Expected Value and Variance of a Binomial Distribution

What the Expected Value Means

You have a fixed number of independent trials, each with the same probability of success, and you want the long-run average number of successes. That is the expected value of a binomial distribution, and it is simply np. For a binomial random variable X ~ Bin(n, p), the expected value E(X) = np. This is the number you would average if you repeated the experiment many times.

Derivation Using Indicators

The derivation uses indicator random variables. Define I_k as 1 if trial k is a success, 0 otherwise. Then X = I_1 + I_2 + ... + I_n. By linearity of expectation, E(X) = E(I_1) + E(I_2) + ... + E(I_n). The expected value of a single indicator is p, because E(I_k) = 1 * p + 0 * (1-p) = p. So E(X) = np. This works even though the indicators are not independent, because linearity of expectation does not require independence. OpenStax Introductory Statistics 2e, Chapter 4, gives the same formula: binomial expected value μ = np. OpenIntro Statistics (4th ed.) section 3.4 reinforces E(X) = Σ[x_i * P(X = x_i)], and when applied to the binomial this sums to np.

A Common Mistake

The failure case: a student computes E(X) = np but then interprets it as the exact number of successes on the next 10 trials. That is wrong. If n = 10 and p = 0.5, E(X) = 5, but you will not get exactly 5 every time. The expected value is the long-run average, not a guarantee for a single experiment.

What Variance Measures

Variance measures the spread of the distribution. For a binomial X ~ Bin(n, p), Var(X) = np(1-p). The standard deviation is the square root of that, √(np(1-p)). OpenStax labels the complement q, giving binomial standard deviation σ = √(npq).

Derivation and Meaning

The derivation follows the same indicator approach. Var(I_k) = p(1-p) for a single trial. Because trials are independent, Var(X) = Var(I_1) + Var(I_2) + ... + Var(I_n) = np(1-p). OpenIntro section 3.4 defines variance as Σ[(x_i, μ)² * P(X = x_i)], which yields the same value. Blitzstein & Hwang (2nd ed.) Chapter 4 gives Var(X) = E(X²), (E(X))², and for the binomial this also reduces to np(1-p).

The practical meaning: higher variance means greater uncertainty. A binomial with n = 100 and p = 0.5 has variance 25 and standard deviation 5. You expect 50 successes, but most outcomes fall between 40 and 60 (roughly ±2 SD). For n = 100 and p = 0.1, variance is 9 and SD is 3; you expect 10 successes, and most fall between 4 and 16.

Example 1: Free Throws

Example 1: Free throws. A basketball player makes 72% of free throws. In a game she takes 12 free throws. What is the expected number of made shots? Here n = 12, p = 0.72. E(X) = 12 × 0.72 = 8.64. Variance = 12 × 0.72 × 0.28 = 2.4192. Standard deviation = √2.4192 ≈ 1.56. Over many games she averages 8.64 made shots per game; most games she makes between about 5 and 12.

Example 2: Quality Control

Example 2: Quality control. A factory produces microchips with a 3% defect rate. Inspect 200 chips per shift. What is the expected number of defects per shift? n = 200, p = 0.03. E(X) = 200 × 0.03 = 6. Variance = 200 × 0.03 × 0.97 = 5.82. Standard deviation ≈ 2.41. The expected number of defects is 6 per shift, and most shifts will see between 1 and 11.

Example 3: Multiple Choice Test

Example 3: Multiple choice test. A 40-question test has four options each. Guessing every answer: n = 40, p = 0.25. E(X) = 40 × 0.25 = 10. Variance = 40 × 0.25 × 0.75 = 7.5. Standard deviation ≈ 2.74. A student guessing averages 10 correct answers, but could easily get 5 or 15.

Expected Value vs. Mode

The failure case: mistaking the expected value for the mode. In Example 3, the expected value is 10, but the most likely single outcome (the mode) is also 10 only if p = 0.5. For p = 0.25, the mode might be 9 or 10 depending on n. Always compute E(X) = np rather than assuming it equals the mode.

Waiting for the First Success

When the question shifts from a fixed number of trials to waiting for the first success, you leave the binomial and enter the geometric distribution. The expected number of trials until the first success is 1/p. If p = 0.2, you wait an average of 5 trials. If p = 0.01, you wait an average of 100 trials.

Derivation and Common Errors

The geometric distribution expected value is derived from the infinite sum E(X) = Σ k * (1-p)^(k-1) * p = 1/p. This assumes independent trials with constant p. The failure case is when a student uses the binomial np for a geometric problem, for example computing E(X) = 10 * 0.2 = 2 for waiting until first success when n is not fixed. The geometric is the correct distribution when the number of trials is not predetermined.

The variance of the geometric distribution is (1-p)/p². For p = 0.2, variance = 0.8/0.04 = 20, standard deviation ≈ 4.47. This large spread means that although the average wait is 5 trials, individual waits vary widely from 1 to well over 20.

Modeling Event Counts

The Poisson distribution models the number of events in a fixed interval when events occur independently at a constant average rate λ. For a Poisson random variable X ~ Poisson(λ), E(X) = λ. The variance is also λ, a unique property shared only with a few other distributions.

A Practical Example

If emergency room arrivals average 2.4 per hour, then λ = 2.4, and the expected number of arrivals in any given hour is 2.4. The standard deviation is √2.4 ≈ 1.55. This means most hourly counts fall between 0 and 6.

Avoiding Confusion

One sentence on related topics: the expected value properties like linearity apply to the Poisson as they do to all distributions, and the expected value for continuous random variables uses integration rather than summation. The failure case: confusing the Poisson expected value λ with the binomial np when the situation involves a rate over time rather than trials.

Distribution	Expected Value E(X)	Variance Var(X)	Key Assumption
Binomial (n, p)	np	np(1-p)	Fixed n independent trials, constant p
Geometric (p)	1/p	(1-p)/p²	Wait for first success, constant p
Poisson (λ)	λ	λ	Events at constant rate, independent
Bernoulli (p)	p	p(1-p)	Single trial, two outcomes
Uniform (discrete, a..b)	(a+b)/2	((b-a+1)² - 1)/12	Each outcome equally likely

Applying Expected Value to Projects

In project management, Expected Monetary Value (EMV) uses the same principle: multiply each possible outcome by its probability and sum. The PMBOK Guide (7th ed.) defines EMV = Σ(probability × impact) for each risk. This is identical to computing expected value for a discrete random variable, but applied to cost or schedule impacts.

A Common Misapplication

The failure case: project managers treat EMV as a precise estimate rather than a long-run average over many projects. A single project either succeeds or fails; the EMV is the average over many similar projects. Using subjective probabilities without data inflates the apparent precision. The PMI Practice Standard for Project Risk Management recommends decision trees for multi-stage risks, but even then the EMV is only as reliable as the probability estimates.

One sentence on related topics: variance of a random variable quantifies the risk around the EMV, and expected value for continuous random variables handles impacts that vary continuously rather than discretely.

Calculating the House Edge

The house edge is the negative of the player's expected value per dollar wagered. For American roulette (double-zero wheel), the house edge is 5.26%. For a $1 bet on red, the expected value is -$0.0526. European roulette (single-zero) has a lower house edge of 2.70%. Craps pass line bets have a house edge of 1.41%. Blackjack with basic strategy has a house edge of 0.5% to 2%, depending on rules.

Why Probability Is Not Expected Value

The failure case: a bettor sees a 1-in-10 chance to win $5 and thinks it is a good bet, ignoring the $1 cost. The expected value is (0.1 × $5) - $1 = -$0.50 per play. Probability of winning is not expected value. Always compute EV = (payout × probability) - cost.

Lotteries and Jackpots

For lotteries like Mega Millions, the advertised jackpot is not the real expected value. The jackpot odds are 1 in 302,575,350, and overall odds of any prize are 1 in 24. The real EV must account for split probability (expected number of winners) and tax. The IRS withholds 24% on large lottery prizes over $5,000. A $1 billion jackpot with a 1-in-302M chance has a raw EV of about $3.30 before taxes and splits. After accounting for an average of 1.5 winners and 24% withholding, the EV drops to roughly $1.70 per $2 ticket. The exact figure changes with each drawing.

Students and bettors alike treat the expected value as a guarantee for a single trial. If the expected value of a bet is positive, they expect to win that amount immediately. When they lose, they conclude the math is wrong. The expected value is the long-run average over many independent repetitions. The Law of Large Numbers guarantees convergence only as the number of trials grows, not on the next flip or the next hand. For a single trial, the outcome is either success or failure, never the expected value. This mismatch between the abstract average and the concrete single event is the most persistent source of confusion. Always ask: how many times will I repeat this? If the answer is once, expected value tells you almost nothing about what happens next.