Calculate expected value E[X], variance and standard deviation for a bet, game or decision from its outcomes and their probabilities or weights.
अद्यतनित तपासलेली उदाहरणे: ४
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.
| Outcome | संभाव्यता | 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.
पडताळणीचा स्रोत: Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with Python fractions = 127.75
पडताळणीचा स्रोत: House edge of a straight-up bet: −2/38 = −5.26% (standard roulette tables)
पडताळणीचा स्रोत: E = 7/2 and Var = 35/12 for a fair die (textbook result; Python fractions)
पडताळणीचा स्रोत: 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.
अचूकता तुमच्या इनपुटवर आणि पद्धतीच्या गृहीतकांवर अवलंबून असते. दशांश गणना 50 सार्थ अंक वापरते, पण अंदाज, संख्यात्मक पद्धती आणि मूळ डेटा कमी अचूक असू शकतात; दाखवलेल्या मूल्यांचे पूर्णांकन केल्याने या मर्यादा दूर होत नाहीत. स्वतंत्र स्रोतांतील सोडवलेल्या उदाहरणांशी पडताळणी: ४. उदाहरणार्थ, “$5 scratch card (defaults)” ची पडताळणी Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with Python fractions = 127.75 याच्याशी केली आहे.
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.
या कॅल्क्युलेटरमध्ये स्वतंत्र स्रोतांमधील उत्तरांसह ४ सोडवलेली उदाहरणे आहेत. ती चाचणी संचात चालतात आणि तुम्ही ती येथेही चालवू शकता.
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