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Expected value of a discrete random variable: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Calculate and interpret the expected value (mean) of a discrete random variable and apply it to decision contexts such as fair games”

8 min read

Quick answer

The expected value E(X)=∑x P(X=x)E(X)=\sum x\,P(X=x) is the probability-weighted average of the values, interpreted as the long-run mean over many repetitions. For a linear function, E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b. A game is fair when the expected gain is zero.

Jump to a section
  1. What this dot point is asking
  2. Definition and interpretation
  3. Expected value of a function
  4. Fair games
  5. Why linearity works
  6. Long-run interpretation in context
  7. Exam-style questions

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

Expected value of a discrete random variable

E(X)=μ=∑xx P(X=x).E(X)=\mu=\sum_x x\,P(X=x).

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 XX; it is the average you would approach over a large number of repetitions. The mean of a fair die is 3.53.5, even though no face shows 3.53.5.

Expected value from a table

A variable XX has P(X=0)=0.4P(X=0)=0.4, P(X=1)=0.35P(X=1)=0.35, P(X=2)=0.25P(X=2)=0.25. Find E(X)E(X).

Step 1: Identify the formula to apply

The expected value weights each outcome by its probability, so we use E(X)=∑x P(X=x)E(X)=\sum x\,P(X=x). Every possible value of XX must appear in the sum, including x=0x=0, 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 XX.

E(X)=0(0.4)+1(0.35)+2(0.25)=0+0.35+0.5=0.85.E(X) = 0(0.4) + 1(0.35) + 2(0.25) = 0 + 0.35 + 0.5 = 0.85.

Final answer: E(X)=0.85E(X) = 0.85

Expected value of a function

If you earn a payoff that is a function of XX, the expected payoff weights each function value by the same probabilities:

E(g(X))=∑xg(x) P(X=x).E(g(X))=\sum_x g(x)\,P(X=x).

For a linear function this simplifies to E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b, because expectation is linear.

Expected value of a linear payoff

A game pays $3\$3 for each unit of XX above, plus a $2\$2 entry refund, so the payout is W=3X+2W=3X+2. Find E(W)E(W).

Step 1: Recognise the linear structure

Because W=3X+2W=3X+2 is a linear function of XX, the linearity rule E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b applies directly. This avoids recomputing a weighted sum from scratch using the full distribution.

Step 2: Substitute the known expected value

We use E(X)=0.85E(X)=0.85 from the previous example, with a=3a=3 and b=2b=2.

E(W)=3E(X)+2=3(0.85)+2=2.55+2=4.55.E(W) = 3E(X) + 2 = 3(0.85) + 2 = 2.55 + 2 = 4.55.

Final answer: E(W)=$4.55E(W) = \$4.55

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.

Deciding if a game is fair

A player pays $2\$2 to play. They win $5\$5 with probability 0.30.3 and nothing otherwise. Determine whether the game is fair.

Step 1: Find the expected winnings

The winnings variable XX takes the value 55 with probability 0.30.3 and 00 with probability 0.70.7. We compute E(X)E(X) as the probability-weighted average of these outcomes.

E(X)=5(0.3)+0(0.7)=1.5.E(X) = 5(0.3) + 0(0.7) = 1.5.

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 $2\$2 fee.

E(gain)=1.5−2=−0.5.E(\text{gain}) = 1.5 - 2 = -0.5.

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: E(gain)=−$0.50E(\text{gain}) = -\$0.50 per play; the game is not fair.

Why linearity works

The rule E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b follows directly from the definition. Each value xx becomes ax+bax+b with the same probability, so

E(aX+b)=∑(ax+b)P(X=x)=a∑xP(X=x)+b∑P(X=x)=aE(X)+b,E(aX+b)=\sum (ax+b)P(X=x)=a\sum xP(X=x)+b\sum P(X=x)=aE(X)+b,

using ∑P(X=x)=1\sum P(X=x)=1. This is why a constant shift bb moves the mean by exactly bb and a scale factor aa multiplies it by aa. The same reasoning does not extend to nonlinear functions, which is why E(X2)E(X^{2}) must be computed term by term rather than as [E(X)]2[E(X)]^{2}.

Long-run interpretation in context

The phrase "expected value" can mislead: E(X)E(X) is the average over very many repetitions, not the value to expect on a single trial. If a raffle ticket has expected winnings of $0.80\$0.80 but costs $1\$1, no single ticket returns 8080 cents, yet buying thousands of tickets loses about 2020 cents each on average. SCSA decision questions hinge on this long-run reading, so always interpret E(X)E(X) as a per-trial average across repeated play.

Common errors
Averaging the values without weighting
E(X)E(X) weights each value by its probability; a plain average of the values is wrong unless all are equally likely.
Expecting E(X)E(X) 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
E(X2)≠[E(X)]2E(X^{2})\neq [E(X)]^{2} in general; only linear functions split as E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b.

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 marks
Calculator-assumed. A spinner pays $1\$1, $2\$2 or $5\$5 with probabilities 0.50.5, 0.30.3 and 0.20.2 respectively. It costs $3\$3 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) E(X)=1(0.5)+2(0.3)+5(0.2)=0.5+0.6+1.0=2.10E(X)=1(0.5)+2(0.3)+5(0.2)=0.5+0.6+1.0=2.10. The expected payout is $2.10\$2.10.

(b) Net expected gain to the player is payout minus cost: 2.10−3.00=−0.902.10-3.00=-0.90. The expected gain is negative, so the game is not fair; it favours the operator by $0.90\$0.90 per game on average.

Markers reward the weighted payout, subtracting the $3\$3 cost, and a clear statement of who the game favours and by how much.

WACE 2023-style4 marks
Calculator-free. A discrete random variable XX has E(X)=2.5E(X)=2.5. Find (a) E(4X−3)E(4X-3) and (b) E(X)E(X) if every value of XX is doubled.
Show worked answer →

A test of the linearity of expectation.

(a) By linearity, E(4X−3)=4E(X)−3=4(2.5)−3=10−3=7E(4X-3)=4E(X)-3=4(2.5)-3=10-3=7.

(b) Doubling every value gives a new variable 2X2X, so E(2X)=2E(X)=2(2.5)=5E(2X)=2E(X)=2(2.5)=5.

Markers reward applying E(aX+b)=aE(X)+bE(aX+b)=aE(X)+b correctly, including the −3-3 in part (a) and the scaling in part (b).

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