Mathematics Advanced • Year 12 • Module 5 • Lesson 5
Discrete Probability Distributions
Apply expected value and variance to gambling, insurance, investment, manufacturing and game-design contexts.
Problem 1 — European roulette
In European roulette, you bet $1 on a single number. You win $35 (net profit) with probability 1/37 and lose $1 with probability 36/37.
Set up: What are we solving for? Let W = your net winnings per spin.
(i) Construct the probability distribution table for W and verify it is valid. 2 marks
(ii) Find E(W). State whether the game is fair, favourable to the player, or favourable to the casino. 2 marks
(iii) Find Var(W) and SD(W). A player who plays 100 spins is approximately expected to lose how much? Comment in one sentence on why the casino loves "low expected value per spin × high variance per spin × many players". 3 marks
Stuck? Revisit lesson § Real-World Anchor — Roulette.Problem 2 — Travel insurance premiums
An insurer offers $5 000 cover for $80. Long-run data shows P(claim) = 0.01, and average payout per claim is $5 000.
Set up: What are we solving for? Let X = insurer's profit per policy.
(i) Construct the distribution table for X. The two outcomes are "no claim" (profit = $80) and "claim" (profit = $80 − $5 000 = −$4 920). 2 marks
(ii) Find E(X). Is the premium structure profitable per policy? 2 marks
(iii) The insurer wants E(X) ≥ $50 per policy on average. What is the minimum premium they can charge (keeping all other figures fixed)? 2 marks
Problem 3 — Investment portfolio choice
Investment A returns +12% with probability 0.5 and +8% with probability 0.5. Investment B returns +25% with probability 0.4 and +0% with probability 0.6.
Set up: What are we solving for?
(i) Find E(R_A) and E(R_B), the expected returns. 2 marks
(ii) Find Var(R_A) and Var(R_B), and hence SD(R_A) and SD(R_B). 3 marks
(iii) A risk-averse investor (e.g. a retiree) would prefer which investment? A risk-tolerant investor (e.g. a young saver with a 30-year horizon) might prefer which? Justify using your computed E(R) and SD(R). 2 marks
Stuck? Same expected return ≠ same risk; SD is the standard measure of "spread / risk".Problem 4 — Number of defective items in a batch
A factory produces items with a 5% defect rate. A quality inspector samples 4 items and records X = number defective.
The distribution (binomial with n = 4, p = 0.05) is:
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X = x) | 0.8145 | 0.1715 | 0.0135 | 0.00048 | 0.000006 |
Set up: What are we solving for?
(i) Verify the distribution sums to (approximately) 1. 1 mark
(ii) Find E(X) using the formula E(X) = Σ x·P(X = x). Compare with the binomial shortcut E(X) = np. 2 marks
(iii) The inspector raises an alarm if ≥ 2 defective items are found in a batch. Find P(alarm) and state the expected number of "false alarms" per 1 000 batches inspected. 2 marks
Problem 5 — Designing a school-fete dice game
You design a stall: customers pay $c to play; they roll a single fair die; you pay them $x dollars where x is the number rolled, capped so that x = 6 pays only $5 (no jackpot).
Set up: What are we solving for?
(i) Construct the distribution for W = payout (in dollars). It has outcomes 1, 2, 3, 4, 5, 5, each with probability 1/6 — note 5 appears twice (the cap collapses 6 onto 5). 2 marks
(ii) Find E(W). Hence find the minimum entry fee c that gives the stall a non-negative expected profit per play. 2 marks
(iii) You set c = $4. Find E(profit per play) and E(profit from 200 plays during the fete). State whether the cap on the jackpot was necessary (i.e. would the game still be viable without it at c = $4?). 3 marks
Stuck? Without the cap, W = X (the roll), so E(W) = 3.5; with cap, replace one of the 6 with a 5.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Roulette
Set up. A two-outcome distribution; we compute E and Var, then scale across many spins.
(i) Distribution: W = 35 with P = 1/37; W = −1 with P = 36/37. Sum: 1/37 + 36/37 = 37/37 = 1 ✓.
(ii) E(W) = 35 × (1/37) + (−1) × (36/37) = (35 − 36)/37 = −1/37 ≈ −$0.027. Negative — favours the casino (a 2.7% house edge).
(iii) E(W²) = 35² × (1/37) + (−1)² × (36/37) = (1225 + 36)/37 = 1261/37 ≈ 34.08. Var(W) = 34.08 − (1/37)² ≈ 34.08 − 0.000731 ≈ 34.08; SD(W) ≈ $5.84.
Over 100 spins, expected loss ≈ 100 × 0.027 = $2.70. Casino loves it because: small per-spin loss feels invisible to the player, high variance gives occasional big wins (which lock in addiction), and 1000s of players average out — the law of large numbers guarantees the casino's edge.
Problem 2 — Travel insurance
Set up. Insurer's profit per policy is a 2-outcome random variable.
(i) X = 80 with P = 0.99; X = −4 920 with P = 0.01.
(ii) E(X) = 80 × 0.99 + (−4 920) × 0.01 = 79.20 − 49.20 = $30 per policy. Profitable on average.
(iii) Let premium be p. Then E(X) = p × 0.99 + (p − 5000) × 0.01 = p − 50. Setting p − 50 = 50 ⇒ p = $100.
Problem 3 — Investments A vs B
Set up. Compare two single-period return distributions on E and SD.
(i) E(R_A) = 0.5(12) + 0.5(8) = 10%; E(R_B) = 0.4(25) + 0.6(0) = 10%. Same expected return.
(ii) E(R_A²) = 0.5(144) + 0.5(64) = 72 + 32 = 104; Var(R_A) = 104 − 100 = 4; SD(R_A) = 2%.
E(R_B²) = 0.4(625) + 0.6(0) = 250; Var(R_B) = 250 − 100 = 150; SD(R_B) ≈ 12.2%.
(iii) Risk-averse → Investment A: same expected return at far lower spread (SD 2% vs 12.2%), so worst-case outcomes are much milder. Risk-tolerant → Investment B: same expectation but with a 40% chance of a 25% win, which compounds over a long horizon if you can stomach the 60% chance of 0%.
Problem 4 — Defective items
Set up. A 5-value binomial-style distribution; we compute E and the "≥ 2" tail probability.
(i) Σ P = 0.8145 + 0.1715 + 0.0135 + 0.00048 + 0.000006 ≈ 1.0000 ✓.
(ii) E(X) = 0(0.8145) + 1(0.1715) + 2(0.0135) + 3(0.00048) + 4(0.000006) ≈ 0 + 0.1715 + 0.027 + 0.00144 + 0.000024 ≈ 0.20. Binomial shortcut: E(X) = np = 4 × 0.05 = 0.20 ✓.
(iii) P(alarm) = P(X ≥ 2) = 0.0135 + 0.00048 + 0.000006 ≈ 0.0140. In 1 000 batches, expected alarms ≈ 1 000 × 0.0140 = 14 alarms.
Problem 5 — Fete dice game
Set up. Capped payout distribution; compute E(payout) and design entry fee.
(i) W: 1, 2, 3, 4, 5, 5 (one for each face), each with probability 1/6. Equivalently W: 1 (P=1/6), 2 (P=1/6), 3 (P=1/6), 4 (P=1/6), 5 (P=2/6).
(ii) E(W) = (1 + 2 + 3 + 4)(1/6) + 5(2/6) = 10/6 + 10/6 = 20/6 ≈ $3.33. Minimum fair entry fee c = $3.33.
(iii) At c = $4: E(profit per play) = 4 − 3.33 = $0.67. E(profit from 200 plays) ≈ $133. Without the cap, E(W) = 3.5, so profit per play at c = 4 would be only $0.50 and the cap is not strictly necessary at c = $4 — it adds about 17¢ per play and helps prevent the perception that one in six players "wins big".