Expected Value

A spinner pays you £1, £2 or £5 depending on where it stops. You can't know what a single spin will give — but if you played all afternoon, what would you win on average per spin? That long-run average is the expected value of the payout, and it is one of the most useful numbers in all of probability.

First, package the situation as a discrete random variable X — a quantity whose value is decided by chance — together with its distribution: a table listing each value X can take and the probability of each.

\begin{array}{c|cccc} x & 1 & 2 & 3 & 4 \\ \hline P(X = x) & 0.1 & 0.2 & 0.4 & 0.3 \end{array}

Because X must take some value, the probabilities always add to 1: 0.1 + 0.2 + 0.4 + 0.3 = 1.

The expected value formula

The expected value E(X) (also written \mu) is each value weighted by how likely it is — multiply every outcome by its probability and add them up:

E(X) = \sum_x x \, P(X = x).
It is the long-run mean: the average value of X over very many independent repetitions.
It is a weighted average — common outcomes pull it toward themselves, rare ones barely move it.

For the table above:

E(X) = 1(0.1) + 2(0.2) + 3(0.4) + 4(0.3) = 0.1 + 0.4 + 1.2 + 1.2 = 2.9.

Notice 2.9 is not one of the outcomes — and that's fine. The expected value is a summary of the whole distribution, not a prediction of any single result.

The mean is the balancing point

Picture the distribution as bars of "probability weight" sitting along a ruler. The expected value is exactly the point where the ruler would balance — heavier (more probable) bars tug the balance point toward them. Here the balance sits at E(X) = 2.9, pulled right by the tall bars at 3 and 4:

This is why E(X) behaves like a centre of mass. Shift probability toward the high values and the balance point slides right; pile it on the low values and it slides left.

Worked examples

A fair die. Each face 1–6 has probability \tfrac{1}{6}, so E(X) = \tfrac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \tfrac{21}{6} = 3.5. A "half" no face shows — the long-run average of many rolls.
Is the game worth it? You pay £2 to spin, and win £5 with probability 0.3, otherwise nothing. Expected winnings are 5(0.3) + 0(0.7) = £1.50. Since £1.50 < £2, on average you lose £0.50 per go — a bad bet.
A weighted spinner. X \in \{0, 10\} with P(X=10) = 0.2. Then E(X) = 0(0.8) + 10(0.2) = 2 — nowhere near the "middle" of 0 and 10, because the big value is rare.

Every casino game is engineered so the player's expected value is slightly negative. On European roulette a £1 bet on a single number pays £35 (plus your stake back) with probability \tfrac{1}{37}: E = 36 \cdot \tfrac{1}{37} - 1 = -\tfrac{1}{37} \approx -£0.027 per pound. Any single player might walk away rich, but expected value is a statement about the long run — and the house plays a very long run indeed.

E(X) is not the most likely value, nor the middle value. You must weight by probability — averaging the outcomes 1,2,3,4 to 2.5 ignores that 3 is four times as likely as 1.
Check your table first: if the probabilities don't sum to 1, something is missing or wrong, and E(X) will be meaningless.