Mathematics · Stochastics

Expected Value

The expected value is the long-run average of a random variable: values times probabilities, summed up.

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Formula

E (X) = μ = i \sum x_{i} \cdot p_{i}

LaTeX: E(X) = \mu = \sum_{i} x_{i} \cdot p_{i}

E(X) in the unit of the random variable X · pᵢ dimensionless

Variables & units – Expected Value

Symbol	Meaning	Unit
E(X), μ	Expected value of the random variable X	same as X
xᵢ	Possible values of the random variable	same as X
pᵢ	Probability P(X = xᵢ)	dimensionless

Derivation & background – Expected Value

Christiaan Huygens introduced the concept in 1657 for games of chance. The expected value is the weighted mean of all possible values; by the law of large numbers the average of many repetitions stabilizes at E(X). A game is called fair if the expected value of the winnings is 0. For the binomial distribution simply E(X) = n·p.

Exam blueprint

Validity range

Applies to discrete random variables with finitely many values; all pᵢ sum to 1. For continuous random variables an integral replaces the sum.

Derivation steps

Weighted mean: each value counts according to its probability.

1In N repetitions, xᵢ occurs about N·pᵢ times; the average is Σxᵢ·N·pᵢ/N.
2Cancelling N gives E(X) = Σxᵢ·pᵢ; the law of large numbers guarantees the stabilization.

Rearrangements

Binomial distribution

E (X) = n \cdot p

Short formula instead of summing over all k.

Linearity

E (a X + b) = a \cdot E (X) + b

Linear transformations pass directly into the expected value.

Fair game

E (G) = 0

G = winnings minus stake; E(G) = 0 means fair.

Task variant

A wheel pays 10 € with p = 0.2, otherwise nothing. Stake 3 €. Is the game worth it?

E(payout) = 10·0.2 = 2 €. Winnings G: E(G) = 2 − 3 = −1 €. On average you lose 1 € per game, the game is not worth it.

X takes the values 0, 1, 2 with p = 0.3/0.5/0.2. Compute E(X).

E(X) = 0·0.3 + 1·0.5 + 2·0.2 = 0.9.

Common mistakes

Simply taking the arithmetic mean of the values.

Correct only for a uniform distribution; otherwise weight with the pᵢ.

Interpreting E(X) as the "most likely value".

E(X) = 3.5 for a die is never rolled; it is the long-run average.

Forgetting the stake in a fair game.

Fair means E(winnings) = 0 including the stake, not E(payout) = 0.

Exam context

Games of chance and payout tasks, distribution parameters, fair stakes.

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Formula cluster

Core stochastics

Location and spread together describe a distribution.

P (X = k) = (k n) \cdot p^{k} \cdot (1 - p)^{n - k}

Binomial Distribution

σ = Var (X) = E ((X - μ)^{2})

Standard Deviation

Worked example

Fair die: E(X) = (1 + 2 + 3 + 4 + 5 + 6)·1/6 = 21/6 = 3.5. Binomial distribution with n = 100 and p = 0.3: E(X) = n·p = 30 successes on average.

Applications

Evaluating games of chance and insurance, fair stakes, mean success counts in exams, decision theory, risk analysis

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Curated exam set for "Expected Value":

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Which formula describes Expected Value?

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How do you rearrange E(X) = Σ xᵢ·pᵢ for Binomial distribution?

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Which common mistake happens with Expected Value?

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Scientific sources

[1]Huygens, C. (1657). De ratiociniis in ludo aleae. Leiden.

Common notations & search queries

E(X)=Summe xi*piErwartungswert berechnenErwartungswert Formelmü Stochastikfaires Spiel ErwartungswertErwartungswert Würfelexpected value formulaErwartungswert Binomialverteilung n*p

Related formulas

P (X = k) = (k n) \cdot p^{k} \cdot (1 - p)^{n - k}

Binomial Distribution

σ = Var (X) = E ((X - μ)^{2})

Standard Deviation

f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}

Normal Distribution (Gaussian Bell Curve)

P (X = k) = \frac{λ ^{k}}{k !} \cdot e^{- λ}

Poisson Distribution

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

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Frequently asked questions about Expected Value

How do I calculate the expected value of a random variable?+

First set up the probability distribution as a table: all possible values xᵢ of the random variable with their probabilities pᵢ. The expected value is then the sum of products: E(X) = x₁·p₁ + x₂·p₂ + ... Die example: E(X) = 1·(1/6) + 2·(1/6) + ... + 6·(1/6) = 21/6 = 3.5. Prize game example: if a game pays 10 € with probability 0.2 and otherwise nothing, E(payout) = 10·0.2 + 0·0.8 = 2 €. Before computing, check that all pᵢ add to 1, otherwise the distribution is incomplete. The expected value need not be a possible value: 3.5 cannot be rolled, yet it is the correct long-run average.

What does the expected value mean intuitively?+

It is the long-run average: if you repeat the random experiment very often and average the results, this mean stabilizes at the expected value; the law of large numbers guarantees it. In 6000 die rolls you expect a total near 6000·3.5 = 21000. Two distinctions matter: E(X) is not a prediction for a single experiment, an individual roll can deviate arbitrarily. And E(X) is not the most likely value (that would be the mode); for skewed distributions the two lie far apart, as in lotteries, where the most frequent prize is 0 but the expected value can be positive. As a physical picture: E(X) is the centre of mass of the distribution, the point where the probability mass balances.

When is a game fair and how do I check it?+

A game is called fair if the expected value of the winnings is zero, where winnings = payout minus stake. So compute the expected payout and subtract the stake. Example: a wheel of fortune pays 10 € with p = 0.2, the stake is 3 €. E(payout) = 2 €, so E(winnings) = 2 − 3 = −1 €: on average the player loses 1 € per round, the game is unfair in the operator's favour. A stake of exactly 2 € would be fair. Typical exam task: "Determine the stake so that the game is fair", the answer is always stake = expected payout. Real gambling (lottery, roulette, slot machines) has a systematically negative expected value; that is what the operator lives on.

Why does E(X) = n·p simply hold for the binomial distribution?+

Because of the additivity of the expected value. A binomially distributed X is the sum of n indicator variables X₁, ..., Xₙ, where Xᵢ = 1 for a success in the i-th trial and 0 otherwise. Each has E(Xᵢ) = 1·p + 0·(1−p) = p. Since the expected value of a sum is always the sum of the expected values (this even holds without independence!), E(X) = n·p follows immediately, without evaluating the sum formula with binomial coefficients. Example: 100 attempts at rolling a six, p = 1/6, give on average 100/6 ≈ 16.7 successes. The formula is intuitive too: a fraction p of n trials succeeds. The same decomposition yields the variance n·p·(1−p), though there independence is required.

What is the difference between expected value and mean?+

The expected value μ = E(X) is a theoretical parameter of the probability model: it is computed from the probabilities before any experiment takes place. The mean x̄ (arithmetic average) is an empirical parameter of concrete data: it is computed from actually observed values after the experiment has run. The two are linked by the law of large numbers: as the number of trials grows, x̄ approaches the expected value μ. If you roll a die 10 times, x̄ might be 3.9; the model still says μ = 3.5. Statistics uses this relation in reverse: x̄ serves as an estimator for an unknown μ. Linguistically: "expected value" for the model, "mean" for the sample.

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How do you calculate with Expected Value?

Here is how to work through a typical Expected Value (E(X) = Σ xᵢ·pᵢ) task step by step:

1
Task
A wheel pays 10 € with p = 0.2, otherwise nothing. Stake 3 €. Is the game worth it?
Solution path
E(payout) = 10·0.2 = 2 €. Winnings G: E(G) = 2 − 3 = −1 €. On average you lose 1 € per game, the game is not worth it.
2
Task
X takes the values 0, 1, 2 with p = 0.3/0.5/0.2. Compute E(X).
Solution path
E(X) = 0·0.3 + 1·0.5 + 2·0.2 = 0.9.