Calculate expected value E[X], variance and standard deviation for a bet, game or decision from its outcomes and their probabilities or weights.
By AdminDrafted with AI writing toolsUpdated Checked examples: 4
Over many repetitions this loses 1.5 per play on average. The standard deviation per play is 11.3; it measures spread, not a guaranteed range. The expected value is a long-run average. Outcomes with positive probability range from −5 to 95. The probability of coming out ahead is 20%.
| Outcome | Probability | x × p | (x − E)² × p |
|---|---|---|---|
| −5 | 0.8 | −4 | 9.8 |
| 5 | 0.15 | 0.75 | 6.3375 |
| 20 | 0.04 | 0.8 | 18.49 |
| 95 | 0.01 | 0.95 | 93.1225 |
Expected value is the probability-weighted average of the outcomes, E[X] = Σ xᵢpᵢ: what a bet, game or decision returns per play over many plays. The variance, Σ(xᵢ − E[X])²pᵢ, and its square root, the standard deviation, measure how far single results swing around that average. Counts or relative weights are rescaled to probabilities before the sums.
The default is a $5 scratch card that loses the stake with probability 0.8 and wins a net $5, $20 or $95 with probabilities 0.15, 0.04 and 0.01. Its expected value is −$1.50 per card, even though 20% of cards come out ahead.
Enter each outcome as a net result, winnings minus the stake: a lost $5 stake is −5 and a $100 prize on a $5 card is 95. The probabilities must add up to 1, or switch to counts and weights.
Multiply each outcome by its probability and add the products: E[X] = Σ xᵢpᵢ. A $5 scratch card that loses the stake with probability 0.8 and wins a net $5, $20 or $95 with probabilities 0.15, 0.04 and 0.01 has E = −4 + 0.75 + 0.8 + 0.95 = −$1.50 per card.
Worked by hand:
The card returns −30% of its price on average, so 100 cards are expected to lose $150. For a fair six-sided die, E = (1 + 2 + 3 + 4 + 5 + 6) ÷ 6 = 3.5.
Expected value is a long-run average per play, and it may not be a possible result at all: no die roll shows 3.5. The scratch card's most likely outcome is losing the full $5, which happens 80% of the time.
Yet 20% of cards come out ahead, since 0.15 + 0.04 + 0.01 = 0.2. In a skewed payoff like this, a few large prizes pull the average above the typical result. A break-even outcome of 0 counts as neither a gain nor a loss.
The spread around the average is Var(X) = Σ (xᵢ − E[X])² pᵢ, and the standard deviation is its square root. For the scratch card, the four terms are 9.8, 6.3375, 18.49 and 93.1225, so the variance is 127.75 and the SD is $11.30.
This is the variance of a stated distribution, so it has no n − 1 correction. It is in squared units, square dollars here, while the SD is back in dollars.
Over many independent plays, the expected total grows with the number of plays and the SD with its square root. 100 cards have an expected total of −$150 with an SD of 11.30 × √100 = $113, so some buyers of 100 cards still come out ahead.
A single-number roulette bet pays 35 to 1. On an American wheel with 38 pockets, the expected value per $1 is 35 × 1/38 − 1 × 37/38 = −2/38 = −$0.0526, a house edge of 5.26%.
A European wheel has 37 pockets, so the value is (35 − 36) ÷ 37 = −1/37, an edge of 2.70%. Almost every other American bet has the same 5.26% edge.
A sure $50 and a 50% chance of $100 both have an expected value of 50, but their standard deviations are 0 and 50. Which one is better depends on how much risk you can carry.
Insurance has a negative expected value for the buyer, yet it is rational when it removes a loss the buyer could not absorb. Expected utility theory (von Neumann and Morgenstern, 1944) formalizes that trade-off.
| Scenario | Expected value | Standard deviation |
|---|---|---|
| Default | -1.500000 | 11.302655 |
| Fair die | 3.500000 | 1.707825 |
| Sure fifty | 50.000000 | 0.000000 |
| Risky fifty | 50.000000 | 50.000000 |
| Zero weight | 5.000000 | 0.000000 |
"Default" is the scratch card and "Fair die" the values 1 to 6 with equal weights. "Sure fifty" is 50 with probability 1, and "Risky fifty" is 0 or 100 at 50% each.
"Zero weight" lists −100, 5 and 500 with probabilities 0, 1 and 0. Outcomes with zero probability cannot happen, so the worst and best outcomes are both 5.
NIST sets out what makes a valid probability distribution, and Grinstead and Snell, chapter 6, develop expectation and variance.
To turn a bookmaker's price into the probability it implies, use the odds converter. For outcome probabilities from a binomial or Poisson model, use the binomial and Poisson distribution calculator.
Checked against: Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with exact fractions = 127.75
Checked against: House edge of a straight-up bet: −2/38 = −5.26% (standard roulette tables)
Checked against: E = 7/2 and Var = 35/12 for a fair die
Checked against: A single outcome with probability 1 has no spread
Multiply each outcome by its probability and add the products: E[X] = Σ xᵢpᵢ. For a fair six-sided die, E = (1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 3.5. For the default scratch card, −5 × 0.8 + 5 × 0.15 + 20 × 0.04 + 95 × 0.01 = −1.5, a loss of $1.50 per $5 card on average.
The bet loses money on average per play. The −$1.50 scratch card returns −30% of its $5 price, so 100 cards are expected to lose $150. Single results vary: with a standard deviation of 11.30 per card, the total over 100 independent cards has a standard deviation of 11.30 × √100 = 113, so some buyers of 100 cards still come out ahead.
5.26% on an American double-zero wheel and 2.70% on a European single-zero wheel. A single-number bet pays 35 to 1. With 38 pockets, the expected value per unit staked is (35 − 37)/38 = −2/38 = −0.0526; with 37 pockets it is (35 − 36)/37 = −1/37 = −0.0270. Almost every other American bet has the same 5.26% edge.
No. It is a long-run average and may not be a possible result at all: a die's expected value is 3.5, which no roll shows. The scratch card's expected value is −1.5, but its most likely outcome is losing the full $5, which happens 80% of the time. In a skewed payoff like this, a few large prizes pull the average above the typical result.
Not always, because expected value ignores risk. A sure $50 and a 50% chance of $100 both have an expected value of 50, but standard deviations of 0 and 50. Insurance has a negative expected value for the buyer yet is rational when it removes a loss the buyer could not absorb. Expected utility theory (von Neumann and Morgenstern, 1944) formalizes this trade-off.
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