B. Probability Distributions · Discrete Probability Distributions

Expected Value: The Long-Run Average

B1.2About 70 minutes

Learning goals

I can calculate the expected value of a discrete random variable with E(X) = Σ x·P(X = x).
I can explain expected value as a weighted mean and as the long-run average of many trials.
I can interpret an expected value in context, even when it is not a possible outcome.
I can predict and verify how adding a constant or multiplying by a constant changes the expected value.

Words to know

expected value
E(X) = Σ x·P(X = x): the probability-weighted mean of a random variable, also written μ. It is the long-run average value over many repetitions.

weighted mean
A mean in which each value is multiplied by a weight, the products are added, and the total is divided by the sum of the weights.

μ (mu)
The Greek letter used for the mean of a probability distribution. μ = E(X).

linearity of expected value
The rule E(aX + b) = a·E(X) + b for constants a and b.

What do you expect?

You roll one die many, many times and write down every result. What is the average of all those numbers? Each of 1, 2, 3, 4, 5, 6 comes up about 1/6 of the time, so the average is close to

(1 + 2 + 3 + 4 + 5 + 6) ÷ 6 = 3.5.

That number, 3.5, is the expected value of a die roll. You can never roll a 3.5. Expected value is not the most likely result, and it may not even be a possible result. It is the long-run average.

The formula

For a discrete random variable X with values x₁, x₂, ..., xₙ:

E(X) = Σ x·P(X = x) = x₁P(x₁) + x₂P(x₂) + ... + xₙP(xₙ)

Multiply each value by its probability, then add. The expected value is also called the mean of the distribution and written μ (the Greek letter mu).

Why it is a weighted mean

You already know weighted means. If your course mark is 70% term work and 30% final evaluation, and you have 80 and 60, your mark is 80(0.7) + 60(0.3) = 74. The weights 0.7 and 0.3 add to 1.

Expected value is exactly this, with the probabilities as the weights. Values that are more likely count more. Since the probabilities add to 1, there is no need to divide at the end.

x	P(X = x)	x·P(X = x)
0	0.15	0
1	0.50	0.50
2	0.20	0.40
3	0.10	0.30
4	0.05	0.20
Total	1.00	E(X) = 1.40

Add a third column, x·P(X = x), and total it. This is the café's hot chocolate distribution from lesson 1.

If each bar were a weight on a seesaw, the seesaw would balance at E(X) = 1.4. The expected value is the balance point (centre of mass) of the probability histogram.

Worked example

Example 1: Expected value of the sum of two dice

Find E(X), where X is the sum of two dice.

Use the distribution: values 2 to 12 with probabilities 1/36, 2/36, ..., 6/36, ..., 1/36.
Σ x·P(x) = [2(1) + 3(2) + 4(3) + 5(4) + 6(5) + 7(6) + 8(5) + 9(4) + 10(3) + 11(2) + 12(1)] ÷ 36.
The numerator is 2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 = 252.
E(X) = 252/36 = 7.
Check with symmetry: the histogram is symmetric about 7, so its balance point is 7. ✓ Also, each die averages 3.5, and 3.5 + 3.5 = 7.

E(X) = 7

Shortcut for symmetric distributions

If a probability histogram is symmetric, the expected value is the centre of symmetry. You can often spot it without computing.

Worked example

Example 2: An expected value that cannot happen

Toss 3 coins. Find the expected number of heads, and interpret it.

The distribution is 1/8, 3/8, 3/8, 1/8 for x = 0, 1, 2, 3.
E(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = (0 + 3 + 6 + 3)/8 = 12/8 = 1.5.

E(X) = 1.5 heads. You can never get 1.5 heads in one toss of 3 coins. It means that over many tosses of 3 coins, the average number of heads per toss gets close to 1.5 (for example, about 150 heads in 100 tosses of three coins).

Changing every value: E(aX + b)

Suppose every value of X is multiplied by a and then b is added. Each product x·P(x) becomes (ax + b)·P(x). So

E(aX + b) = Σ (ax + b)P(x) = a·Σ x·P(x) + b·Σ P(x) = a·E(X) + b,

because Σ P(x) = 1. In words: shifting every value shifts the mean by the same amount, and scaling every value scales the mean by the same factor. This is called the linearity of expected value.

Six prize cases, two of each amount. Adding $10 000 to every case adds $10 000 to the expected value; multiplying every case by 10 multiplies the expected value by 10.

Worked example

Example 3: Prize cases (from the curriculum)

Six closed cases hold prizes: two hold $1, two hold $1000 and two hold $100 000. You choose one case at random. (a) Find the expected value of your prize. (b) Conjecture, then verify, what happens if $10 000 is added to each prize. (c) What if each prize is multiplied by 10?

(a) Each case has probability 1/6. E(X) = 1(2/6) + 1000(2/6) + 100 000(2/6) = 202 002/6 = 33 667.
(b) Conjecture: the mean also goes up by $10 000. Verify: the new prizes are $10 001, $11 000 and $110 000, each with probability 2/6. E = (2)(10 001 + 11 000 + 110 000)/6 = 262 002/6 = 43 667. ✓
(c) Conjecture: the mean is multiplied by 10. New prizes $10, $10 000, $1 000 000. E = 2(1 010 010)/6 = 336 670. ✓
Both agree with E(X + b) = E(X) + b and E(aX) = a·E(X).

(a) $33 667; (b) $43 667; (c) $336 670. Notice that no case actually contains $33 667: two-thirds of the time you get far less.

Worked example

Example 4: Finding an unknown value from the expected value

A spinner gives 2 points with probability 0.5, 5 points with probability 0.3 and k points with probability 0.2. The expected score is 4.5. Find k.

E(X) = 2(0.5) + 5(0.3) + k(0.2) = 1 + 1.5 + 0.2k = 2.5 + 0.2k.
Set equal to 4.5: 0.2k = 2, so k = 10.
Check: 1 + 1.5 + 2 = 4.5. ✓

k = 10 points

Expected value in a business

Insurers, airlines and stores use expected value every day. If a store expects an average of 1.4 hot chocolates per customer and serves 300 customers, it plans for about 300 × 1.4 = 420 hot chocolates. Expected value turns probabilities into planning numbers.

Just for fun

If a survey finds that families on a street have an average of 1.5 dogs, nobody on that street owns half a dog. An average does not have to be a possible value, and expected value is no different.

Hands-on activity

Weighted means at home

You need: one die, a pencil, paper, and a calculator (or a spreadsheet).

Roll the die 50 times and record each result.
Find the ordinary mean of your 50 results (add them and divide by 50).
Now find the mean a second way: count how many 1s, 2s, ..., 6s you got, divide each count by 50 (relative frequencies), and compute Σ x × (relative frequency). Check that both ways give the same number. They always will: this is why expected value is a weighted mean.
Compare with E(X) = 3.5.
Extension: make a "weird die" by covering the faces with tape labels 1, 1, 1, 2, 2, 6 (or imagine it). Compute its expected value before rolling, then roll it 50 times and check.

What to look for: your mean for 50 rolls will probably be within about 0.5 of 3.5. The weird die has E = (1 + 1 + 1 + 2 + 2 + 6)/6 = 13/6 ≈ 2.17, even though its largest value is 6.

Your turn

Find E(X): P(X = 1) = 0.2, P(X = 2) = 0.5, P(X = 3) = 0.3.

A spinner has equal sectors 1, 1, 2, 5. Find the expected value.

Toss 4 coins. Find the expected number of tails. (Distribution: 1/16, 4/16, 6/16, 4/16, 1/16.)

A bag has 3 red and 2 blue marbles. Draw 2 without replacement; X = number of red. (P = 0.1, 0.6, 0.3.) Find E(X).

E(X) = 6. Find E(3X − 4) and E(X + 10).

Your marks: tests 85 (weight 50%), assignments 90 (weight 20%), exam 75 (weight 30%). Find your weighted mean.

A random variable has values 0, 10 and 20 with probabilities 0.5, 0.3 and 0.2. Over 200 trials, about what total do you expect?

More practice: warm-up

What is the expected value of one roll of a fair die?

What is the expected sum of two dice?

P(X = 0) = 0.5 and P(X = 10) = 0.5. Find E(X).

What symbol is often used for the mean of a distribution?

E(X) = 4. Find E(2X).

More practice: on your own

Roll two dice; D = absolute difference (P = 6/36, 10/36, 8/36, 6/36, 4/36, 2/36 for d = 0 to 5). Find E(D).

Roll two dice; X = the larger number (P = 1/36, 3/36, 5/36, 7/36, 9/36, 11/36). Find E(X).

P(X = 1) = 0.4, P(X = 4) = p and P(X = 6) = q. E(X) = 3.2. Find p and q.

A delivery driver's number of stops X has E(X) = 25. The driver earns $40 per day plus $3 per stop. Find the expected daily pay.

Explain why E(X) = 1.5 heads in 3 tosses does not mean that 1.5 is the most likely result.

Challenge

Show that the expected value of a single roll of an n-sided die (faces 1 to n) is (n + 1)/2.

A box has 4 batteries, one dead. You test until you find it (P = 1/4 each for 1 to 4 tests). Each test takes 2 minutes, plus 1 minute to set up. Find the expected time.

Six cases hold $1, $1, $1000, $1000, $100 000, $100 000. You pick one case at random, without opening it, and are then offered $30 000 to walk away instead of keeping it. Use expected value to compare, then give one reason a sensible person might take the $30 000 anyway.

Check your understanding

Quiz

P(X = 0) = 0.3, P(X = 1) = 0.4, P(X = 2) = 0.3. What is E(X)?

Which statement about expected value is true?

E(X) = 12. What is E(2X + 5)?

A game pays $0 with probability 0.7, $5 with probability 0.2 and $20 with probability 0.1. What is the expected payout?

Expected value is a weighted mean in which the weights are...

Toss 3 coins. What is the expected number of heads?

For parents and teachers

Big idea (B1.2): expected value E(X) = Σ x·P(X = x) is the probability-weighted average of a random variable, the long-run average over many trials. It is the balance point of the probability histogram. The rule E(aX + b) = a·E(X) + b lets students handle prices, bonuses and changes of units quickly.

How to help: connect it to report cards: a weighted course mark is the same calculation. Ask your teen what an expected value of 1.5 heads "means" when you can't toss half a head. A good answer mentions the long-run average.

Common difficulties:

Averaging the values without weighting by probability (for example, (0 + 10 + 20)/3 when the values are not equally likely).
Thinking the expected value must be a possible or a most likely result.
Forgetting to add the constant in E(aX + b), or multiplying it by a.