Comparing with Expected Value — I can compare choices (games, decisions) using expected value and justify a decision at my level. - Grade 11 Math

SECTION 1

Historical Context & Motivation

For most of human history, people made risky decisions based on gut feelings, superstition, or sheer luck. Gamblers tossed dice hoping fortune would favor them, and merchants wagered fortunes on trade voyages without any mathematical framework for evaluating whether the risk was worthwhile. The concept of expected value changed all of that by giving us a single number that captures the long-run average outcome of any decision involving chance.

1654 • The Problem of Points

Blaise Pascal and Pierre de Fermat exchanged letters about how to fairly divide the stakes of an interrupted gambling game. Their correspondence laid the foundation for probability theory and the idea of weighting outcomes by likelihood.

1657 • Huygens Formalizes Expected Value

Dutch mathematician Christiaan Huygens published the first treatise on probability, introducing the concept of expectatio — the expected value of a game — as a way to judge whether a wager is fair.

1713 • Bernoulli's Ars Conjectandi

Jakob Bernoulli's posthumous work proved the Law of Large Numbers, showing that observed frequencies converge to theoretical probabilities as trials increase — the mathematical backbone of expected value.

1944 • Game Theory & Decision Science

John von Neumann and Oskar Morgenstern published Theory of Games and Economic Behavior, extending expected value into strategic decision-making in economics, politics, and everyday life.

The central question expected value answers is deceptively simple: If I could repeat this decision thousands of times, which option would leave me better off on average? Whether you are choosing between carnival games, insurance plans, or investment strategies, expected value gives you a rational, mathematically grounded way to compare your options.

SECTION 2

Core Principles & Definitions

Before you can compare choices using expected value, you need to understand the building blocks. Every decision under uncertainty has outcomes (the possible results), probabilities (how likely each result is), and values (what each result is worth to you, often in dollars). Expected value combines all three into one number.

Outcome & Probability

List every possible result and assign each a probability between 0 and 1. All probabilities must sum to exactly 1, accounting for every scenario.

Value (Payoff)

Assign a numerical value to each outcome — gains are positive, losses are negative. Include any cost to play (entry fee) as a negative component.

Expected Value (EV)

Multiply each outcome's value by its probability and sum the products. The result is the long-run average gain or loss per trial.

Comparing Choices

Calculate the EV for each option. The option with the higher (or less negative) EV is the mathematically better choice over many repetitions.

Fair Game

A game is called fair when its EV equals zero — neither the player nor the house has a long-run advantage.

✦

KEY TAKEAWAY

Think of expected value like a GPS for decisions. You might hit traffic on any single trip (a bad outcome), but over hundreds of commutes the GPS route with the lowest average travel time will save you the most time overall. Expected value tells you which 'route' through a risky decision saves the most money (or loses the least) in the long run.

SECTION 3

Visual Explanation — Decision Tree

A decision tree is a powerful visual tool for comparing expected values. The tree starts at a decision node (your choice), branches into chance nodes (random outcomes), and ends at payoff leaves (dollar values). By working backward — multiplying payoffs by probabilities and summing — you can see at a glance which branch gives a higher expected value.

A decision tree comparing Game A (30 % chance of winning $10, 70 % chance of losing $3) with Game B (50-50 chance of winning $6 or losing $2). Game B has a higher expected value of $2.00 compared to Game A's $0.90.

In the diagram above, the square node on the left represents your decision: which game to play. Each circular node is a chance event controlled by probability, not by you. The rectangles on the right show payoffs. By multiplying each payoff by its probability and summing the products, you arrive at the expected value for each game. Game B's EV of $2.00 beats Game A's EV of $0.90, so a rational decision-maker would choose Game B when the goal is to maximize average winnings over many plays.

SECTION 4

Mathematical Framework

The formula for expected value is straightforward once you see it. For a random variable with n possible outcomes, you multiply each outcome's value by its probability and add them all together.

EXPECTED VALUE FORMULA

E(X) = x₁ · P(x₁) + x₂ · P(x₂) + … + xₙ · P(xₙ)

Where xᵢ is the value (payoff) of outcome i and P(xᵢ) is the probability that outcome i occurs. All probabilities must satisfy P(x₁) + P(x₂) + … + P(xₙ) = 1.

SUMMATION NOTATION

E(X) = Σ [xᵢ × P(xᵢ)] for i = 1 to n

This compact form says the same thing: sum all the products of each outcome value and its probability.

NET EXPECTED VALUE (WITH COST)

Net EV = E(X) − Cost to Play

When a game or decision has an entry fee (cost to play), subtract that cost from the expected winnings. A positive Net EV means you profit on average; a negative Net EV means you lose on average.

To compare two choices, compute the expected value for each and pick the one with the higher EV. If both EVs are negative (both options lose money on average), the better choice is the one that loses less. Remember that EV is a long-run concept: on any single trial, anything can happen, but over many trials, results cluster around the expected value.

⚠

IMPORTANT NOTE

Expected value does not tell you what will happen on a single play. It tells you what happens on average over many repetitions. A game with a positive EV can still result in a loss on any given round.

SECTION 5

Detailed Comparison — Visualizing Multiple Options

When you have more than two choices, a bar comparison chart is one of the clearest ways to see which option has the highest expected value at a glance. The diagram below compares three hypothetical carnival games, each with a $2 entry fee. The heights of the bars represent the net expected value (expected winnings minus the $2 cost).

Bar chart showing net expected values for three carnival games. The Wheel Spin has the highest net EV of $1.20, the Ring Toss has a modest net EV of $0.30, and the Duck Pond has a negative net EV of −$0.50, meaning you lose money on average.

Game	Cost	Outcomes & Probabilities	Gross EV	Net EV
Ring Toss	$2	Win $10 (P = 0.20), Win $1 (P = 0.10), Win $0 (P = 0.70)	$2.30	+$0.30
Wheel Spin	$2	Win $8 (P = 0.40), Win $0 (P = 0.60)	$3.20	+$1.20
Duck Pond	$2	Win $5 (P = 0.15), Win $1 (P = 0.35), Win $0 (P = 0.50)	$1.50	−$0.50

Detailed breakdown of three carnival games and their expected values

The table and chart make the comparison crystal clear. Even though the Ring Toss offers the biggest single prize ($10), its low probability of winning (20 %) drags its expected value down. The Wheel Spin has a smaller top prize ($8) but a much better win probability (40 %), which makes its expected value the highest. The Duck Pond actually costs you money on average because the prizes don't compensate for the entry fee.

SECTION 6

Worked Example — Choosing the Better Game

Let's walk through a full comparison step by step. Imagine you're at a school fundraiser and can play one of two games. Game X costs $3 to play: you roll a standard six-sided die, and if you roll a 6 you win $15, otherwise you win nothing. Game Y costs $2 to play: you flip two coins, and if both are heads you win $10, otherwise you win nothing. Which game should you play?

Comparing Game X vs. Game Y

1
Step 1 — List Outcomes for Game X

Game X has two outcomes. Outcome 1: Roll a 6 (win $15). Outcome 2: Roll anything else (win $0). The probability of rolling a 6 on a fair die is 1/6, and the probability of not rolling a 6 is 5/6.
2
Step 2 — Calculate EV for Game X

E(X) = $15 × (1/6) + $0 × (5/6) = $15 × 0.1667 + $0 = $2.50. Now subtract the $3 entry fee: Net EV(X) = $2.50 − $3.00 = −$0.50.

Net EV(X) = −$0.50
3
Step 3 — List Outcomes for Game Y

Game Y also has two outcomes. Outcome 1: Both coins are heads (win $10). Outcome 2: Any other combination (win $0). With two fair coins, each flip has a 1/2 chance of heads, so the probability of both heads is (1/2) × (1/2) = 1/4. The probability of not getting both heads is 3/4.
4
Step 4 — Calculate EV for Game Y

E(Y) = $10 × (1/4) + $0 × (3/4) = $10 × 0.25 + $0 = $2.50. Now subtract the $2 entry fee: Net EV(Y) = $2.50 − $2.00 = +$0.50.

Net EV(Y) = +$0.50
5
Step 5 — Compare and Justify

Game X has a net EV of −$0.50, meaning you lose 50 cents on average per play. Game Y has a net EV of +$0.50, meaning you gain 50 cents on average per play. Even though both games have the same gross expected winnings ($2.50), Game Y's lower entry fee makes it the clear winner. You should choose Game Y.

Game Y is the better choice (Net EV = +$0.50 vs. −$0.50)

SECTION 7

Strengths & Limitations of Expected Value

Expected value is one of the most powerful tools in probability, but it isn't perfect for every situation. Understanding when it works well and when it falls short will help you become a more thoughtful decision-maker.

Strengths	Limitations
Provides a single, objective number for comparison — no guesswork involved.	Ignores risk tolerance: a −$1 EV game with a chance to win $1,000,000 might still be worth playing once (like the lottery).
Works perfectly for repeated decisions — the more you repeat, the closer your average gets to the EV.	For one-time, high-stakes decisions, EV alone may not capture the full picture (e.g., buying a house, choosing a career).
Easy to calculate with basic arithmetic — no advanced math required.	Requires accurate probabilities and payoffs, which can be hard to estimate in real life.
Universally applicable — works for games, business, insurance, sports, and more.	Doesn't account for how spread out outcomes are — two games can have the same EV but very different risk profiles.

Strengths vs. limitations of using expected value for decision-making

✦

KEY TAKEAWAY

Expected value is like a compass for decision-making — it reliably points you toward the best average outcome. However, just as a compass can't tell you about the terrain ahead (cliffs, rivers, storms), EV can't tell you about the variability or emotional impact of outcomes. For repeated, low-stakes decisions, EV is all you need. For rare, high-stakes decisions, consider additional factors like risk and personal circumstances.

SECTION 8

Connection to Advanced Theory

Expected value is the foundation for several advanced topics you may encounter in future math or economics courses. Once you're comfortable comparing choices with EV, you'll be ready to explore concepts like variance (how spread out outcomes are), expected utility (which accounts for how much you personally value gains and losses), and decision theory (a formal framework for rational choice under uncertainty).

Concept	Expected Value (This Lesson)	Advanced Extension
What it measures	Long-run average outcome (center of the distribution)	Variance / Standard deviation: how much outcomes spread around the EV
How it values money	Every dollar is worth the same ($1 = $1)	Expected utility: the 1000th dollar may be worth less to you than the 1st (diminishing returns)
Decision rule	Pick the option with the highest EV	Decision theory: weigh EV against risk, personal values, and context
Best used for	Repeated, low-to-moderate-stakes decisions	One-time, high-stakes, or emotionally significant decisions

Expected value compared to advanced decision-making concepts

As you move into AP Statistics, college probability courses, or fields like actuarial science and data science, you'll combine expected value with measures of spread (like standard deviation) to make even more nuanced comparisons. For now, mastering EV comparison gives you the single most important tool for rational decision-making under uncertainty.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A game has an expected value of −$0.75. What does this mean in practical terms? Would you recommend playing this game 100 times? Explain your reasoning.

Show answer

An expected value of −$0.75 means that, on average, a player loses 75 cents each time the game is played. Over 100 plays, you would expect to lose approximately 100 × $0.75 = $75 total. This game is not favorable for the player because the house (game operator) has a built-in advantage. You would not recommend playing 100 times because the negative EV guarantees long-run losses that converge toward $75.

Question 2: BASIC CALCULATION

A raffle ticket costs $5. There are 200 tickets sold. The prizes are: 1st place wins $400, 2nd place wins $100, and 3rd place wins $50. Calculate the net expected value of buying one ticket.

Show answer

First, find the gross expected value of winning. E(X) = $400 × (1/200) + $100 × (1/200) + $50 × (1/200) + $0 × (197/200) = $2.00 + $0.50 + $0.25 + $0 = $2.75. Then subtract the $5 ticket cost: Net EV = $2.75 − $5.00 = −$2.25. On average, you lose $2.25 per ticket purchased. This is typical of raffle games — the organizers make money to fund their cause.

Question 3: INTERMEDIATE

You can play Game M or Game N. Game M costs $4: you draw a card from a standard 52-card deck, and if it's a face card (J, Q, K) you win $20, otherwise you win $1. Game N costs $3: you roll two dice, and if the sum is 7 or 11 you win $15, otherwise you win nothing. Which game should you choose? Show all work.

Show answer

For Game M: There are 12 face cards in a 52-card deck, so P(face card) = 12/52 ≈ 0.2308 and P(other) = 40/52 ≈ 0.7692. Gross EV(M) = $20 × (12/52) + $1 × (40/52) = $4.615 + $0.769 = $5.385. Net EV(M) = $5.385 − $4 = +$1.385. For Game N: With two dice, P(sum = 7) = 6/36 and P(sum = 11) = 2/36, so P(7 or 11) = 8/36 ≈ 0.2222 and P(other) = 28/36. Gross EV(N) = $15 × (8/36) + $0 × (28/36) = $3.333. Net EV(N) = $3.333 − $3 = +$0.333. Since Net EV(M) = +$1.39 > Net EV(N) = +$0.33, you should choose Game M.

Question 4: APPLIED

A student is deciding between two after-school jobs. Job A pays $12/hour with a 90 % chance of getting hired, and if not hired they earn $0. Job B pays $18/hour but only has a 50 % chance of being offered. Both jobs are 10 hours per week. Using expected value, which job should the student pursue? What other factors might matter beyond EV?

Show answer

Calculate weekly expected earnings. EV(A) = ($12 × 10) × 0.90 + $0 × 0.10 = $120 × 0.90 = $108 per week. EV(B) = ($18 × 10) × 0.50 + $0 × 0.50 = $180 × 0.50 = $90 per week. By expected value, Job A is better ($108 vs. $90). However, factors beyond EV include: risk tolerance (Job B has a 50 % chance of earning nothing), whether the student can apply to both simultaneously, the value of experience or resume building, schedule flexibility, and whether a failed Job B application allows time to still get Job A. EV gives a clear mathematical answer, but real-world context adds nuance.

Question 5: CRITICAL THINKING

Two games have identical expected values of +$1.00. Game P pays $2 with probability 0.5 and $0 with probability 0.5. Game Q pays $100 with probability 0.01 and $0 with probability 0.99. Explain why someone might prefer one game over the other despite their equal EVs. What additional statistical measure would help distinguish between them?

Show answer

Both games have the same EV: Game P: $2 × 0.5 + $0 × 0.5 = $1.00; Game Q: $100 × 0.01 + $0 × 0.99 = $1.00. However, they feel very different. Game P gives you a 50 % chance of winning every time — consistent, predictable results. Game Q almost never pays out (99 % of the time you get nothing), but when it does, you get a big payout. A risk-averse person would prefer Game P because of its reliability. A risk-seeking person might prefer Game Q for the thrill of a large win. The statistical measure that distinguishes them is variance (or standard deviation). Game Q has a much higher variance, meaning its outcomes are more spread out from the mean. Variance quantifies the 'riskiness' that expected value alone cannot capture.

SUMMARY

Lesson Summary

Summary

Expected value (EV) is the weighted average of all possible outcomes of a decision, where each outcome is multiplied by its probability. To compute it, use the formula E(X) = Σ [xᵢ × P(xᵢ)], then subtract any cost to play to get the net expected value. When comparing two or more choices, the option with the highest net EV is mathematically the best long-run decision.

A game or decision is fair when its EV equals zero. Use decision trees and comparison charts to visualize your options. Remember that EV is a long-run average — it doesn't guarantee any single outcome. For high-stakes or one-time decisions, also consider variance and personal risk tolerance beyond the raw numbers.