Expected Value — I can compute expected value in a simple discrete setting and interpret it as a long-run average. - Grade 11 Math

SECTION 1

Historical Context & Motivation

Imagine you're offered a bet: pay $5 to roll a die, and you win $20 if you roll a 6. Should you take it? Your gut might say "maybe," but mathematicians wanted a precise way to answer questions like this. The concept of expected value was born from exactly this kind of question — how do we assign a fair numerical value to uncertain outcomes? The story begins with gamblers, letters between friends, and a puzzle about dice.

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 groundwork for probability theory and the idea of weighting outcomes by their likelihood.

1657 • Huygens Formalizes Expectation

Christiaan Huygens published De Ratiociniis in Ludo Aleae, the first formal treatment of expected value. He defined it as the fair price one should pay to play a game of chance.

1713 • Bernoulli's Law of Large Numbers

Jacob Bernoulli proved that as you repeat a random experiment more and more times, the average result converges to the expected value. This gave expected value its powerful interpretation as a long-run average.

1900s–Today • Modern Applications

Expected value now drives decisions in insurance, finance, game design, medicine, and artificial intelligence. Any time you need to summarize what a random process yields on average, expected value is the tool.

The central question that expected value answers is deceptively simple: if I could repeat a random experiment forever, what single number would my average result settle toward? That number might not even be a possible outcome of the experiment — and that's perfectly okay. The rest of this lesson shows you how to compute it and why it matters.

SECTION 2

Core Principles & Definitions

Before diving into calculations, let's nail down the key ideas that make expected value work. Each principle below builds on the previous one, moving from basic probability language to the full concept of expected value.

Discrete Random Variable

A discrete random variable is a variable whose possible values can be listed out (counted). Examples: the number showing on a die, the number of heads in 10 coin flips, or the number of questions you get right on a quiz.

Probability Distribution

A probability distribution is a table or rule that assigns a probability to each possible outcome. Every probability is between 0 and 1, and all the probabilities must add up to exactly 1.

Weighted Average

Expected value is not a simple average of the outcomes. Instead, it is a weighted average — each outcome is multiplied by how likely it is before summing. More probable outcomes pull the expected value toward them more strongly.

Long-Run Average Interpretation

If you repeat the experiment a huge number of times, the average of all results will approach the expected value. One trial can be far from it, but across many trials the average stabilizes — this is the Law of Large Numbers in action.

Expected Value May Not Be a Possible Outcome

The expected value of a standard die roll is 3.5, but you can never actually roll a 3.5. Expected value is a theoretical center of gravity, not necessarily something you'll ever observe in a single trial.

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KEY TAKEAWAY

Think of expected value like the balance point of a seesaw. If you placed weights on a number line — heavier weights for more probable outcomes — the expected value is the exact point where the number line would balance perfectly. It's the center of gravity of the probability distribution.

SECTION 3

Visual Explanation — The Balance Point

The diagram below shows the probability distribution for a single roll of a fair six-sided die. Each outcome (1 through 6) has an equal probability of 1/6. The bars represent those equal probabilities, and the dashed line marks the expected value of 3.5 — the exact balance point of the distribution. Notice that 3.5 sits right in the middle because the distribution is perfectly symmetric.

Each cyan bar represents the probability of rolling that number on a fair die (all equal at 1/6). The dashed pink line at 3.5 shows the expected value — the balance point of the distribution.

Now consider what happens when a distribution is not symmetric. If certain outcomes are more probable, the balance point shifts toward those heavier outcomes. That's why expected value is a weighted average — the probabilities act as weights. In Section 5 you'll see a diagram where the probabilities are unequal, and the expected value is pulled away from the center.

SECTION 4

Mathematical Framework

Let's formalize the idea. Suppose a discrete random variable X can take on the values x₁, x₂, …, xₙ with corresponding probabilities P(x₁), P(x₂), …, P(xₙ). The expected value of X, written E(X) or sometimes μ (the Greek letter mu), is defined by the formula below.

EXPECTED VALUE FORMULA

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

Each outcome xᵢ is multiplied by its probability P(xᵢ), and then you add up all the products. The result is the theoretical long-run average.

Using summation notation, this can also be written more compactly.

SUMMATION NOTATION

E(X) = Σ xᵢ · P(xᵢ) for i = 1 to n

The capital sigma (Σ) means "add up" all the terms from i = 1 through i = n. This is just a compact way to write the same formula.

There's one important requirement for the formula to be valid: the probabilities must form a legitimate distribution. That means every P(xᵢ) must satisfy 0 ≤ P(xᵢ) ≤ 1, and all probabilities must sum to 1.

VALIDITY CHECK

P(x₁) + P(x₂) + … + P(xₙ) = 1

Always verify this condition before computing E(X). If the probabilities don't sum to 1, something is missing or incorrect in your distribution.

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Quick Tip

When all outcomes are equally likely (like a fair die), the formula simplifies to the ordinary arithmetic mean: E(X) = (x₁ + x₂ + … + xₙ) / n. The general formula is the go-to tool when outcomes are not equally likely.

SECTION 5

Detailed Breakdown — Unequal Probabilities

Expected value gets really interesting when the probabilities aren't all the same. Consider a carnival game where you spin a wheel divided into unequal sections. The table below shows the prize amounts and their probabilities.

Prize (x)	Probability P(x)	Product x · P(x)
$0	0.50	$0 × 0.50 = $0.00
$2	0.30	$2 × 0.30 = $0.60
$5	0.15	$5 × 0.15 = $0.75
$20	0.05	$20 × 0.05 = $1.00
Total	1.00 ✓	E(X) = $2.35

Carnival wheel prize distribution and expected value calculation

Even though $20 is the biggest prize, it only contributes $1.00 to the expected value because it's so unlikely (5% chance). Meanwhile, the $0 outcome has the largest probability and contributes nothing. The expected value of $2.35 tells you that if you played this game hundreds of times, your average prize would settle near $2.35 per spin. If the game costs $3 to play, you'd lose money in the long run — a fact the carnival is counting on.

The $0 bar dominates due to its 50% probability, pulling the expected value (dashed green line at $2.35) far to the left, closer to $0 and $2 than to the larger prizes.

Notice how the dashed green line is pulled strongly to the left — toward the outcomes with the biggest probabilities. The rare $20 prize barely shifts the balance. This is the power of the weighted-average perspective: probability determines influence.

SECTION 6

Worked Example — Should You Play the Game?

A school fundraiser sells raffle tickets for $5 each. The prizes and probabilities are shown below. Let's compute the expected value of the raffle ticket and decide whether it's a good deal for the buyer.

Prize	Value	Probability
Grand Prize	$100	1/200 = 0.005
Second Prize	$25	5/200 = 0.025
Small Prize	$5	20/200 = 0.10
No Prize	$0	174/200 = 0.87

Raffle prize structure (200 tickets sold)

Computing Expected Value of a Raffle Ticket

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Step 1 — List Outcomes and Probabilities

The random variable X represents the prize amount. The possible values are $100, $25, $5, and $0, with probabilities 0.005, 0.025, 0.10, and 0.87, respectively. Check: 0.005 + 0.025 + 0.10 + 0.87 = 1.00 ✓
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Step 2 — Multiply Each Outcome by Its Probability

$100 × 0.005 = $0.50 $25 × 0.025 = $0.625 $5 × 0.10 = $0.50 $0 × 0.87 = $0.00
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Step 3 — Sum the Products

E(X) = $0.50 + $0.625 + $0.50 + $0.00 = $1.625

E(X) = $1.625
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Step 4 — Interpret the Result

On average, a raffle ticket returns $1.63 in prizes. Since each ticket costs $5, the buyer loses about $5.00 − $1.63 = $3.37 on average per ticket. The fundraiser keeps the difference — which is the whole point!

Average net loss per ticket ≈ $3.37
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Step 5 — Long-Run Verification

If all 200 tickets are sold, the fundraiser collects 200 × $5 = $1,000 and pays out $100 + 5($25) + 20($5) = $100 + $125 + $100 = $325 in prizes. That leaves $675 for the school. Per ticket, the average prize is $325 / 200 = $1.625, matching our E(X).

Confirmed: $1.625 per ticket

SECTION 7

Strengths and Limitations of Expected Value

Expected value is one of the most useful numbers in all of probability, but it doesn't tell you everything. Understanding both its strengths and its limitations will help you use it wisely.

Strengths	Limitations
Summarizes a distribution in a single number, making comparisons easy.	Ignores spread — two distributions can have the same E(X) but very different variability.
Accurately predicts long-run averages over many repetitions.	Doesn't predict any single trial — your result could be far from E(X).
Works for any discrete random variable, regardless of the shape of its distribution.	Can be misleading when distributions are highly skewed (e.g., lotteries).
Foundation for decision-making in business, insurance, and game theory.	Doesn't account for risk tolerance — losing $1,000 once may matter more than gaining $10 one hundred times.

Strengths vs. limitations of expected value

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KEY TAKEAWAY

Expected value is like a weather forecast that says "the average temperature this month will be 68°F." That's incredibly useful for planning, but it doesn't tell you whether any particular day will be 55°F or 80°F. To understand how much individual outcomes deviate from E(X), you need a companion measure called standard deviation, which you'll study soon.

SECTION 8

Connection to Advanced Theory

The expected value you're learning now is the foundation for a tower of more advanced ideas in statistics and probability. Here's a preview of how this concept grows as you move into AP Statistics, college courses, and real-world applications.

What You Know Now	Where It Leads
E(X) for finite, discrete outcomes	E(X) for continuous random variables, computed using integrals (calculus-based probability)
Long-run average interpretation	The Law of Large Numbers (formal proof that the sample mean converges to E(X))
Expected value of a single variable	Linearity of expectation: E(aX + bY) = aE(X) + bE(Y), a powerful rule for combining random variables
Deciding if a game is fair (E(X) = 0)	Expected utility theory in economics — adjusting for risk preferences, not just averages
Variance as a companion to E(X)	Var(X) = E(X²) − [E(X)]², a formula that itself relies on expected value

How expected value connects to advanced topics

One particularly elegant result is the linearity of expectation. It says that the expected value of a sum of random variables is always the sum of their expected values — even if the variables are dependent. This property makes expected value far easier to work with than many other statistical measures. It's the reason you can compute the expected total of rolling five dice by simply multiplying 3.5 × 5 = 17.5.

SECTION 9

Practice Problems

Test your understanding with these five problems, arranged from foundational reasoning to critical thinking. Show your work and check that probabilities sum to 1 before computing E(X).

Question 1: CONCEPTUAL

The expected value of rolling a standard six-sided die is 3.5. Explain why you can never actually roll a 3.5, and describe what 3.5 means in terms of a long-run average.

Show answer

A die can only land on 1, 2, 3, 4, 5, or 6, so 3.5 is not a possible individual outcome. The value 3.5 represents the theoretical long-run average: if you rolled the die thousands of times and averaged all the results, that average would approach 3.5 as the number of rolls increased. This is the Law of Large Numbers in action. The expected value serves as the "balance point" of the distribution, not as a prediction for any single roll.

Question 2: BASIC CALCULATION

A spinner has three sections: "Win $10" with probability 0.2, "Win $3" with probability 0.5, and "Win $0" with probability 0.3. Compute E(X).

Show answer

First verify: 0.2 + 0.5 + 0.3 = 1.0 ✓. Now apply the formula: E(X) = $10 × 0.2 + $3 × 0.5 + $0 × 0.3 = $2.00 + $1.50 + $0.00 = $3.50. The expected value of one spin is $3.50. If you spun this spinner many times, your average winnings would converge to $3.50 per spin.

Question 3: INTERMEDIATE

A game costs $4 to play. You draw one card from a standard 52-card deck. If you draw an Ace, you win $20. If you draw a face card (J, Q, K), you win $5. Otherwise, you win nothing. Find the expected net gain or loss per game.

Show answer

There are 4 Aces (probability 4/52 = 1/13 ≈ 0.0769), 12 face cards (probability 12/52 = 3/13 ≈ 0.2308), and 36 other cards (probability 36/52 = 9/13 ≈ 0.6923). Expected prize: E(prize) = $20 × (1/13) + $5 × (3/13) + $0 × (9/13) = $20/13 + $15/13 + $0 = $35/13 ≈ $2.69. Since the game costs $4, the expected net gain is $2.69 − $4.00 = −$1.31. On average, you lose about $1.31 per game. This is a losing proposition for the player in the long run.

Question 4: APPLIED

An insurance company sells a one-year phone protection plan for $120. Based on historical data, there is a 15% chance the customer files a small claim (company pays $200), a 5% chance they file a large claim (company pays $800), and an 80% chance no claim is filed. What is the expected profit per policy for the insurance company?

Show answer

First compute the expected payout: E(payout) = $200 × 0.15 + $800 × 0.05 + $0 × 0.80 = $30 + $40 + $0 = $70. The company collects $120 in premiums per policy, so the expected profit = $120 − $70 = $50 per policy. This means that across a large portfolio of customers, the company expects to earn roughly $50 per policy after paying claims. This is exactly how insurance companies set premiums — they use expected value to ensure long-run profitability.

Question 5: CRITICAL THINKING

Two games have the same expected value of $5. Game A pays $5 every single time. Game B pays $0 half the time and $10 half the time. Both have E(X) = $5. If you need exactly $5 to buy a bus ticket home, which game would you prefer and why? What does this reveal about the limitations of expected value as a decision-making tool?

Show answer

You should prefer Game A because it guarantees the $5 you need. Game B has a 50% chance of leaving you with $0 and stranded. Even though both games have E(X) = $5, they differ in their variability (spread). Game A has zero variance, while Game B has substantial variance. This reveals a key limitation: expected value alone ignores risk. When the consequences of bad outcomes are severe (missing the bus), the spread of the distribution matters as much as the center. In advanced courses, this idea leads to expected utility theory, where people factor in their risk tolerance when making decisions under uncertainty.

SUMMARY

Lesson Summary

Summary

The expected value of a discrete random variable is computed by multiplying each possible outcome by its probability and summing all the products: E(X) = Σ xᵢ · P(xᵢ). This weighted average gives outcomes with higher probabilities more influence over the result. Before computing, always verify that the probabilities in your distribution sum to 1.

The expected value represents the long-run average — the value your sample mean approaches as the number of trials grows, a result guaranteed by the Law of Large Numbers. It may not equal any actual possible outcome (like 3.5 on a die). While E(X) is powerful for comparing options and evaluating fairness, it doesn't capture variability or risk — for that, you'll soon learn about standard deviation and variance.