Expected Value Calculator

How it works

The expected value of a random quantity is its average over the long run: each possible outcome weighted by how likely it is, then added up. A fair die averages 3.5 even though no single roll shows 3.5. For a bet, the expected value is what you win or lose per play on average, so a negative expected value loses money over many plays even if any one play can win.

The variance measures how spread out the outcomes are around that average: the expected squared distance from the mean. The standard deviation is its square root, back in the original units, and is the typical size of a single result's departure from the expected value.

The probabilities you enter must cover every possibility, so they must add up to 1. The calculator accepts rounding (0.1667 for 1/6, six times, adds to 1.0002) and rescales the probabilities to sum to exactly 1 before computing; the entered sum is shown so you can see it was accepted.

Formula

E[X]   = Σ xᵢ · pᵢ
E[X²]  = Σ xᵢ² · pᵢ
Var(X) = E[X²] − (E[X])²
SD(X)  = √Var(X)
with Σ pᵢ = 1 (entries within 0.01 of 1 are rescaled to 1)

Example

A fair eight-sided die: outcomes 1 to 8, each with probability 0.125. E[X] = (1 + 2 + … + 8) × 0.125 = 36 ÷ 8 = 4.5. E[X²] = (1 + 4 + 9 + … + 64) × 0.125 = 204 ÷ 8 = 25.5, so the variance is 25.5 − 4.5² = 5.25 and the standard deviation is √5.25 = 2.2913.

A fair six-sided die (outcomes 1 to 6, each at 1/6) gives E[X] = 3.5, variance 35/12 = 2.9167 and standard deviation 1.7078. A raffle ticket that pays 500 with probability 0.001 and 0 otherwise has E[X] = 0.5: pay more than 50 cents for it and you lose on average.

Assumptions and limitations

• The outcomes you list must be the only ones possible; that is why the probabilities must add up to 1.
• The expected value is a long-run average per play. It is not what happens on any single play, and a positive expected value can still come with a large chance of losing.
• The variance is the population variance of the distribution you entered, not a sample estimate; no n − 1 correction applies.
• Outcomes and probabilities are paired by position: the third probability belongs to the third outcome, so the two lists must be the same length and in the same order. A missing value shifts every later pair, so check both lists have the same count and order.

Frequently asked questions