Expected value & games: interview questions and key results
Linearity, optimal stopping, fair prices. The results to know for quant trading interviews, then a worked example of each of the 7 question types. Free, no sign-up.
Games with choices: solve backwards from the last decision.
A geometric waiting time with success probability p has mean 1/p.
Worked examples (7)
Try each one before opening the solution. In the app, every type generates fresh numbers each time.
Reroll games (backward induction)
You roll a fair 8-sided die and are paid its face value in dollars. After seeing the roll you may reroll once (and must accept the second roll). What is the expected payoff under optimal play?
Answer and worked solution
Answer: 5.5
A reroll is worth 4.5. Keep any roll above 4.5 (i.e. 5–8) and reroll the other 4 faces. EV = (1/8)·[(5 + … + 8) + 4·4.5] = 5.5. This is backward induction, the same logic you use to price American options.
Expected maximum (tail sum)
You roll 4 fair dice. What is the expected value of the maximum?
Answer and worked solution
Answer: 5.2446
P(max ≤ m) = (m/6)4. Using the tail-sum formula E[X] = Σ P(X ≥ m) = Σm=1..6 [1 − ((m−1)/6)4] ≈ 5.2446.
Coupon collector
You roll a fair 4-sided die until every face has appeared at least once. What is the expected number of rolls?
Answer and worked solution
Answer: 25/3 ≈ 8.3333
After i distinct faces, the wait for a new one is geometric with success probability (4−i)/4, so mean 4/(4−i). Summing gives 4(1 + ½ + … + 1/4) = 4·H4 ≈ 8.3333. (This is the coupon collector problem.)
Indicator variables
You roll a fair die 12 times. What is the expected number of distinct faces that appear?
Answer and worked solution
Answer: 5.3271
Linearity of expectation with indicators: face j appears with probability 1 − (5/6)12. Six faces: 6·(1 − (5/6)12) ≈ 5.3271.
Runs of heads
You flip a fair coin until you get 3 heads in a row. What is the expected number of flips?
Answer and worked solution
Answer: 14
Let Ek be the expected flips for k in a row. You need k−1 in a row, then one more flip, which fails half the time and sends you back to the start: Ek = Ek−1 + 1 + ½Ek ⇒ Ek = 2Ek−1 + 2. With E0 = 0, Ek = 2k+1 − 2 = 14.
Convex payoffs
You roll a fair 6-sided die and are paid the square of the face in dollars. What is the fair price of this game?
Answer and worked solution
Answer: 91/6 ≈ 15.167
E[X²] = (1² + … + 6²)/6 = (6+1)(2·6+1)/6 = 15.167. This is more than (E[X])² = 12.25; the gap is Var(X). Convex payoffs are worth more with more variance, which is why options have value.
Broken stick
A stick of length 20 is broken at a uniformly random point. What is the expected length of the longer piece?
Answer and worked solution
Answer: 15
If the break is at U ~ U(0,20), the longer piece is max(U, 20−U), which is uniform on [10, 20]. Its mean is ¾·20 = 15.