11.2 Expected Values, Expected Value Tables and Their Limitations
Key Takeaways
The expected value is EV = Σ(outcome × probability), the long-run average result if the same decision were repeated many times.
An expected value table lists each possible decision against each possible outcome, calculates the payoff in every cell, and selects the decision with the highest expected value.
The expected value may not be a possible outcome of a single decision, so it is most reliable for decisions that are repeated many times.
Expected values ignore the spread of outcomes and assume the decision-maker is risk-neutral, so two options with equal EVs can carry very different risks.
Why this topic is examined
Syllabus area D1(b) asks you to demonstrate the use of expected values and joint probabilities in decision-making, including expected values and expected value tables and the limitations of expected values. Expect number-entry questions (calculate an EV or the best option from a payoff table) and conceptual questions about when an EV is a sensible basis for a decision. Section 11.1 covered the probability rules this section uses.
The expected value
where is each possible outcome and its probability. The probabilities must sum to 1.
Worked example
A company is considering a new subscription service. Its estimates of net present value (NPV) under three demand scenarios are:
| Demand | Probability | NPV $ | NPV × probability $ |
|---|---|---|---|
| Weak | 0.25 | (40,000) | (10,000) |
| Moderate | 0.50 | 50,000 | 25,000 |
| Strong | 0.25 | 120,000 | 30,000 |
| Expected value | 1.00 | 45,000 |
The expected NPV is positive ($45,000), so on an expected value basis the project is worthwhile. Note that $45,000 is not one of the possible outcomes: the actual result will be a $40,000 loss, a $50,000 gain or a $120,000 gain.
Expected values from joint probabilities
In Section 11.1, sales volume and contribution per unit were combined into four joint outcomes. The expected contribution is:
| Joint outcome | Joint probability | Contribution $ | Weighted $ |
|---|---|---|---|
| 10,000 units at $5 | 0.42 | 50,000 | 21,000 |
| 10,000 units at $4 | 0.18 | 40,000 | 7,200 |
| 8,000 units at $5 | 0.28 | 40,000 | 11,200 |
| 8,000 units at $4 | 0.12 | 32,000 | 3,840 |
| Expected contribution | 1.00 | 43,240 |
Because the two variables are independent, the same answer comes from expected volume × expected contribution per unit: (0.6 × 10,000 + 0.4 × 8,000) × (0.7 × $5 + 0.3 × $4) = 9,200 × $4.70 = $43,240. After fixed costs of $41,000, the expected profit is $2,240, even though the probability of a loss is 0.58. This shows why expected values should not be read in isolation.
Expected value tables (payoff tables)
An expected value table sets out the payoff for every combination of decision (what management chooses) and outcome (what actually happens), then calculates the expected value of each decision.
Worked example: how many cakes to bake
A bakery buys cakes for $4 each and sells them for $10. Unsold cakes are thrown away at the end of the day. Daily demand is either 100, 200 or 300 cakes, with probabilities 0.3, 0.5 and 0.2. The bakery must decide in the morning whether to buy 100, 200 or 300 cakes.
Step 1: calculate each payoff. Payoff = $10 × cakes sold − $4 × cakes bought, where cakes sold is the lower of demand and cakes bought.
| Cakes bought | Demand 100 (p = 0.3) | Demand 200 (p = 0.5) | Demand 300 (p = 0.2) |
|---|---|---|---|
| Buy 100 | 1,000 − 400 = 600 | 600 | 600 |
| Buy 200 | 1,000 − 800 = 200 | 2,000 − 800 = 1,200 | 1,200 |
| Buy 300 | 1,000 − 1,200 = (200) | 2,000 − 1,200 = 800 | 3,000 − 1,200 = 1,800 |
Step 2: calculate the expected value of each decision.
| Decision | Calculation | Expected payoff $ |
|---|---|---|
| Buy 100 | (0.3 × 600) + (0.5 × 600) + (0.2 × 600) | 600 |
| Buy 200 | (0.3 × 200) + (0.5 × 1,200) + (0.2 × 1,200) | 900 |
| Buy 300 | (0.3 × −200) + (0.5 × 800) + (0.2 × 1,800) | 700 |
Step 3: choose. On an expected value basis the bakery should buy 200 cakes, giving an expected daily profit of $900.
Tip
Work down each row (decision) of the payoff table to find its EV, and compare the EVs. A common error is to average across a column (an outcome), which mixes different decisions together.
Because the bakery makes the same decision every day, the expected value is a sensible guide: over many days, the average daily profit from buying 200 cakes should be close to $900.
Limitations of expected values
| Limitation | Explanation |
|---|---|
| Long-run average | The EV is the average result over many repetitions. For a one-off decision the EV may never actually occur, as in the subscription example. |
| Ignores risk (spread) | Two options can have the same EV but very different ranges of outcomes. The EV says nothing about the chance of a large loss. |
| Assumes risk neutrality | Using EV alone treats a decision-maker as indifferent to risk. A risk-averse manager might prefer a lower but safer outcome. |
| Depends on probability estimates | Probabilities are often estimates or based on past data that may not hold in future. |
| Simplified outcomes | Real outcomes usually form a continuous range, not the two or three values in a table. |
Same EV, different risk
| Option | Outcomes $ | Probabilities | EV $ |
|---|---|---|---|
| A | 50,000 or 70,000 | 0.5 / 0.5 | 60,000 |
| B | (40,000) or 160,000 | 0.5 / 0.5 | 60,000 |
Both options have an EV of $60,000, but Option B has a 50% chance of a $40,000 loss. A small company might not survive that loss. The standard deviation (Section 11.3) measures this spread.
Note
Techniques for decisions under uncertainty, where probabilities cannot be estimated (such as maximin, maximax and minimax regret), are not examined in BA2: the syllabus states that candidates will not be asked to apply techniques to deal with uncertainty. Focus on expected values, joint probabilities, measures of dispersion and the normal distribution.
A project's possible profits are $20,000 with probability 0.3, $50,000 with probability 0.5 and $80,000 with probability 0.2. What is the expected profit?
$50,000
$47,000
$41,000
$150,000
A newsagent buys magazines for $2 and sells them for $5; unsold copies have no value. Demand is either 40 copies or 60 copies, each with probability 0.5. What is the expected profit if the newsagent orders 60 copies?
$180
$120
$80
$130
Why is an expected value a less reliable basis for a one-off strategic investment than for a decision repeated every day?
Expected values cannot be calculated when some outcomes are losses
The expected value is a long-run average that may not match the result of a single decision, and it ignores the spread of possible outcomes
Expected values can only be used when all probabilities are equal
The expected value always overstates the most likely outcome
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