Expected value of a discrete random variable: WACE Year 12 Mathematics Methods Unit 3
“Calculate and interpret the expected value (mean) of a discrete random variable and apply it to decision contexts such as fair games”
The expected value is the probability-weighted average of the values, interpreted as the long-run mean over many repetitions. For a linear function, . A game is fair when the expected gain is zero.
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What this dot point is asking
SCSA Unit 3 defines the expected value of a discrete random variable as its mean. This dot point asks you to compute it, interpret it as a long-run average, transform it under a linear change, and apply it to decisions such as whether a game is fair. It is examined in both sections.
Definition and interpretation
Each value is weighted by its probability, so likely outcomes pull the mean toward them. The expected value is not necessarily an attainable value of ; it is the average you would approach over a large number of repetitions. The mean of a fair die is , even though no face shows .
A variable has , , . Find .
Step 1: Identify the formula to apply
The expected value weights each outcome by its probability, so we use . Every possible value of must appear in the sum, including , because multiplying by its probability correctly contributes zero.
Step 2: Compute the weighted sum and interpret
Multiplying each value by its probability and adding gives the long-run average. The result need not equal any actual value of .
Final answer:
Expected value of a function
If you earn a payoff that is a function of , the expected payoff weights each function value by the same probabilities:
For a linear function this simplifies to , because expectation is linear.
A game pays for each unit of above, plus a entry refund, so the payout is . Find .
Step 1: Recognise the linear structure
Because is a linear function of , the linearity rule applies directly. This avoids recomputing a weighted sum from scratch using the full distribution.
Step 2: Substitute the known expected value
We use from the previous example, with and .
Final answer:
Fair games
A game is fair if the expected net gain to the player is zero, meaning the expected winnings equal the cost to play. If the expected gain is negative, the game favours the operator; if positive, it favours the player. Commercial games are designed with a small negative expected gain to the player (the "house edge"), which guarantees the operator a profit over many plays even though individual players sometimes win. SCSA decision questions often ask you to set the expected gain to zero and solve for a fair entry price or a fair prize.
A player pays to play. They win with probability and nothing otherwise. Determine whether the game is fair.
Step 1: Find the expected winnings
The winnings variable takes the value with probability and with probability . We compute as the probability-weighted average of these outcomes.
Step 2: Subtract the cost to find the expected gain
A fair game requires the expected net gain to be zero. The net gain per play is the winnings minus the entry cost, so we subtract the fee.
Step 3: Interpret the sign
A negative expected gain means the player loses money on average over many plays. The game therefore favours the operator, not the player.
Final answer: per play; the game is not fair.
Why linearity works
The rule follows directly from the definition. Each value becomes with the same probability, so
using . This is why a constant shift moves the mean by exactly and a scale factor multiplies it by . The same reasoning does not extend to nonlinear functions, which is why must be computed term by term rather than as .
Long-run interpretation in context
The phrase "expected value" can mislead: is the average over very many repetitions, not the value to expect on a single trial. If a raffle ticket has expected winnings of but costs , no single ticket returns cents, yet buying thousands of tickets loses about cents each on average. SCSA decision questions hinge on this long-run reading, so always interpret as a per-trial average across repeated play.
- Averaging the values without weighting
- weights each value by its probability; a plain average of the values is wrong unless all are equally likely.
- Expecting to be an attainable value
- The mean need not be one of the possible outcomes.
- Forgetting the entry cost in a fair-game question
- Net gain is expected winnings minus the cost to play.
- Misusing linearity for nonlinear functions
- in general; only linear functions split as .
Exam-style questions
Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
WACE 2021-style6 marksCalculator-assumed. A spinner pays , or with probabilities , and respectively. It costs to play. (a) Find the expected payout per spin. (b) Determine whether the game is fair, and if not, by how much it favours the operator per game.Show worked answer →
A fair-game decision built on expected value.
(a) . The expected payout is .
(b) Net expected gain to the player is payout minus cost: . The expected gain is negative, so the game is not fair; it favours the operator by per game on average.
Markers reward the weighted payout, subtracting the cost, and a clear statement of who the game favours and by how much.
WACE 2023-style4 marksCalculator-free. A discrete random variable has . Find (a) and (b) if every value of is doubled.Show worked answer →
A test of the linearity of expectation.
(a) By linearity, .
(b) Doubling every value gives a new variable , so .
Markers reward applying correctly, including the in part (a) and the scaling in part (b).