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Expected value calculator

Calculate expected value E[X], variance and standard deviation for a bet, game or decision from its outcomes and their probabilities or weights.

更新于 已验证的示例：4

Net result of each outcome, e.g. −5 for losing a $5 stake.
One per outcome, in the same order. Fractions such as 1/38 work.
试一试
Expected value E[X]
Expected value E[X]: −1.5
最大小数位数：6；取最近值，等距时远离零
Variance
127.75
Standard deviation
11.302655
Probability of a positive outcome
0.2
Worst outcome
−5
Best outcome
95

Over many repetitions this loses 1.5 per play on average, with a typical swing of ±11.3 on any single play. The expected value is a long-run average: a single play can land anywhere from −5 to 95, and 20% of plays come out ahead.

Probability of each outcome

−5: 0.80.8−55: 0.150.15520: 0.040.042095: 0.010.0195
Outcomes 行数：4
Outcome概率x × p(x − E)² × p
−50.8−49.8
50.150.756.3375
200.040.818.49
950.010.9593.1225
计算方法 S

  1. 期望值

    E[X]=∑xipi=−5×0.8+5×0.15+20×0.04+95×0.01=−1.5E[X] = \sum x_i p_i = -5 \times 0.8 + 5 \times 0.15 + 20 \times 0.04 + 95 \times 0.01 = -1.5

  2. Variance

    Var⁡(X)=∑(xi−−1.5)2pi=127.75\operatorname{Var}(X) = \sum (x_i - -1.5)^2 p_i = 127.75

  3. Standard deviation

    σ=127.75=11.302655\sigma = \sqrt{127.75} = 11.302655

关于Expected value calculator

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.

计算示例

$5 scratch card (defaults)

Outcomes (values or payoffs)
-5, 5, 20, 95
Probabilities
0.8, 0.15, 0.04, 0.01
Second list holds
Probabilities (sum to 1)
Expected value E[X]
-1.5
Variance
127.75
Probability of a positive outcome
0.2

核验来源：Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with Python fractions = 127.75

American roulette, single-number bet

Outcomes (values or payoffs)
35, -1
Probabilities
1/38, 37/38
Second list holds
Probabilities (sum to 1)
Expected value E[X]
-0.052632

核验来源：House edge of a straight-up bet: −2/38 = −5.26% (standard roulette tables)

Fair die

Outcomes (values or payoffs)
1 2 3 4 5 6
Probabilities
1 1 1 1 1 1
Second list holds
Counts or weights
Expected value E[X]
3.5
Variance
2.916667

核验来源：E = 7/2 and Var = 35/12 for a fair die (textbook result; Python fractions)

Edge case: a certain outcome

Outcomes (values or payoffs)
42
Probabilities
1
Second list holds
Probabilities (sum to 1)
Expected value E[X]
42
Variance
0
Standard deviation
0

核验来源：A single outcome with probability 1 has no spread

常见问题

How do you calculate expected value?

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.

What does a negative expected value mean?

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.

What is the house edge in roulette?

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.

Is expected value the most likely outcome?

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.

Is the option with the highest expected value always the best choice?

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.

“Expected value calculator”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例：4。 例如，“$5 scratch card (defaults)”根据Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with Python fractions = 127.75进行核验。

这种方法出自哪里？

Grinstead & Snell, Introduction to Probability (AMS), chapter 6 Expected value and variance; NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.1 What is a probability distribution.

关于此计算器

E[X]=∑ixi pi,Var⁡(X)=∑i(xi−E[X])2 piE[X] = \sum_i x_i\,p_i,\qquad \operatorname{Var}(X) = \sum_i (x_i - E[X])^2\,p_i

来源

  1. Grinstead & Snell, Introduction to Probability (AMS), chapter 6 Expected value and variance
  2. NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.1 What is a probability distribution

已对照来源验证

此计算器包含 4 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。

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