Discrete Probability Distributions
A casino designs a new dice game. To know if the house will profit over thousands of plays, they need $E(X)$ — the expected value. By the end of this lesson you'll build probability distributions from scratch, calculate expected value and variance, and understand exactly why "the house always wins."
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
You roll a fair die and win the number of dollars showing on the face. Is your expected winnings always one of the possible outcomes ($1, $2, $3, $4, $5$, or $6$)? Make a prediction before reading on.
Everything in this lesson flows from two formulas. Memorise them now and the worked examples will feel obvious.
Expected value is the weighted average of all outcomes — multiply each value by its probability and sum them up. Variance uses the computational shortcut: find $E(X^2)$ then subtract $[E(X)]^2$.
Key facts
- $E(X) = \sum x \cdot P(X = x)$
- $\text{Var}(X) = E(X^2) - \mu^2$
- $\text{SD}(X) = \sqrt{\text{Var}(X)}$
Concepts
- A random variable assigns numbers to outcomes
- Expected value is a weighted average — not always achievable
- Variance measures spread; higher variance means more uncertainty
Skills
- Construct a probability distribution table from a scenario
- Calculate $E(X)$, $\text{Var}(X)$, and $\text{SD}(X)$
- Use expected value to analyse games and decisions
A random variable $X$ assigns a numerical value to each outcome of a random experiment. We use capital $X$ for the variable and lowercase $x$ for the values it can take. A probability distribution lists every possible value together with its probability.
Probability mass function: bar heights show $P(X=x)$ for each value. All bars must sum to 1.
Requirements for a valid distribution:
- Every probability is between 0 and 1: $0 \leq P(X = x) \leq 1$
- The probabilities sum to 1: $\displaystyle\sum P(X = x) = 1$
Example — fair die:
| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X=x)$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ |
This is a uniform distribution — all outcomes equally likely.
A random variable $X$ assigns a number to each outcome of a random experiment; A valid probability distribution requires $0 \le P(X=x) \le 1$ and $\sum P(X=x) = 1$
Pause — copy the two validity conditions for a probability distribution: $0 \leq P(X=x) \leq 1$ for each value, and $\sum P(X=x) = 1$ (all probabilities sum to exactly 1) into your book.
Did you get this? True or false: a probability distribution is valid if all individual probabilities are between 0 and 1 and they sum to exactly 1.
Expected value · the long-run average
We just saw that a probability distribution lists all possible values of $X$ with their probabilities, and the probabilities must sum to 1. That raises a question: given this distribution, what single number best summarises "where the distribution is centred" in the long run? This card answers it → the expected value $E(X) = \sum x \cdot P(X=x)$: the probability-weighted average of all possible outcomes.
The expected value $E(X)$, also denoted $\mu$, is the weighted average of all possible values, weighted by their probabilities:
Example — fair die:
$E(X) = 1 \times \tfrac{1}{6} + 2 \times \tfrac{1}{6} + 3 \times \tfrac{1}{6} + 4 \times \tfrac{1}{6} + 5 \times \tfrac{1}{6} + 6 \times \tfrac{1}{6} = \tfrac{21}{6} = 3.5$
Critical insight: $E(X) = 3.5$ is not a possible outcome — you cannot roll 3.5. The expected value is the average over infinitely many rolls. It represents the centre of mass of the distribution, not a guaranteed or achievable result.
Linearity of expectation: $E(aX + b) = aE(X) + b$. This works even when variables are dependent, making it one of the most powerful tools in probability.
$E(X) = \mu = \sum x \cdot P(X=x)$ — multiply each value by its probability and sum; $E(X)$ is the long-run average, not necessarily an achievable value
Pause — copy the expected value formula $E(X) = \sum x \cdot P(X=x)$ and the key note that $E(X)$ is the long-run average, not necessarily an achievable value of $X$ into your book.
Quick check: A random variable has $P(X=0)=0.3$, $P(X=2)=0.5$, $P(X=5)=0.2$. What is $E(X)$?
Worked examples · 3 in a row, reveal as you go
A biased die has distribution: $P(1)=0.1$, $P(2)=0.15$, $P(3)=0.2$, $P(4)=0.2$, $P(5)=0.15$, $P(6)=0.2$. Find (a) $E(X)$, (b) $\text{Var}(X)$, (c) $\text{SD}(X)$.
A game: roll two fair dice. Win $10 if sum is 7, win $5 if sum is 11, lose $2 otherwise. Find $E(W)$ and state whether to play repeatedly.
$X$ is temperature in Celsius with $E(X) = 20$ and $\text{SD}(X) = 5$. Let $Y = \tfrac{9}{5}X + 32$ (Fahrenheit). Find $E(Y)$ and $\text{SD}(Y)$.
Fill the gap: If $E(X) = 4$ and $E(X^2) = 20$, then $\text{Var}(X) = E(X^2) - [E(X)]^2 = 20 - \underline{\quad} = \underline{\quad}$.
Common errors · the 3 traps that cost marks
Odd one out: Which statement is the odd one out — the one that is incorrect?
Quick-fire activities
$P(X=0)=0.2$, $P(X=2)=0.5$, $P(X=5)=0.3$. Find $E(X)$, $\text{Var}(X)$, $\text{SD}(X)$.
Biased coin: $P(H)=0.6$, win $2 for heads, lose $1 for tails. Find $E(W)$ and $\text{Var}(W)$.
If $E(X)=5$ and $\text{Var}(X)=4$, find $E(3X-2)$ and $\text{Var}(3X-2)$.
A game costs $3 to play. You roll a die and win $x$ dollars. Find the expected profit/loss.
Explain in one sentence why a casino prefers low-variance games even when expected value for players is equally negative.
$E(X) = 3.5$ for a fair die — which is not one of the possible outcomes ($1, 2, 3, 4, 5, 6$). Expected value is the long-run average, not a guaranteed result. Roll once: never 3.5. Roll 1,000 times: the average converges to 3.5. This is why $E(X)$ is called the "mean" of the distribution — it describes the centre of mass, not an achievable value.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. A random variable $X$ has the probability distribution below. (a) Find the value of $k$. (b) Find $E(X)$. (c) Find $\text{Var}(X)$. (3 marks)
| $x$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $P(X=x)$ | 0.1 | 0.25 | $k$ | 0.2 | 0.15 |
Q2. A game: roll two fair dice. Win $10 if sum is 7, win $5 if sum is 11, lose $2 otherwise. (a) Construct the probability distribution for net winnings $W$. (b) Calculate $E(W)$. (c) Calculate $\text{Var}(W)$. (d) Would you play repeatedly? Justify using expected value. (3 marks)
Q3. A charity raffle sells 1,000 tickets at $2 each. One prize of $500, two prizes of $200, five prizes of $50. Let $X$ = net winnings for one ticket holder. (a) Construct the probability distribution for $X$. (b) Calculate $E(X)$ and interpret in context. (c) The charity claims "Every ticket could win $500!" Analyse this from both a mathematical and ethical perspective. Does expected value tell the whole story? (3 marks)
Comprehensive answers (click to reveal)
Drill 1: $E(X) = 0(0.2)+2(0.5)+5(0.3) = 2.5$; $E(X^2) = 0(0.2)+4(0.5)+25(0.3) = 9.5$; $\text{Var}(X) = 9.5-6.25 = 3.25$; $\text{SD}(X) \approx 1.803$.
Drill 2: $E(W) = 2(0.6)+(-1)(0.4) = 0.8$; $E(W^2) = 4(0.6)+1(0.4) = 2.8$; $\text{Var}(W) = 2.8-0.64 = 2.16$.
Drill 3: $E(3X-2) = 3(5)-2 = 13$; $\text{Var}(3X-2) = 9(4) = 36$.
Drill 4: Profit = $x-3$, $E(\text{profit}) = 3.5-3 = 0.50$. Expected profit of 50¢ per game.
Drill 5: Low variance makes outcomes predictable; the casino can reliably plan revenue without risk of large sudden payouts.
Q1 (3 marks): (a) $0.1+0.25+k+0.2+0.15 = 1$, so $k = 0.3$. (b) $E(X) = 1(0.1)+2(0.25)+3(0.3)+4(0.2)+5(0.15) = 3.05$. (c) $E(X^2) = 1(0.1)+4(0.25)+9(0.3)+16(0.2)+25(0.15) = 10.75$; $\text{Var}(X) = 10.75 - 3.05^2 = 10.75 - 9.3025 = 1.4475$.
Q2 (3 marks): (b) $E(W) = 10(\tfrac{1}{6})+5(\tfrac{1}{18})+(-2)(\tfrac{7}{9}) = \tfrac{30+5-28}{18} = \tfrac{7}{18} \approx 0.39$. (c) $E(W^2) = 100(\tfrac{1}{6})+25(\tfrac{1}{18})+4(\tfrac{7}{9}) = \tfrac{300+25+56}{18} \approx 21.17$; $\text{Var}(W) \approx 21.17 - 0.39^2 \approx 21.02$. (d) $E(W) > 0$: yes, play repeatedly — expected gain ≈ 39¢ per game.
Q3 (3 marks): (b) $E(X) = 498(0.001)+198(0.002)+48(0.005)+(-2)(0.992) = 0.498+0.396+0.24-1.984 = -0.85$. Each ticket buyer loses on average 85¢. (c) Mathematically misleading — highlights the maximum while hiding the negative expected value. Ethically, the charitable purpose adds non-monetary value, but presenting a raffle primarily as a chance to "win big" exploits optimism bias. Expected value alone doesn't capture entertainment or social value.
Five timed questions on expected value and variance. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms by answering probability distribution questions. Lighter alternative to the boss.
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