IB Maths AA SL
Topic 4 — Probability Distributions
Paper 1 & 2
~9 min read
Expected Values
If you played a game forever, what would your average outcome be? That’s the expected value — the long-run mean of a random variable. It’s the maths casinos use to set odds, and the maths you use to decide if a game is fair.
📘 What you need to know
- The expected value E(X) is the long-run average of a random variable.
- Formula: E(X) = ∑ x · P(X = x) (in the formula booklet)
- E(X) does not have to be a value X can actually take! e.g. expected number of heads in 5 flips is 2.5.
- For a symmetric distribution, E(X) = the centre of symmetry.
- A fair game has E(gain) = 0 — players neither win nor lose on average.
- If E(gain) > 0 → expected gain. If E(gain) < 0 → expected loss.
What does E(X) mean?
E(X) is the expected value — also called the mean — of a discrete random variable. Think of it as the average outcome you’d get if you repeated the experiment over and over.
The formula is in the formula booklet:
In plain English: multiply each value by its probability, then add them all up.
🤔 Why does this work as an “average”?
If a value has a high probability, it’ll happen often — so it pulls the average toward itself. Multiplying by the probability gives that value its proper “weight”. The result is a weighted average — exactly what a long-run mean should be.
The expected value doesn’t have to be a real outcome!
This trips students up: the expected value can be impossible in any single trial.
- Flip a coin 5 times — expected number of heads is 5 × 0.5 = 2.5. You’ll never actually get 2.5 heads in one go!
- Roll a fair dice — expected value is 3.5. There’s no 3.5 face on the dice.
E(X) is what you’d average to if you ran the experiment millions of times. In any individual trial, you get a real value (like 2 heads). Across many trials, the average creeps closer and closer to E(X).
How to calculate E(X)
The 3-step method
- Get the probability distribution as a table. Build one if needed.
- Multiply each value of x by its probability P(X = x).
- Add up all the products. The result is E(X).
A quick example
Say X has the distribution:
| x | 1 | 2 | 3 | 4 |
|---|
| P(X = x) | 0.1 | 0.3 | 0.4 | 0.2 |
Then E(X) = 1×0.1 + 2×0.3 + 3×0.4 + 4×0.2 = 0.1 + 0.6 + 1.2 + 0.8 = 2.7.
📍Lay out your work as x × P first
Write out each multiplication clearly. It avoids skipping a term and makes it easy to spot errors. Examiners also award method marks for showing the products even if you arithmetic-fail at the end.
Spotting symmetric distributions
If a distribution is symmetric (the values and their probabilities are mirror images), the expected value is just the centre of symmetry. No need to add anything!
Example: if X takes values 1, 5, 9 with probabilities 0.3, 0.4, 0.3 — the values and the probabilities mirror around 5, so E(X) = 5.
🧠Memory trick: “Symmetric? Find the middle”
If both the values AND the probabilities are symmetric, the mean is just the centre. This saves you from doing the full calculation when distributions look “balanced” around a single point.
Always double-check that BOTH the values and probabilities are symmetric. If the values are 1, 5, 9 but the probabilities are 0.5, 0.3, 0.2, the distribution isn’t symmetric — and you’d need the full formula.
Is a game fair?
One of the most common exam patterns: someone pays to play a game and wins a prize based on chance. Is it fair?
Let X be the player’s net gain or loss (prize − cost). Then:
Expected gain
E(X) > 0
Player wins on average. Game favours the player.
Fair game
E(X) = 0
Player breaks even on average.
Expected loss
E(X) < 0
Player loses on average. Game favours the house.
Two ways to set up a fairness question
You can either:
- Method A: Calculate the expected prize, then subtract the cost. Fair if expected prize = cost.
- Method B: Build a “net gain” distribution where each value already has the cost subtracted. Then fair if E(net gain) = 0.
Both work, but Method A is usually quicker — calculate E(prize), compare to the cost.
🤔 Why isn’t a fair game profitable for the seller?
If E(gain) = 0 for the player, the house breaks even too — they don’t make money. That’s why real-world casinos and lotteries always set odds so E(gain) is negative for the player. The “house edge” is just the size of that negative expected value.
Expected number of occurrences
If you do an experiment n times and the probability of a particular outcome is p, the expected number of times it happens is just np.
Example: roll a fair dice 60 times. Expected number of 6’s = 60 × 16 = 10.
📍Two different “expected” formulas — don’t confuse them
E(X) = ∑ x·P(X = x) → for the average value of a random variable.
Expected count = np → for the average number of times an event happens. Choose the right one based on what’s being asked.
Worked examples
WE 1Calculate E(W) and decide if a game is fair
Daphne pays $15 to play a game where she wins a prize of $1, $5, $10 or $100. The random variable W represents the prize, with the distribution:
| w | 1 | 5 | 10 | 100 |
|---|
| P(W = w) | 0.35 | 0.5 | 0.05 | 0.1 |
(a) Calculate the expected value of Daphne’s prize. (b) Determine whether the game is fair.
Use formula E(W) = ∑ w · P(W = w). Then compare expected prize to the $15 cost.part (a) — expected prize
Multiply each prize by its probability:
E(W) = 1 × 0.35 + 5 × 0.5 + 10 × 0.05 + 100 × 0.1
= 0.35 + 2.5 + 0.5 + 10
= 13.35
E(W) = $13.35part (b) — is the game fair?
A game is fair when expected gain = 0 (prize = cost).
Expected gain: prize − cost = 13.35 − 15 = −1.65
Expected loss = $1.65 per play.
NOT fair — Daphne expects to lose $1.65
always state the expected gain/loss number — it earns marks!
WE 2Expected value of a fair dice roll
A fair 6-sided dice is rolled. Let X be the number shown. Find E(X).
Each face has probability 1/6 — uniform distribution.
Apply formula:
E(X) = 1 × 16 + 2 × 16 + 3 × 16 + 4 × 16 + 5 × 16 + 6 × 16
= 1+2+3+4+5+66 = 216 = 3.5
E(X) = 3.5
3.5 isn’t a value on the dice — but it’s the long-run average roll!
WE 3Use E(X) to find an unknown
The discrete random variable X has the distribution:
Given that E(X) = 1.5, find p.
Two equations: ∑P = 1 AND E(X) = 1.5. Use either (or both) to find p.
Method 1 — use ∑P = 1:
0.2 + p + 0.3 + 0.1 = 1
p = 1 − 0.6 = 0.4
Check with E(X):
0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 0 + 0.4 + 0.6 + 0.3 = 1.3
But E(X) should be 1.5 — contradiction!
So the table must have ANOTHER unknown too. Re-check assumptions.
Use the E(X) equation directly:
0(0.2) + 1·p + 2(0.3) + 3(0.1) = 1.5
p + 0.6 + 0.3 = 1.5
p = 0.6
p = 0.6 (using the E(X) equation)
when given E(X), use it directly! (the table likely has a 2nd unknown elsewhere)
WE 4Expected number of occurrences
A spinner has 5 equal sectors: 2 red, 2 blue, 1 green. The spinner is spun 100 times.
(a) Find the expected number of greens. (b) Find the expected number of reds.
Expected occurrences = n × p.part (a) — green
P(green): 15 = 0.2
Expected = 100 × 0.2 = 20
Expected greens = 20part (b) — red
P(red): 25 = 0.4
Expected = 100 × 0.4 = 40
Expected reds = 40
“expected” is a long-run average — actual count may vary!
WE 5Find the cost that makes a game fair
A game has prizes $2, $5, $20 with probabilities 0.5, 0.4, 0.1. What entry fee makes the game fair?
A fair game means E(prize) = entry fee. Calculate E(prize) first.
Apply formula:
E(prize) = 2 × 0.5 + 5 × 0.4 + 20 × 0.1
= 1 + 2 + 2 = 5
For fairness, fee = E(prize):
Fair entry fee = $5
if charge = expected prize, the game is fair (E(gain) = 0)
💡 Top tips
- Use the formula carefully: E(X) = ∑ x × P(X = x). Multiply each value by its probability, then add.
- Lay out the products clearly so the examiner can see your method — earns method marks even if final answer is wrong.
- For symmetric distributions, E(X) = the centre of symmetry — saves time.
- For fairness questions, fair = E(gain) = 0. Compare expected prize to the cost.
- Expected occurrences = np — different from E(X). Pick the right formula based on what’s asked.
- E(X) doesn’t have to be a possible value of X — it’s an average, not a specific outcome.
- Always check probabilities sum to 1 before computing E(X).
- If E(X) is given, use it as an equation to find unknowns — it’s often quicker than ∑P = 1.
⚠ Common mistakes
- Forgetting to multiply by probability. E(X) is NOT just the average of the values — each value gets weighted by its probability.
- Confusing E(X) with np. They’re different: E(X) is the average value of X; np is the expected count over n trials.
- Saying “E(X) = 2.5 is impossible”. It’s not a contradiction — averages can be non-integer even when X only takes integer values.
- Forgetting to subtract the cost in fair-game questions. Expected prize alone doesn’t tell you if a game is fair — compare to the entry fee.
- Saying a game is “fair” because the cost equals one of the prizes. Fair means E(gain) = 0, NOT just that “you can win back your money”.
- Skipping the table when given a function-defined distribution. Build the table first.
- Including impossible outcomes in the sum. Only multiply values that have non-zero probability.
- Adding the probabilities together by mistake. You add the products x × P(x), not the probabilities themselves.
🎉 You’ve finished the entire Probability Distributions subtopic! You can now describe a discrete random variable, compute probabilities for any inequality, and find the long-run average outcome. The next chunk of Topic 4 looks at two specific discrete distributions — the Binomial and the Normal distribution — both of which use the same fundamentals you’ve just learned.
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