High School Maths · Year 2 · Unit 13 · Lesson 5 of 5

Expected Value and Fair Games: Should You Play This Game?

About 15 minutes · probability distributions · expected value E(X) = Σ x × P(X = x) · expected profit and the cost to play · fair games · finding a missing prize, cost or probability · mixed exam problems · no calculator · exit quiz

Hook

$2 to play · roll a 6 to win $9

A stall charges $2 a go. You roll a fair dice. A 6 wins $9; anything else wins nothing. Should you play?

On average you win 16 × $9 = $1.50 a go, but each go costs $2. So in the long run you lose about 50 cents a go, and the stall makes money. $1.50 is the expected value of your prize. The game would be fair only if the expected prize equalled the $2 cost.

Learn

Master the essentials

x1234
P(X = x)0.10.3k0.2

A probability distribution with one missing value.

Random variable

A number decided by chance, such as a score, a prize or a profit. It is often called X.

X = the score on a dice

Probability distribution

A table of every value of X with its probability. The probabilities add up to 1.

P(X = 0) + P(X = 1) + … = 1

Expected value

E(X) = the sum of (each value × its probability).

0 × 0.5 + 2 × 0.3 + 5 × 0.2 = 1.6

What it means

E(X) is the average result per go over very many goes. It need not be a value X can take.

A fair dice: E(X) = 3.5

Profit

Profit = prize − cost to play. Expected profit = expected prize − cost.

Pay $3, win $10: profit $7. Win nothing: profit −$3.

Fair game

A game is fair when the expected profit is 0: the expected prize equals the cost.

Cost $2 and expected prize $2: fair

Many goes

Expected total = number of goes × expected value of one go.

200 × $0.40 = $80

A missing value

Write E(X) with the unknown in it, set it equal to what you know, and solve.

Win $x with P = 14, cost $3, fair: x4 = 3, x = 12

Worked examples

See it done

1. The table shows the probability distribution of X. Find k and E(X).

x1234
P(X = x)0.10.3k0.2

A probability distribution with one missing value.

  1. The probabilities add up to 1: 0.1 + 0.3 + k + 0.2 = 1, so k = 0.4.
  2. E(X) = 1 × 0.1 + 2 × 0.3 + 3 × 0.4 + 4 × 0.2.
  3. = 0.1 + 0.6 + 1.2 + 0.8 = 2.7.

k = 0.4 and E(X) = 2.7

2. A fair spinner has 8 equal sections. One section wins $12, three sections win $4 and the rest win nothing. A go costs $3. Is the game fair?

  1. Expected prize = $12 × 18 + $4 × 38 + $0 × 48.
  2. = $1.50 + $1.50 = $3.
  3. Expected profit = $3 − $3 = $0.
  4. The expected profit is 0, so the game is fair.

Yes, it is fair

3. A game costs $2. Two fair coins are thrown. Two heads win $x, exactly one head wins $1, and no heads wins nothing. Find x so that the game is fair.

  1. P(two heads) = 14, P(exactly one head) = 12, P(no heads) = 14.
  2. Expected prize = 14x + 1 × 12.
  3. Fair: 14x + 12 = 2, so 14x = 32.
  4. x = 6, so two heads should win $6.

x = 6

Common mistake: Forgetting the cost to play, or taking it away twice. Either work out the expected PRIZE and take the cost away once at the end, or write the PROFIT for every outcome (prize − cost, so −$2 for losing) and find its expected value. Do not mix the two.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Multiply each value by its probability and add. Take the cost away once. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.The table shows the probability distribution of X. Find E(X).

x1234
P(X = x)0.40.30.20.1

The probability distribution of X.

Correct.

Solution

  1. E(X) = 1 × 0.4 + 2 × 0.3 + 3 × 0.2 + 4 × 0.1.
  2. = 0.4 + 0.6 + 0.6 + 0.4 = 2.

Not quite. This is the mean of the four values. It ignores the probabilities.

Solution

  1. E(X) = 1 × 0.4 + 2 × 0.3 + 3 × 0.2 + 4 × 0.1.
  2. = 0.4 + 0.6 + 0.6 + 0.4 = 2.

Answer: A. 2.

Not quite. This adds the values without multiplying by the probabilities.

Solution

  1. E(X) = 1 × 0.4 + 2 × 0.3 + 3 × 0.2 + 4 × 0.1.
  2. = 0.4 + 0.6 + 0.6 + 0.4 = 2.

Answer: A. 2.

Not quite. This pairs the values with the probabilities in the wrong order.

Solution

  1. E(X) = 1 × 0.4 + 2 × 0.3 + 3 × 0.2 + 4 × 0.1.
  2. = 0.4 + 0.6 + 0.6 + 0.4 = 2.

Answer: A. 2.

2.A fair dice is rolled. A 6 wins $6, a 4 or a 5 wins $3, and anything else wins nothing. Find the expected amount won.

Not quite. This averages $6, $3 and $0. They are not equally likely.

Solution

  1. P($6) = 16, P($3) = 26, P($0) = 36.
  2. $6 × 16 + $3 × 26 = $1 + $1.
  3. = $2.

Answer: C. $2.

Not quite. This adds the prizes.

Solution

  1. P($6) = 16, P($3) = 26, P($0) = 36.
  2. $6 × 16 + $3 × 26 = $1 + $1.
  3. = $2.

Answer: C. $2.

Correct.

Solution

  1. P($6) = 16, P($3) = 26, P($0) = 36.
  2. $6 × 16 + $3 × 26 = $1 + $1.
  3. = $2.

Not quite. This uses 16 for the $3 prize. Two scores win $3.

Solution

  1. P($6) = 16, P($3) = 26, P($0) = 36.
  2. $6 × 16 + $3 × 26 = $1 + $1.
  3. = $2.

Answer: C. $2.

3.A raffle sells 500 tickets at $2 each. There is one prize of $400 and two prizes of $100. Find the expected profit for a person who buys one ticket.

Not quite. This is the expected prize. Take away the $2 ticket.

Solution

  1. Expected prize = $400 × 1500 + $100 × 2500.
  2. = $0.80 + $0.40 = $1.20.
  3. Expected profit = $1.20 − $2 = −$0.80.

Answer: B. −$0.80.

Correct.

Solution

  1. Expected prize = $400 × 1500 + $100 × 2500.
  2. = $0.80 + $0.40 = $1.20.
  3. Expected profit = $1.20 − $2 = −$0.80.

Not quite. This works out cost − prize. Profit is prize − cost.

Solution

  1. Expected prize = $400 × 1500 + $100 × 2500.
  2. = $0.80 + $0.40 = $1.20.
  3. Expected profit = $1.20 − $2 = −$0.80.

Answer: B. −$0.80.

Not quite. This counts only one $100 prize.

Solution

  1. Expected prize = $400 × 1500 + $100 × 2500.
  2. = $0.80 + $0.40 = $1.20.
  3. Expected profit = $1.20 − $2 = −$0.80.

Answer: B. −$0.80.

Exam standard (Several steps, like real exam questions.)

4.X takes the values 0, 10, 20 and 50 with probabilities 0.5, 0.3, k and 0.05. Find E(X).

Not quite. This is the mean of the four values. It ignores the probabilities.

Solution

  1. k = 1 − 0.5 − 0.3 − 0.05 = 0.15.
  2. E(X) = 0 + 10 × 0.3 + 20 × 0.15 + 50 × 0.05.
  3. = 3 + 3 + 2.5 = 8.5.

Answer: C. 8.5.

Not quite. This leaves out the value 20. Find k first.

Solution

  1. k = 1 − 0.5 − 0.3 − 0.05 = 0.15.
  2. E(X) = 0 + 10 × 0.3 + 20 × 0.15 + 50 × 0.05.
  3. = 3 + 3 + 2.5 = 8.5.

Answer: C. 8.5.

Correct.

Solution

  1. k = 1 − 0.5 − 0.3 − 0.05 = 0.15.
  2. E(X) = 0 + 10 × 0.3 + 20 × 0.15 + 50 × 0.05.
  3. = 3 + 3 + 2.5 = 8.5.

Not quite. This is k, not E(X).

Solution

  1. k = 1 − 0.5 − 0.3 − 0.05 = 0.15.
  2. E(X) = 0 + 10 × 0.3 + 20 × 0.15 + 50 × 0.05.
  3. = 3 + 3 + 2.5 = 8.5.

Answer: C. 8.5.

5.A fair spinner has 10 equal sections. One section wins $x, two sections win $5 and the rest win nothing. A go costs $4. Find x so that the game is fair.

Not quite. This leaves out the two $5 sections.

Solution

  1. Expected prize = x10 + $5 × 210 = x10 + $1.
  2. Fair: x10 + 1 = 4, so x10 = 3.
  3. x = $30.

Answer: D. $30.

Not quite. This counts only one $5 section.

Solution

  1. Expected prize = x10 + $5 × 210 = x10 + $1.
  2. Fair: x10 + 1 = 4, so x10 = 3.
  3. x = $30.

Answer: D. $30.

Not quite. This forgets that $x is won only 110 of the time.

Solution

  1. Expected prize = x10 + $5 × 210 = x10 + $1.
  2. Fair: x10 + 1 = 4, so x10 = 3.
  3. x = $30.

Answer: D. $30.

Correct.

Solution

  1. Expected prize = x10 + $5 × 210 = x10 + $1.
  2. Fair: x10 + 1 = 4, so x10 = 3.
  3. x = $30.

6.A game costs $1. A fair spinner has 4 equal sections: one wins $2, one wins $0.50 and two win nothing. Which working gives the expected profit for a player?

Not quite. This treats the three prizes as equally likely. "Nothing" has two sections.

Solution

  1. Expected prize = $2 × 14 + $0.50 × 14 = $0.625.
  2. The player pays $1 every go, so take away 1.
  3. Expected profit = 0.625 − 1 = −$0.375.

Answer: D. 2 × 14 + 0.5 × 14 − 1.

Not quite. This is the expected prize. It forgets the $1 cost.

Solution

  1. Expected prize = $2 × 14 + $0.50 × 14 = $0.625.
  2. The player pays $1 every go, so take away 1.
  3. Expected profit = 0.625 − 1 = −$0.375.

Answer: D. 2 × 14 + 0.5 × 14 − 1.

Not quite. This takes the cost away only on a losing go. The player pays $1 every go.

Solution

  1. Expected prize = $2 × 14 + $0.50 × 14 = $0.625.
  2. The player pays $1 every go, so take away 1.
  3. Expected profit = 0.625 − 1 = −$0.375.

Answer: D. 2 × 14 + 0.5 × 14 − 1.

Correct.

Solution

  1. Expected prize = $2 × 14 + $0.50 × 14 = $0.625.
  2. The player pays $1 every go, so take away 1.
  3. Expected profit = 0.625 − 1 = −$0.375.

7.A charity runs a spinner game. A go costs $2. The spinner wins $0 with probability 0.7, $2 with probability 0.2 and $10 with probability 0.1. 300 people play once each. How much money should the charity expect to make?

Correct.

Solution

  1. Expected prize per go = $2 × 0.2 + $10 × 0.1 = $0.40 + $1 = $1.40.
  2. Charity profit per go = $2 − $1.40 = $0.60.
  3. 300 goes: 300 × $0.60 = $180.

Not quite. This is the expected amount paid out in prizes.

Solution

  1. Expected prize per go = $2 × 0.2 + $10 × 0.1 = $0.40 + $1 = $1.40.
  2. Charity profit per go = $2 − $1.40 = $0.60.
  3. 300 goes: 300 × $0.60 = $180.

Answer: A. $180.

Not quite. This is the money taken in. Prizes must be paid out of it.

Solution

  1. Expected prize per go = $2 × 0.2 + $10 × 0.1 = $0.40 + $1 = $1.40.
  2. Charity profit per go = $2 − $1.40 = $0.60.
  3. 300 goes: 300 × $0.60 = $180.

Answer: A. $180.

Not quite. This is the expected profit from one go. 300 people play.

Solution

  1. Expected prize per go = $2 × 0.2 + $10 × 0.1 = $0.40 + $1 = $1.40.
  2. Charity profit per go = $2 − $1.40 = $0.60.
  3. 300 goes: 300 × $0.60 = $180.

Answer: A. $180.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A bag holds 3 red and 2 blue counters. Two counters are taken at random, without replacement. You win $5 for each red counter. Find the expected amount won.

Not quite. This averages $0, $5 and $10. They are not equally likely.

Solution

  1. P(no red) = 25 × 14 = 110. P(two red) = 35 × 24 = 310.
  2. P(one red) = 1 − 110 − 310 = 35.
  3. Expected win = $5 × 35 + $10 × 310 = $3 + $3 = $6.

Answer: C. $6.

Not quite. This counts red then blue only. Blue then red also gives one red.

Solution

  1. P(no red) = 25 × 14 = 110. P(two red) = 35 × 24 = 310.
  2. P(one red) = 1 − 110 − 310 = 35.
  3. Expected win = $5 × 35 + $10 × 310 = $3 + $3 = $6.

Answer: C. $6.

Correct.

Solution

  1. P(no red) = 25 × 14 = 110. P(two red) = 35 × 24 = 310.
  2. P(one red) = 1 − 110 − 310 = 35.
  3. Expected win = $5 × 35 + $10 × 310 = $3 + $3 = $6.

Not quite. This leaves out the outcome with two reds.

Solution

  1. P(no red) = 25 × 14 = 110. P(two red) = 35 × 24 = 310.
  2. P(one red) = 1 − 110 − 310 = 35.
  3. Expected win = $5 × 35 + $10 × 310 = $3 + $3 = $6.

Answer: C. $6.

9.A game costs $3. A player wins $10 with probability p, wins $2 with probability 2p, and otherwise wins nothing. The game is fair. Find the probability that a player wins nothing.

Not quite. This is p, the probability of winning $10.

Solution

  1. Fair: 10p + 2 × 2p = 3.
  2. 14p = 3, so p = 314.
  3. P(nothing) = 1 − p − 2p = 1 − 914 = 514.

Answer: B. 514.

Correct.

Solution

  1. Fair: 10p + 2 × 2p = 3.
  2. 14p = 3, so p = 314.
  3. P(nothing) = 1 − p − 2p = 1 − 914 = 514.

Not quite. This is 1 − p. The player can also win $2, with probability 2p.

Solution

  1. Fair: 10p + 2 × 2p = 3.
  2. 14p = 3, so p = 314.
  3. P(nothing) = 1 − p − 2p = 1 − 914 = 514.

Answer: B. 514.

Not quite. This is 3p, the probability of winning something.

Solution

  1. Fair: 10p + 2 × 2p = 3.
  2. 14p = 3, so p = 314.
  3. P(nothing) = 1 − p − 2p = 1 − 914 = 514.

Answer: B. 514.

10.Game A costs $2 and wins $7 with probability 14. Game B costs $3 and wins $10 with probability 15 or $5 with probability 110. Which statement is correct?

Not quite. The biggest prize does not decide it. Compare the expected profits.

Solution

  1. Game A: expected prize = $7 × 14 = $1.75, so expected profit = −$0.25.
  2. Game B: $10 × 15 + $5 × 110 = $2.50, so expected profit = −$0.50.
  3. Both are negative, and Game A loses less per go.

Answer: D. Both games lose money on average, but Game A loses less: $0.25 a go against $0.50.

Not quite. This forgets the costs. B costs $1 more.

Solution

  1. Game A: expected prize = $7 × 14 = $1.75, so expected profit = −$0.25.
  2. Game B: $10 × 15 + $5 × 110 = $2.50, so expected profit = −$0.50.
  3. Both are negative, and Game A loses less per go.

Answer: D. Both games lose money on average, but Game A loses less: $0.25 a go against $0.50.

Not quite. This is Game A's expected prize. Take away the $2 cost.

Solution

  1. Game A: expected prize = $7 × 14 = $1.75, so expected profit = −$0.25.
  2. Game B: $10 × 15 + $5 × 110 = $2.50, so expected profit = −$0.50.
  3. Both are negative, and Game A loses less per go.

Answer: D. Both games lose money on average, but Game A loses less: $0.25 a go against $0.50.

Correct.

Solution

  1. Game A: expected prize = $7 × 14 = $1.75, so expected profit = −$0.25.
  2. Game B: $10 × 15 + $5 × 110 = $2.50, so expected profit = −$0.50.
  3. Both are negative, and Game A loses less per go.

Apply

Run a fair stall

At a school fair, a player pays $1 and rolls two fair dice. A double six wins $18, any other double wins $3, and anything else wins nothing.

1. Find the probability of a double six and of any other double.

2. Show that the expected prize is 1112 of a dollar.

3. 120 people play. How much should the stall expect to make?

4. The stall wants the game to be fair. What should the double-six prize be, if the other prizes stay the same?

  1. There are 36 outcomes. P(double six) = 136. The other doubles are 1-1 to 5-5: P = 536.
  2. Expected prize = 18 × 136 + 3 × 536 = 1836 + 1536 = 3336 = 1112 of a dollar ✓.
  3. The stall makes 1 − 1112 = 112 of a dollar per go on average. 120 × 112 = $10.
  4. Fair: x36 + 1536 = 1, so x = 36 − 15 = $21.

Next: the Lesson 5 exit quiz, then the unit test. The unit test has 18 multiple-choice questions and five structured problems, 44 marks in all. No calculator. Allow 70 minutes.

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