Expected Value Formula with Worked Examples

Key Points

• The expected value formula is $E(X) = \sum x \cdot P(x)$, summing every outcome multiplied by its probability.

• It is a weighted average in which the weights are probabilities, so the probabilities must total 1.

• The expected value is often not an attainable outcome, as a single fair die has an expected value of 3.5.

• A game is fair when its expected value is zero, meaning neither side profits in the long run.

• Expected value describes the long-run average over many repetitions, not what will happen on any single trial.

What Is the Expected Value Formula?

The expected value formula gives the long-run average of a random quantity:

$$E(X) = \sum_{i} x_i \cdot P(x_i)$$

Each possible value is multiplied by its probability and the products are added. Because the probabilities sum to 1, no separate division is needed.

Why No Division Is Needed

An ordinary weighted mean divides by the total weight. Here the weights are probabilities, and a complete probability distribution always sums to exactly 1.

Dividing by 1 changes nothing, so the formula reduces to a plain sum of $x$ times $P(x)$. That is also a useful check: if the probabilities in a problem do not add to 1, something has been missed before any expected value is computed.

The same reasoning explains why expected value is not usually an achievable outcome. It is an average position among the outcomes rather than one of them, in the same way that an average family size of 2.4 children describes no actual family.

Variable Key

Examples of the Expected Value Formula

Example 1

Find the expected value of one roll of a fair six-sided die.

$$E(X) = \frac{1}{6}(1+2+3+4+5+6) = \frac{21}{6} = 3.5$$

Final answer: 3.5

Example 2

A game pays 10 with probability 0.2 and 2 with probability 0.8. Find the expected payout by averaging the two payouts.

Averaging gives $\frac{10+2}{2} = 6$, which sits between the two values.

But the two outcomes are not equally likely. The payout of 2 happens four times as often as the payout of 10, so the average must sit much closer to 2 than to 10.

Weighting by the probabilities:

$$E(X) = 10(0.2) + 2(0.8) = 2 + 1.6 = 3.6$$

Final answer: 3.6. A plain average ignores the weights, which is exactly what the formula exists to supply.

Example 3

A lottery ticket costs 5. It pays 100 with probability 0.02 and nothing otherwise. Find the expected profit per ticket.

Work with profit, so the winning outcome is $100 - 5 = 95$ and the losing outcome is $-5$:

$$E(X) = 95(0.02) + (-5)(0.98) = 1.9 - 4.9 = -3$$

Final answer: a loss of 3 per ticket on average.

A negative expected value means the game favours the operator, which is how lotteries are designed.

Example 4

A fair game charges a stake $s$ and pays 20 with probability 0.25, otherwise nothing. Find the stake that makes it fair.

A fair game has expected value zero:

$$(20 - s)(0.25) + (-s)(0.75) = 0 \implies 5 - 0.25s - 0.75s = 0 \implies s = 5$$

Final answer: a stake of 5

Example 5

A random variable takes values 0, 1 and 2 with probabilities 0.5, 0.3 and 0.2. Find $E(X)$ and check the distribution.

First confirm the probabilities sum to 1: $0.5 + 0.3 + 0.2 = 1$. Then

$$E(X) = 0(0.5) + 1(0.3) + 2(0.2) = 0 + 0.3 + 0.4 = 0.7$$

Final answer: 0.7

Expected Value and Its Relatives

Expected value and sample mean are often confused. The first is a property of a distribution, computed from probabilities; the second is a property of data you actually collected, and it approaches the expected value as the sample grows.

Where Do Students Trip Up on Expected Value?

Mistake 1: Averaging the outcomes without weighting

Where it slips in: Expected value is described as an average, and averaging is normally done by dividing by the count.

Don't do this: $E(X) = \frac{10+2}{2}$

The correct way: $E(X) = \sum x P(x)$, so $10(0.2) + 2(0.8) = 3.6$. Unequal probabilities must pull the answer towards the likelier outcomes.

Students who check that the probabilities sum to 1 before computing anything rarely fall into this, because the act of adding the probabilities makes their inequality obvious. The check costs one line and validates the distribution at the same time.

Mistake 2: Expecting the expected value to be achievable

Where it slips in: The name suggests a value one should expect to see.

Don't do this: Conclude that a die roll of 3.5 is possible because $E(X) = 3.5$.

The correct way: Read it as a long-run average. It is a balance point of the distribution and frequently falls between the attainable outcomes.

Mistake 3: Forgetting the cost in a profit calculation

Where it slips in: The prize is the salient number, and the stake gets handled separately or not at all.

Don't do this: Compute $100(0.02) = 2$ and call it the expected profit on a 5 ticket.

The correct way: Subtract the cost from each outcome, giving 95 and $-5$ as in Example 3. The sign of the answer then tells you who the game favours.

Key Takeaways

• The expected value formula is $E(X) = \sum x \cdot P(x)$.

• It is a weighted average whose weights are probabilities, which is why no division appears.

• The result is a long-run average and frequently not an attainable outcome.

• Checking that the probabilities sum to 1 validates the distribution and exposes unequal weights at once.

• To work through probability with a teacher, explore a high school math tutor or math tutoring.

Practice These Three Before Moving On

Find the expected value of a variable taking 1, 3 and 9 with probabilities 0.5, 0.4 and 0.1. Then find the fair stake for a game paying 50 with probability 0.1, and say whether a game with expected value $-2$ favours the player.

Want a live Bhanzu trainer to walk through more probability problems? Book a free demo class.

Read More

• Weighted Mean Formula

• Conditional Probability Formula

• What Is Probability

• Mean Median Mode Formula

• Average Formula