Da Vinci Derivatives interview questions

Graduate trader · Online test · Expected value · reported by a candidate

A dataset has 102 samples, 7 of which are outliers. You pick samples at random without replacement.

1. On average, how many samples do you pick until you find the third outlier?
   Answer and worked solution

   Answer: 38.625

   Key idea. Symmetry: the outliers split the normal samples into equal gaps on average.

   1. 7 outliers split 95 normal samples into 8 gaps of 95/8 on average.
   2. The third outlier comes after 3 gaps and 3 outliers.
   3. 3 × 95/8 + 3 = 3 × 103/8 = 38.625.

   Check. General formula: the k-th of K special items among N appears at position k(N + 1)/(K + 1): 3 × 103/8.

Interviewers may push further: And until the last outlier?

Source: Glassdoor: Da Vinci graduate trader interviews · Try it in the app →

Trader intern · Interview · Market making · reported by a candidate

Draw 10 cards with replacement (Ace = 1, …, King = 13). Make a market on how far their total is from 70, i.e. |total − 70|.

Cards count Ace = 1 up to King = 13, which is what makes 70 the centre.

1. (a) What is the standard deviation of the total?
   Answer and worked solution

   Answer: 11.832

   Key idea. Variance of one card, then add over 10 independent cards.

   1. Uniform on 1–13: variance (13² − 1)/12 = 14.
   2. 10 cards: 140.
   3. SD = √140 ≈ 11.8.
2. (b) What is the expected value of |total − 70| (your mid)?
   Answer and worked solution

   Answer: 9.4839

   Key idea. The sum is close to normal, centred at 70.

   1. For a normal, E|X − μ| = σ√(2/π) ≈ 0.8σ.
   2. 0.8 × 11.8 ≈ 9.4 (exactly 9.48).

   Check. Quote around it, e.g. 8.5 @ 10.5.

Source: Glassdoor · Try it in the app →

Graduate Trader · Interview · Statistics · reported by a candidate

A standard deck with Ace through King worth 1 to 13. You draw three cards.

The report doesn't say whether the three cards are replaced. It doesn't matter for the mean; it does for the spread.

1. (a) What is the expected total?
   Answer and worked solution

   Answer: 21

   Key idea. Linearity, with or without replacement.

   1. Each card averages 7: 3 × 7 = 21.
2. (b) Without replacement, what is the standard deviation of the total?
   Answer and worked solution

   Answer: 6.3524

   Key idea. Without replacement, apply the finite-population correction.

   1. With replacement: variance 3 × 14 = 42.
   2. Multiply by (N − n)/(N − 1) = 49/51: 40.35.
   3. SD ≈ 6.35.

   Check. Slightly less than √42 ≈ 6.48: drawing without replacement reduces variability.

Source: Glassdoor · Try it in the app →

Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

• Why Da Vinci, and why trading, using your past experience?
• What do you expect to struggle with most in your first year?

Source: Glassdoor · Try it in the app →

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

You roll a fair die repeatedly and keep a running total.

1. (a) What is the probability the running total ever equals exactly 3?
   Answer and worked solution

   Answer: 0.22685

   Key idea. Condition on the last roll: p(n) = (1/6)(p(n − 1) + … + p(n − 6)), with p(0) = 1.

   1. p(1) = 1/6, p(2) = (1 + 1/6)/6 = 7/36.
   2. p(3) = (1 + 1/6 + 7/36)/6 = 49/216 ≈ 0.227.

   Check. Pattern for n ≤ 6: p(n) = (7/6)ⁿ⁻¹/6.

2. (b) Exactly 4?
   Answer and worked solution

   Answer: 0.26466

   Key idea. One more step of the recursion.

   1. p(4) = (1 + 1/6 + 7/36 + 49/216)/6 = 343/1296 ≈ 0.265.
3. (c) Which total x is most likely to be hit?
   Answer and worked solution

   Answer: 6

   Key idea. Up to n = 6, the start (p(0) = 1) is among the six terms feeding p(n), so p keeps growing.

   1. p(6) = (7/6)⁵/6 = 16807/46656 ≈ 0.360.
   2. p(7) drops to ≈ 0.254 because p(0) leaves the sum, and later values settle around 2/7 ≈ 0.286.
   3. Most likely: 6.

   Check. Long run: you visit one total in every 3.5 on average, so p(n) → 1/3.5.

Interviewers may push further: What does the probability settle to for large x, and why? (1/3.5 = 2/7: on average one total in every 3.5 is visited.)

Source: Wall Street Oasis: Da Vinci interview reports · Try it in the app →

Graduate Quant Trader · Interview · Options · reported by a candidate

An at-the-money call option. Its implied volatility rises without limit.

1. What does its delta tend to?
   Answer and worked solution

   Answer: 1

   Key idea. Black–Scholes call delta is N(d₁), and at the money d₁ = (r/σ + σ/2)√T.

   1. As σ → ∞, d₁ → ∞.
   2. N(∞) = 1, so delta → 1.

   Check. Intuition: with enormous volatility the call is worth almost the whole stock (the strike barely matters), so it moves one-for-one with it.

   Trap. Answering 0.5 because "ATM delta is about a half". That's only for low volatility and short expiry.

Interviewers may push further: And the put's delta? (Tends to 0, by put–call parity.)

Source: Wall Street Oasis: Da Vinci interview reports · Try it in the app →

Quant Trading intern · Interview · Mental maths · reported by a candidate

Quick-fire mental maths.

1. 47 ÷ 13, to two decimal places
   Answer and worked solution

   Answer: 47/13 ≈ 3.6154

   Key idea. Whole part first, then the remainder.

   1. 13 × 3 = 39, remainder 8.
   2. 8/13 ≈ 0.615.
   3. ≈ 3.62.

   Check. 13 × 3.6 = 46.8, close to 47.

Source: Wall Street Oasis: Da Vinci interview reports · Try it in the app →