Expected Value

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.

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.

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.

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.

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.