Learning Objectives
- Construct probability distribution tables for discrete variables.
- Understand the concept of Expected Value, E(X), as a long-run average.
- Calculate the Expected Value of an event using the formula E(X) = Σ x • P(x).
- Apply Expected Value to real-world scenarios like financial investments and insurance.
- Determine if a game of chance is mathematically "fair".
Interactive Study Tools
These tools are here to help students slow down and think about the method before rushing to an answer.
Study Planner
Use this to break a topic into small study sessions.
Method Reminder
This quick guide reminds students what to do when they feel stuck.
- Read the story: what is the question really asking?
- Choose the method: identify the correct idea before calculating.
- Show the reason: do not skip the logic.
- Check the answer: make sure it fits the situation.
1. Probability Distributions
A probability distribution shows all the possible outcomes of an experiment along with the probability of each outcome occurring. It is usually laid out in a table format.
| Outcome (x) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Probability P(x) | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
| Number of Heads (x) | 0 | 1 | 2 |
|---|---|---|---|
| Probability P(x) | 1/4 | 2/4 (or 1/2) | 1/4 |
2. Expected Value (Mean)
The Expected Value, denoted as E(X), is the long-term average outcome of a random process. If you played a game thousands of times, what would your average score/payout be per game? To find it, you multiply each outcome by its probability, and add them all together.
E(X) = (1 × 1/6) + (2 × 1/6) + (3 × 1/6) + (4 × 1/6) + (5 × 1/6) + (6 × 1/6)
E(X) = 21/6 = 3.5. (Note: It's completely fine that 3.5 is not an actual number on the die!)
E(X) = (€50 × 1/100) + (€10 × 2/100) + (€0 × 97/100)
E(X) = 0.50 + 0.20 + 0 = €0.70 expected payout.
3. Fairness in Games of Chance
In mathematics, a game is considered FAIR if the Expected Value of your net profit is exactly zero.
Alternatively, a game is fair if: Cost to Play = Expected Payout.
First, find Expected Payout: E(X) = (€5 × 1/2) + (€0 × 1/2) = €2.50.
Compare to Cost: Cost (€2) ≠ Expected Payout (€2.50). The game is NOT fair (it actually favours the player!).
If Heads: Net Profit = €5 (win) - €2 (cost) = +€3.
If Tails: Net Profit = €0 (win) - €2 (cost) = -€2.
E(Net Profit) = (+€3 × 1/2) + (-€2 × 1/2) = 1.5 - 1 = +€0.50.
Because E(Net Profit) ≠ 0, the game is not fair. (Positive means good for player, negative means good for the house/casino).
📝 Exam Style Questions
Easy Q1: A spinner is divided into four equal sections: €10, €20, €30, and €40. What is the Expected Value of one spin?
Medium Q2: A lottery has 1,000 tickets. One ticket wins €500, five tickets win €100, and the rest win nothing. It costs €2 to buy a ticket. What is the Expected Net Profit for someone buying one ticket? Is the lottery fair?
Hard Q3: In a game, a player rolls a fair six-sided die. If they roll a 6, they win €X. If they roll a 1, 2, 3, 4, or 5, they lose €3. Find the value of X that makes this a fair game.
A1: Each section has a probability of 1/4.
E(X) = (10 × 0.25) + (20 × 0.25) + (30 × 0.25) + (40 × 0.25) = 2.5 + 5 + 7.5 + 10 = €25.
A2: Let's use the Net Profit method.
Win Top Prize (Prob 1/1000): Net = €500 - €2 = +€498
Win Small Prize (Prob 5/1000): Net = €100 - €2 = +€98
Lose (Prob 994/1000): Net = €0 - €2 = -€2
E(Net Profit) = (498 × 0.001) + (98 × 0.005) + (-2 × 0.994) = 0.498 + 0.490 - 1.988 = -€1.00.
The game is not fair because the Expected Net Profit is not €0 (the house wins €1 per ticket on average).
A3: For a fair game, Expected Net Profit must equal 0.
Win X (Prob 1/6). Lose €3 (Prob 5/6).
E(Net Profit) = (X × 1/6) + (-3 × 5/6) = 0
X/6 - 15/6 = 0
X/6 = 15/6
X = €15. The payout must be €15 for rolling a 6.
🔥 Challenge Questions (No Peeking!)
Easy C1: A probability distribution is given as: P(x=1)=0.2, P(x=2)=k, P(x=3)=2k, P(x=4)=0.2. Find the value of k, and then find E(X).
Medium C2: An insurance company sells a policy for €500. There is a 0.5% chance the policyholder will make a claim of €20,000, and a 2% chance they will make a claim of €5,000. What is the expected profit for the insurance company per policy sold?
Hard (HL) C3: Two fair dice are rolled and the sum of the scores is recorded. A player pays €5 to play. They win €A if the sum is a prime number, and win nothing otherwise. If the game is strictly mathematically fair, find the value of A.
(Answers at the very bottom of the page)
📌 Quick Summary
1. Probability Distribution: A table listing all outcomes ($x$) and their probabilities ($P(x)$). The sum of $P(x)$ is always 1.
2. Expected Value: The long term average. Multiply each top row by its bottom row and add them up: $E(X) = \sum x P(x)$.
3. Fair Game Rule: A game is only fair if the Expected Payout is exactly equal to the Cost of playing (meaning E(Net Profit) = 0).
4. House Edge: If E(Net Profit) is negative, the casino/house has the advantage!
🎓 Final Examiner Advice
When dealing with games of chance, read the question like a lawyer! Does the "payout" include getting your initial stake returned, or is the stake lost once placed? A game where you pay €10, win, and get handed €20 total means your net profit is only €10. Always establish the Net Profit table before calculating E(X) if the question asks about fairness.
Challenge Answers:
C1: Probabilities must sum to 1. So, 0.2 + k + 2k + 0.2 = 1. => 3k + 0.4 = 1 => 3k = 0.6 => k = 0.2. The distribution is 0.2, 0.2, 0.4, 0.2. E(X) = (1×0.2)+(2×0.2)+(3×0.4)+(4×0.2) = 0.2+0.4+1.2+0.8 = 2.6.
C2: The company takes in €500 100% of the time. Expected Payout = (€20,000 × 0.005) + (€5,000 × 0.02) + (€0 × 0.975) = €100 + €100 = €200. Expected Profit = Income - Expected Payout = €500 - €200 = €300 per policy.
C3: Prime sums from 2 dice (2, 3, 5, 7, 11). Outcomes mapping to primes: 2(1/36), 3(2/36), 5(4/36), 7(6/36), 11(2/36). Total Probability of prime = 15/36 = 5/12. Fair game means Expected Payout = Cost. So, (A × 5/12) + (0 × 7/12) = 5. => A(5/12) = 5. => A = 5 × (12/5) => A = €12. The payout must be €12.