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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.

Cập nhật Ví dụ đã kiểm tra: 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.
Thử
Expected value E[X]
Expected value E[X]: −1.5
Số chữ số thập phân tối đa: 6; Gần nhất; nếu cách đều thì ra xa số không
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 Số hàng: 4
OutcomeXác suấtx × p(x − E)² × p
−50.8−49.8
50.150.756.3375
200.040.818.49
950.010.9593.1225
Cách tính S

  1. Giá trị kỳ vọng

    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

Giới thiệu 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.

Ví dụ có lời giải

$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

Nguồn đối chiếu: 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

Nguồn đối chiếu: 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

Nguồn đối chiếu: 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

Nguồn đối chiếu: A single outcome with probability 1 has no spread

Câu hỏi

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” chính xác đến mức nào?

Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 4. Ví dụ, “$5 scratch card (defaults)” được kiểm tra bằng Hand calculation: −4 + 0.75 + 0.8 + 0.95 = −1.5; Σp(x − μ)² with Python fractions = 127.75.

Phương pháp này lấy từ đâu?

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.

Về công cụ tính này

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

Nguồn

  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

Đã đối chiếu nguồn tham khảo

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