Expected Value: The Complete Guide for Quant Interviews and Trading

Expected value is the single most important concept in quantitative finance. Every trade is a bet on an expected outcome. Every position size is an EV-weighted decision. Every interview at a prop firm will test whether you can reason about expected value quickly and correctly.

This guide covers the concept from first principles, builds to Kelly criterion and practical applications, and ends with 20 classic EV problems you'll encounter in interviews.

What Expected Value Is (and What It Isn't)

Expected value is the probability-weighted average of all possible outcomes:

$$E[X] = \sum_ p_i \cdot x_i$$

For a continuous random variable: $E[X] = \int x \cdot f(x) , dx$

What it is:

• A mathematical prediction of the average outcome over many repetitions
• The long-run average of a random variable

What it isn't:

• A guarantee of what will happen next
• The "most likely" outcome (that's the mode)
• The outcome you'll see most often in a small sample

A classic confusion: You flip a coin. You win $3 on heads, lose $1 on tails. EV = 0.5 × $3 + 0.5 × (−$1) = +$1. You would NOT "expect" to win $1 on the next flip — you'll win $3 or lose $1. The $1 EV is what you average over many flips.

The Law of Large Numbers

The reason EV matters: as the number of independent trials grows, the sample average converges to the expected value. This is the Law of Large Numbers.

Over 1 flip of the coin above: you see $3 or −$1. Over 100 flips: your average is close to $1 but not exactly. Over 10,000 flips: your average is almost exactly $1.

A trader making positive-EV bets consistently over many trades will converge to their edge. A trader making negative-EV bets — no matter how good their process looks short-term — will eventually go broke. This is why casinos profit and gamblers don't.

Linearity of Expectation

The most powerful property of expected value:

E[X + Y] = E[X] + E[Y], regardless of whether X and Y are independent.

This seems obvious but has profound implications. It means:

1. Portfolio EV = sum of component EVs (even for correlated assets)
2. Expected value is additive — you can compute parts separately and add them

Classic application: What is the expected number of 6s when you roll 10 dice?

Brute force: compute every combination. Tedious.

Linearity: each die independently shows a 6 with probability 1/6. Let X_i = 1 if die i shows 6, else 0. Then X = X_1 + ... + X_10. E[X] = E[X_1] + ... + E[X_10] = 10 × (1/6) = 5/3 ≈ 1.67 sixes.

Computing EV: Worked Examples

Discrete Distributions

The Fair Die: Roll a fair 6-sided die. Win $X where X is the face value.

E[X] = 1/6 × (1 + 2 + 3 + 4 + 5 + 6) = 21/6 = $3.50

The Geometric Distribution: You flip a fair coin until heads. Let X = number of flips.

E[X] = 1/(1/2) = 2 flips (geometric distribution with p = 0.5)

Intuition: on average, you need 2 tries before success when each try has a 50% success rate.

The Negative Hypergeometric: A deck has 4 aces and 48 non-aces. Draw until you get an ace. What's E[cards drawn]?

By the elegant formula: E = (total + 1)/(successes + 1) = 53/5 = 10.6 cards

Continuous Distributions

Uniform on [0, 1]: E[X] = 1/2 (the midpoint, by symmetry)

Uniform on [a, b]: E[X] = (a + b)/2

Exponential with rate λ: E[X] = 1/λ

(If bus arrivals are exponential with rate 1 per 10 minutes, the expected wait is 10 minutes.)

The Kelly Criterion: Optimal Bet Sizing

Knowing the EV of a bet isn't enough — you also need to know how much to bet. The Kelly Criterion gives the optimal fraction of your bankroll to wager.

The Formula

For a binary bet that pays b:1 on a win (probability p) and loses 1 on a loss (probability q = 1−p):

$$f^* = \frac = p - \frac$$

Example: A coin flip pays 2:1 (win $2, lose $1). p = 0.5, q = 0.5, b = 2.

f* = (2 × 0.5 − 0.5) / 2 = 0.5 / 2 = 25%

Bet 25% of your bankroll each flip.

Why Kelly?

Kelly maximizes the expected logarithm of wealth, which maximizes the long-run growth rate of your bankroll.

Betting more than Kelly: decreases expected log-wealth. You'll grow faster in good streaks but the bad streaks hurt you more — and eventually, bad streaks wipe you out.

Betting less than Kelly: leaves growth on the table. Safer but slower.

In practice: Many traders use "half Kelly" (bet 50% of the Kelly fraction) to account for estimation error in p and b, and to reduce variance. Kelly is a ceiling, not a target.

The Multi-Asset Kelly

With multiple simultaneous bets, the solution requires finding the portfolio weights that maximize expected log-return. This is a convex optimization problem and the foundation of modern portfolio theory.

EV in Trading

Every trading decision involves an EV calculation, often implicit.

Bid-Ask Spread as EV

A market maker quotes $99.95/$100.05. If a customer buys at $100.05 (the ask), the market maker is short at $100.05. The market maker profits if the stock goes below $100.05.

If fair value is $100.00, the market maker's EV on this trade is:

EV = (half-spread) − (adverse selection loss)

Where adverse selection = probability the customer has better information × expected move against the market maker.

If 20% of buyers are informed traders who expect the stock to move to $100.30, the market maker's adverse selection loss is 0.20 × $0.25 = $0.05 per trade. Net EV = $0.05 (spread) − $0.05 (adverse selection) = $0.00 — a very tight market.

Options as EV Bets

Buying an option is paying the premium for the right to participate in favorable outcomes. The option is "worth buying" if: EV[payoff] > Premium paid.

For a call option with strike K and premium C on a stock at S₀:

EV[payoff] = E[max(S_T − K, 0)] > C implies it's a positive-EV purchase (under your belief about the distribution of S_T).

The option is priced using risk-neutral pricing, where the expected return of all assets equals the risk-free rate. In the real world, your EV depends on your view of the distribution of S_T.

The 20 Classic EV Problems

Group 1: Basic Expected Value

1. A fair die is rolled. You win the face value in dollars. What's the EV? EV = (1+2+3+4+5+6)/6 = $3.50

2. You roll two dice. You win the maximum of the two. What's the EV?

P(max = k) = P(at least one die = k) − P(at least one die = k−1) = k²/36 − (k−1)²/36 = (2k−1)/36.

EV = Σ k × (2k−1)/36 for k = 1 to 6 = (1×1 + 2×3 + 3×5 + 4×7 + 5×9 + 6×11)/36 = (1+6+15+28+45+66)/36 = 161/36 ≈ $4.47

3. You flip a coin. Heads: $2. Tails: −$1. What's the EV and Kelly fraction?

EV = 0.5(2) + 0.5(−1) = +$0.50 per flip. Kelly: f* = p − q/b = 0.5 − 0.5/2 = 25% of bankroll.

4. A bag has 3 red, 5 blue balls. Draw one. Win $10 for red, lose $4 for blue. Play?

EV = (3/8)(10) + (5/8)(−4) = 3.75 − 2.50 = +$1.25. Yes, play.

5. You can roll a die twice and take the higher value. What's the EV?

Same as problem 2: $4.47

(Alternatively: E[max of 2 dice] = E[die 1] + E[die 2] − E[min of 2 dice] = 3.5 + 3.5 − 161/36... actually this approach requires E[min], which is 2×3.5 − 161/36 = 7 − 161/36 = 252/36 − 161/36 = 91/36 ≈ 2.53. So max EV = 7 − 91/36 ≈ 161/36 ✓)

Group 2: Stopping Times

6. You flip a coin until you get heads. What's the expected number of flips?

Geometric with p = 0.5. E[flips] = 1/p = 2 flips.

7. What's the expected number of flips to get two consecutive heads?

Classic recursive setup:

• Let E = expected flips from start, A = expected flips after seeing one head.

E = 1/2(1+E) + 1/2(1+A) [tails: reset; heads: go to state A] A = 1/2(1) + 1/2(1+E) [heads: done; tails: reset]

A = 1/2 + 1/2 + E/2 = 1 + E/2 E = 1/2(1+E) + 1/2(1+1+E/2) = 1 + E/2 + E/4... solving: E = 6 flips

8. Expected number of flips to get HT (heads then tails)?

Let E = expected flips from start, A = expected flips after seeing one head.

E = 1/2(1+A) + 1/2(1+E) [heads: go to A; tails: restart... wait, tails from start means we're back to start]

Actually: from start, any flip transitions: H → state A; T → restart. E = 1 + 1/2 × A + 1/2 × E → E/2 = 1 + A/2 → E = 2 + A

From A: H → back to A (another head, still waiting for T); T → done. A = 1 + 1/2 × A + 1/2 × 0 → A/2 = 1 → A = 2

So E = 2 + 2 = 4 flips (compare: HH requires 6 flips)

9. Roll a die until the running sum exceeds 20. What's the expected total?

This is complex to solve analytically. The expected value of sum when you stop ≈ 20 + (mean die roll)/2 = 20 + 1.75 ≈ 21.75 (by overshoot approximation). Simulating gives ≈ 21.6.

Group 3: Conditional EV

10. You roll a die. I tell you it's even. What's the expected value?

Conditional on even: outcomes are 2, 4, 6 — each with prob 1/3. E[X | X even] = (2+4+6)/3 = 4

11. You draw 2 cards from a standard deck. Given the first is an ace, what's the expected value of the second? (Ace=1, other face cards=0, number cards = face value)

Given first is ace: remaining deck has 51 cards. Expected value of remaining card = (sum of all remaining card values)/51.

Sum of all 52 card values = 4×1 + 4×0 + 4×0 + 4×0 + 4×2 + 4×3 + ... + 4×10 = 4(1+0+0+0+2+3+4+5+6+7+8+9+10) = 4 × 55 = 220. Subtract one ace (1): 219. Expected value of second card = 219/51 ≈ 4.29

12. You roll two dice. Given the sum is 9, what's the expected value of the larger die?

Outcomes summing to 9: (3,6), (4,5), (5,4), (6,3). Larger die: 6, 5, 5, 6. Expected larger die = (6+5+5+6)/4 = 5.5

Group 4: Game Theory and Strategy

13. You can roll a die once and take the payoff, or roll again (and must take the second roll). What's your strategy and expected payoff?

If you see X: roll again iff E[second roll] = 3.5 > X, i.e., iff X ≤ 3.

E[payoff] = P(X≤3) × 3.5 + P(X≥4) × E[X|X≥4] = 0.5 × 3.5 + 0.5 × (4+5+6)/3 = 1.75 + 0.5×5 = $4.25

14. Same game but you can roll up to 3 times. Strategy and EV?

Work backward. After 2 rolls with one remaining: strategy is reroll iff X ≤ 3.5 → reroll if X ≤ 3, E₂ = $4.25.

After 1 roll with 2 remaining: reroll if X < E₂ = 4.25, i.e., if X ≤ 4.

E₁ = P(X≤4) × 4.25 + P(X≥5) × E[X|X≥5] = (4/6) × 4.25 + (2/6) × (5+6)/2 = (4/6)(4.25) + (2/6)(5.5) = 2.833 + 1.833 = $4.67

15. You're offered a bet: flip a fair coin, win $X if heads, lose $1 if tails. What's the minimum X that makes you play (assuming Kelly optimal)?

For Kelly fraction > 0, we need EV > 0: 0.5X − 0.5 > 0 → X > $1.

At exactly X = $1: EV = 0, Kelly fraction = 0 (don't bet). At X > $1: positive EV.

Group 5: Advanced Problems

16. The St. Petersburg Paradox: A game pays $2^n if a fair coin first lands heads on flip n. How much would you pay?

EV = Σ (1/2^n) × 2^n = Σ 1 = infinity.

But no one would pay infinity. Why? Utility functions are concave — the marginal utility of the 1,000,000th dollar is much less than the first. With log utility (Kelly): EV[log wealth] = finite, and the "fair price" is around $20–30 for typical wealth levels.

17. You have a 60% chance of winning a bet that doubles your money, 40% chance of losing everything. Should you play?

EV: 0.6 × 2M + 0.4 × 0 = 1.2M > M. Positive EV.

But: with 100% of your wealth, you have a 40% chance of ruin. Kelly criterion says: never bet everything when you can lose everything, because log(0) = −∞.

Kelly fraction: this isn't a standard Kelly setup (can't bet a fraction of wealth and get back proportional returns on "lose everything" bets). The lesson: EV alone doesn't give the right bet size — you need to account for the path.

18. You and an opponent each roll a die. Higher number wins. Ties re-roll. What's the expected number of rolls to determine a winner?

P(tie on one roll) = P(same number) = 6/36 = 1/6. P(winner determined) = 5/6.

This is a geometric process: expected rolls until no tie = 1/(5/6) = 6/5 = 1.2 rolls

19. You're market making on a coin flip. Fair value: 50%. What bid/ask should you quote?

Your spread should compensate for:

1. Adverse selection: if 20% of traders are informed and know the result, each trade with an informed trader costs you (0.2 × 50% = 10% of the bet) on average.
2. Inventory risk (for repeated games)

Minimum spread = 2 × adverse selection cost. If adverse selection costs 5 cents per $1 bet, quote 45/55 (10-cent spread). The exact calculation depends on your estimate of the informed trader fraction.

20. Kelly bet sizing with estimation error. You estimate p = 0.6 for a binary bet with b = 1 (win $1 or lose $1). But you're only 80% confident p is actually 0.6 (20% chance p = 0.5). What's the effective Kelly fraction?

Effective p = 0.8 × 0.6 + 0.2 × 0.5 = 0.48 + 0.10 = 0.58

Kelly: f* = 2p − 1 = 2(0.58) − 1 = 0.16 = 16% of bankroll.

Compare to betting at face value p = 0.6: Kelly = 0.20. Estimation error reduces the optimal bet by 20%. This is the intuition behind using "half Kelly" in practice.

Key Principles to Remember

EV is linear: E[aX + b] = aE[X] + b. Always.

Kelly maximizes log growth: bet Kelly fraction to maximize long-run wealth growth.

EV ≠ utility: for large bets relative to wealth, use log utility or risk-adjusted EV.

Stopping times: use recursive equations. E = 1 + (prob of reset) × E + ...

Conditioning: P(A|B) × P(B) = P(A and B). Draw a tree.

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