more likely, and that effect is stronger in a small deck.

   Trap. Assuming the deck size doesn't matter.

Interviewers may push further: Would a double deck of 104 cards be better or worse for this, and why?

Source: Wall Street Oasis: Old Mission interview reports · Try it in the app →