Direct EV Calculation

Easy2 hrsExpected ValueGamesPayoffs

Overview

Direct EV is the foundation. The key insight is that expected value is a probability-weighted average of all outcomes. For interview games: (1) list outcomes, (2) find each probability, (3) multiply and sum. The hardest part is being exhaustive — missing an outcome is the most common error. The philosophical point: EV tells you what to do in the long run, not necessarily in a single trial.

How to Recognize

→'What is the expected value/return/payoff of this game?'
→Asked whether a bet is 'fair' or whether to play
→Payoff matrix is given or easily enumerable
→Simple probability × outcome problems

Step-by-Step Approach

1.List ALL outcomes and their probabilities — be exhaustive
2.E[X] = Σ xᵢ·P(X=xᵢ) — weight each outcome by its probability
3.Check: all probabilities must sum to 1
4.If EV > 0 and you can play repeatedly, the rational choice is to play
5.Ask: is the game being offered at its fair price? EV > cost → take it

Key Formulas

E[X] = \sum_i x_i \cdot P(X=x_i)

(discrete)

E[X] = \int x\,f(x)\,dx

(continuous)

E[aX+b] = a\,E[X] + b

(linearity)

E[X+Y] = E[X] + E[Y]

(ALWAYS — even if X, Y are dependent)

Worked Examples

0Quick Example

Roll a die, win

n if n≥4, pay

1 if n≤3. E = (4+5+6)/6 − 3·(1/6) = 15/6 − 3/6 = $2. Play it.

Problem

A game costs $3 to play. Roll a fair die: win$ 10 if you roll a 6, win $2 if you roll a 4 or 5, win$ 0 otherwise. Should you play?

Solution

List outcomes and calculate their expected payout:

Roll 6 (P=1/6): win $10. Roll 4 or 5 (P=2/6): win$ 2. Roll 1–3 (P=3/6): win $0.

E[payout] = (1/6)·10 + (2/6)·2 + (3/6)·0 = 10/6 + 4/6 + 0 = 14/6 ≈ $2.33.

Cost to play = $3. Net EV =$ 2.33 − $3 = −$ 0.67 per game.

Don't play — negative EV.

Answer

$E[\text{payout}] = \$ 2.33 < \ $3$ cost. This is a losing game. Negative EV means you'd lose $0.67 per play on average.

Common Mistakes

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Not checking that probabilities sum to 1 before computing EV.
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Confusing EV with the most likely outcome. If EV = $3.50, no single die roll gives exactly$ 3.50.
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Forgetting that E[X+Y] = E[X] + E[Y] even when X and Y are correlated. This is the unconditional linearity property.
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Not accounting for the cost to play when computing whether a game is profitable.

Practice Problems

Click "Show Answer" to reveal

You flip 3 fair coins. Win $8 if all heads, lose$ 1 otherwise. What is the EV?

Hint: P(3 heads) = 1/8. P(otherwise) = 7/8.

A raffle: 1000 tickets at $5 each. First prize$ 2000, second prize $500, third prize$ 100. What is the expected profit/loss from one ticket?

Hint: E[prize] = sum of (prize × probability). Net = E[prize] − ticket cost.

You pay $10 to play. Draw 2 cards without replacement. If both are aces, you win$ 100. Otherwise you win $0. Should you play?

Hint: P(2 aces) = C(4,2)/C(52,2).

Back to Study Plan

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