4.7 Discrete random variables

Introduction

Many uncertain situations produce a number: the score on a die, the number of faulty phones in a box, the amount of money won or lost in a game. A discrete random variable gives a name to that number, and its probability distribution lists every value it can take with the probability of each. Once the distribution is known, the single most useful summary is the expected value: the mean value the variable would take over a very large number of repetitions.

Expected value is the mathematics behind every casino, lottery and insurance company. A roulette wheel in a European casino pays 35 to 1 on a single number but has 37 pockets, so a player loses about 2.7 cents for every dollar staked in the long run, and that small expected loss, multiplied over millions of bets, is the casino's income. The same calculation shows a business whether a decision pays off on average and shows a player whether a game is fair.

Random variables

From outcomes to numbers

A trial such as rolling two dice has many possible outcomes, and the sample space lists them all. Often, though, interest lies not in the outcome itself but in a number calculated from it: the total of the two dice, the larger of the two, or the difference between them. A random variable is the rule that turns each outcome into a number. It is "random" because its value is not known before the trial, and a "variable" because it can take different values on different trials.

Random variables are named with capital letters such as $X$, $Y$ or $T$. The matching lower-case letter stands for a particular value the variable can take. So $X$ is "the number of heads", while $x = 2$ is one possible value, and the statement $\mathrm{P}(X = 2)$ is read as "the probability that the number of heads is 2". Keeping the capital letter for the variable and the lower-case letter for its values makes probability statements precise: $\mathrm{P}(X = x)$ is a general expression for the probability that $X$ takes the value $x$, whatever $x$ is.

Discrete and continuous random variables

The first five variables in the table take separate values that can be listed, with gaps between them: a box can hold 2 or 3 broken eggs but not 2.4. Such variables usually come from counting, and they match discrete data in statistics. The list of values may be finite (0 to 12 eggs) or may continue without a fixed end (0, 1, 2, … cars), but the values can still be listed one by one.

The runner's time is different. It is measured, not counted, and can take any value in an interval, so it is a continuous random variable. Continuous random variables need different tools, because the probability of any single exact value such as 11.0000… seconds is zero; the normal distribution in subtopic 4.9 is the main example. Everything that follows concerns discrete random variables.

Probability distributions

What a probability distribution shows

Knowing the possible values of a random variable is only half the description. The other half is how likely each value is. When three fair coins are tossed, the sample space has eight equally likely outcomes, and counting the heads in each gives the probability of each value of $H$.

The bottom two rows are the probability distribution of $H$. Each value of the variable is paired with exactly one probability, and the probabilities share out the whole sample space between the values, so they add to 1.

A probability distribution is a mathematical model of an uncertain quantity. It plays the same role for a random variable that a frequency distribution plays for a set of data: a frequency table records what did happen in a set of observations, while a probability distribution describes what is expected to happen in the long run.

The two conditions every distribution satisfies

Every probability distribution of a discrete random variable obeys two rules.

The first rule holds because each $\mathrm{P}(X = x)$ is a probability. The second holds because the events $X = x$ for the different values of $x$ are mutually exclusive (the variable cannot take two values at once) and together they cover every outcome of the trial, so their probabilities add to the probability of the whole sample space, which is 1. The sum-to-1 rule is the usual way to find an unknown probability in a distribution.

Building a distribution from a sample space diagram

When two fair six-sided dice are rolled, let $X$ be the difference between the two scores (the larger minus the smaller, or 0 for a double). A sample space diagram shows all 36 equally likely outcomes, with the value of $X$ written in each cell.

Counting cells in the grid gives the distribution. The value 0 appears on the leading diagonal (the six doubles), and each step away from that diagonal increases the difference by 1 and removes two cells, so the counts fall from 10 to 8, 6, 4 and 2.

The numerators add to 36, so the probabilities add to 1, which is a quick check that no cell was missed or counted twice. Fractions are usually left with the denominator 36 rather than simplified, because a common denominator makes the table easier to read and to check.

A probability distribution can be drawn as a bar chart, with the values of $X$ on the horizontal axis and their probabilities as the heights of separate bars (or vertical lines). The gaps between the bars show that the variable is discrete.

Inequality words need care. "At least 3" and "3 or more" mean $X \ge 3$; "more than 3" means $X > 3$, which for whole-number values is $X \ge 4$; "at most 1" and "no more than 1" mean $X \le 1$; "fewer than 1" means $X = 0$ here. With a discrete variable, $\le$ and $<$ give different events, so writing the event out as a list of values before adding is the safest method.

Building a distribution from a tree diagram

When a random process happens in stages, a tree diagram organises the outcomes and their probabilities. Each complete path through the tree is one outcome, the probability of a path is the product of the probabilities on its branches, and the value of the random variable is read from each path.

Distributions given as a function

A distribution can also be given as a formula for $\mathrm{P}(X = x)$, together with the set of values for which it applies. For example,

means that $X$ takes only the values 1, 2, 3 and 4, and the probability of each is found by substituting into the formula. Substituting each value turns the function into a table, which is usually the clearest first step.

The discrete uniform distribution

When every value of a random variable is equally likely, the distribution is called uniform. The score on a fair six-sided die is the standard example: each of the six values has probability $\frac{1}{6}$, and the bar chart is a row of equal bars.

A uniform distribution applies only when the outcomes are equally likely and each outcome gives a different value. The difference between two dice is not uniform, even though each of the 36 outcomes is equally likely, because different numbers of outcomes give each value. Many real situations are modelled as uniform when nothing makes one value more likely than another: the last digit of a randomly chosen phone number, or the number on a ball drawn in a lottery.

Finding unknown probabilities

The sum-to-1 rule gives one equation, so it can find one unknown. When the probabilities contain the unknown in more than one form, the equation may be a quadratic, which is solved with the GDC's polynomial solver or by graphing, and then each solution is tested against the rule $0 \le \mathrm{P}(X = x) \le 1$.

A solution can also fail the other way: a positive value of $k$ may make some probability greater than 1, or make the remaining probabilities negative. Every solution has to be substituted back into every entry of the table before it is accepted.

Expected value

The long-run mean of a random variable

Suppose the two dice are rolled 3600 times and the difference recorded each time. The results would not match the probabilities exactly, but the relative frequencies would be close to them: roughly 600 zeros, 1000 ones, 800 twos, 600 threes, 400 fours and 200 fives. The mean of these 3600 results, from the frequency-table formula of subtopic 4.3, would be about

Dividing each frequency by 3600 before adding gives the same result, and each fraction $\frac{f}{3600}$ is a relative frequency, which is close to the probability $\mathrm{P}(X = x)$. As the number of repetitions grows, the relative frequencies settle down to the probabilities, and the mean settles down to a fixed value. That limiting value is the expected value.

The formula is a weighted mean: each value of $X$ is weighted by how likely it is. Values with high probability pull the expected value towards them, and values with tiny probability have little effect unless they are very large. The expected value is also written $\mu$, the same symbol as a population mean, because it is the mean of the whole (infinite) population of possible results.

For the dice difference:

On the bar chart of the distribution, the expected value is the balance point: if the bars were weights on a light rod, the rod would balance on a pivot placed at $x \approx 1.94$. This picture explains why a long right tail pulls the expected value to the right of the most likely value.

An expected value need not be a possible value

The score on a fair die has expected value

No single roll can score 3.5. The expected value describes the average over many rolls, not the result of one roll: 1000 rolls should total about 3500. In the same way, a family cannot have 1.7 children, but the expected number of children per family in a population can be 1.7. Expected values should therefore not be rounded to a possible value of $X$; they are given exactly or to 3 significant figures like any other mean.

Using the GDC

Expected values are usually short enough to calculate by writing out the sum. The GDC gives a fast check, using one-variable statistics with the probabilities as frequencies: enter the values of $x$ in one list and the probabilities in a second list, then run one-variable statistics with the second list as the frequency list. The mean it reports, $\bar{x}$, is $\mathrm{E}(X)$, because the frequencies (the probabilities) add to 1. On a TI-84 Plus CE this is STAT → CALC → 1-Var Stats with List: L1 and FreqList: L2.

The other statistics the GDC prints at the same time, such as the standard deviation, describe the spread of the distribution; only $\bar{x}$ is used for the expected value.

Unknown probabilities from a given expected value

When a distribution contains two unknowns, the sum-to-1 rule gives one equation and a given expected value gives a second. The two linear equations form a system, solved by hand or with the GDC's simultaneous equation solver (subtopic 1.8, Equations and systems using technology).

Expected value and the expected number of occurrences

Subtopic 4.5 introduced the expected number of occurrences: in $n$ trials, an event of probability $p$ is expected to happen $np$ times. The expected value extends the same idea from counting one event to averaging a whole variable. If a game is played $n$ times, the expected number of times each value $x$ occurs is $n\,\mathrm{P}(X = x)$, and the expected total is

For example, the expected total of 50 rolls of the dice difference is $50 \times \frac{35}{18} \approx 97.2$, and the expected number of those 50 rolls that give a difference of 1 is $50 \times \frac{10}{36} \approx 13.9$.

Applications: games, fairness and decisions

Gain and expected gain

In a game that costs money to play, the random variable of interest is usually the player's gain: what the player receives minus what the player paid. A prize of $10 in a game that costs $3 is a gain of $7, and a game with no prize is a gain of −$3, a loss. Working with gains rather than prizes means the cost to play is counted on every outcome, including the losing ones.

The expected gain $\mathrm{E}(X)$ is the average amount the player wins (if positive) or loses (if negative) per game in the long run. Because every dollar a player gains is a dollar the organiser loses, the organiser's expected profit per game is $-\mathrm{E}(X)$.

Fair games

A game is called fair when neither the player nor the organiser is expected to make money from it in the long run. This is a precise mathematical meaning of "fair", which differs from the everyday one: a fair game need not give the player a good chance of winning, and a game in which every outcome is equally likely need not be fair.

If $\mathrm{E}(X) > 0$, the game favours the player; if $\mathrm{E}(X) < 0$, it favours the organiser. Raffles, fair stalls, casino games and lotteries are all designed with $\mathrm{E}(X) < 0$, because they must cover their costs or raise money.

The alternative method in the last card works because every value is reduced by the same amount, so the long-run average is reduced by the same amount too. In the dice game, the expected prize is $20 \times \frac{1}{36} + 6 \times \frac{5}{36} = \frac{50}{36} \approx 1.39$ dollars, and $1.39 - 4 = -2.61$, as before. A game is fair exactly when the cost to play equals the expected prize.

House edge: roulette

A roulette wheel has numbered pockets, and a ball lands in one of them at random. A European wheel has 37 pockets (the numbers 1 to 36 and a single zero); an American wheel has 38 (1 to 36, 0 and 00). A bet of one unit on a single number pays 35 to 1: if the number comes up, the player keeps the stake and wins 35 units; otherwise the stake is lost.

On a European wheel, the gain $X$ on a one-unit single-number bet is 35 with probability $\frac{1}{37}$ and −1 with probability $\frac{36}{37}$.

If the payout matched the odds, a winning bet would pay 36 to 1 and the game would be fair. Paying only 35 to 1 leaves the casino an expected profit of 2.70% of everything staked. On an American wheel the same bet gives $35 \times \frac{1}{38} - \frac{37}{38} = -\frac{2}{38} \approx -0.0526$, a 5.26% edge.

The European wheel has the same edge on an even-money bet such as "red": 18 of the 37 pockets are red, so $\mathrm{E}(X) = 1 \times \frac{18}{37} - 1 \times \frac{19}{37} = -\frac{1}{37}$. The zero is what makes every standard bet unfair, because it is a pocket that loses for the player without being paid for.

Source: wheel layouts and payouts from Wikipedia, "Roulette", accessed 2026; expected values calculated.

Insurance and extended warranties

Insurance turns a small chance of a large loss into a certain small payment, the premium. The insurer's payout is a random variable, and its expected value is the average amount the insurer pays per policy. To cover running costs and make a profit, the premium is set above the expected payout, so for the customer an insurance policy is an unfair game in the mathematical sense.

Expected value in business decisions

Expected value is a tool for comparing decisions under uncertainty: for each option, find the distribution of the profit, calculate its expected value, and prefer the option with the higher expected profit if the decision is repeated many times. The essential step is weighting each outcome by its probability; a list of what could happen, without the probabilities, cannot rank the options.

Limitations of expected value

The expected value is a long-run average. It describes what happens over many repetitions and says nothing directly about how much one result can vary. Two games can have the same expected gain of −$1 while one risks $2 and the other risks $10,000. People and businesses that cannot survive a large loss often accept a lower expected value for less risk, which is exactly why insurance exists.

A second limitation appears when very large values have very small probabilities. In the St Petersburg paradox, described by Nicolas Bernoulli in a letter of 1713, a fair coin is tossed until the first head appears. If the first head appears on toss $k$, the player wins $2^k$ dollars. The probability of that is $\left(\frac{1}{2}\right)^k$, so each possible outcome contributes the same amount to the expected value.

The sum $1 + 1 + 1 + \ldots$ never stops growing, so the expected prize is infinite, and by the fair-game rule a player should pay any price to play. Almost nobody would pay even $50, because the prize is at most $8 with probability $\frac{7}{8}$. Daniel Bernoulli's 1738 resolution argued that the value of money to a person grows more slowly than the amount of money, an idea now called utility. Real casinos also could not pay the enormous prizes, so the game's long tail could never actually be offered.

Source: Wikipedia, "St. Petersburg paradox", accessed 2026.

Case study: the expected value of a Powerball ticket

Powerball is a lottery played in most of the United States. A $2 ticket chooses five different "white" numbers from 1 to 69 and one red "Powerball" number from 1 to 26, and the draw does the same. Prizes depend on how many white numbers match and whether the Powerball matches. The jackpot varies from draw to draw; every other prize is a fixed amount.

The probabilities can be found by counting. The number of possible draws is the number of ways to choose 5 numbers from 69 multiplied by 26, which is 292,201,338. Counting the draws that give each match, and dividing, gives the probabilities in the table; they agree with the odds the lottery publishes.

Source: Powerball prize structure in force since October 2015 (Multi-State Lottery Association), via Wikipedia, "Powerball", accessed 2026; probabilities and expected values calculated.

Adding the last column for the fixed prizes gives an expected fixed prize of about $0.320 per ticket. The expected prize from a ticket is therefore

and the expected gain is this minus the $2 price. With a jackpot of $20 million, the jackpot adds only $\frac{20\,000\,000}{292\,201\,338} \approx 0.068$ dollars, so

A player loses on average about $1.61 of every $2 ticket, so about 81% of the money spent is not returned as prizes. The fixed prizes alone return only 16% of the price.

Setting the expected gain to zero shows how large the jackpot would have to be for a fair ticket:

Advertised jackpots occasionally exceed this, which seems to make a ticket a fair game or better. Three features of the real game are left out of the simple model, and each of them works against the player.

• The advertised jackpot is the total of 30 annual payments spread over 29 years, and a winner who takes the money at once receives roughly half of it.

• Prizes are taxed.

• Large jackpots attract many more players, so a winning ticket is more likely to share the jackpot with others.

Each of these lowers the true value of $J$ in the formula, and the ticket stays an unfair game.

The Powerball calculation illustrates both the strength and the limits of expected value. It is exact, given the prize table, and it shows clearly where the money goes. But a lottery ticket is bought once, not millions of times, and the gap between a near-certain loss of $2 and a tiny chance of an enormous prize is a matter of how people value risk and hope, which the expected value does not measure.

Exam focus

Discrete random variables appear in Paper 1 as short questions of 4–7 marks and in Paper 2 as parts of longer questions on games, business or the binomial distribution. Common shapes are:

• a table with one unknown: find it from the sum-to-1 rule, then calculate $\mathrm{E}(X)$, sometimes the expected number of times a value occurs in $n$ trials;

• a table with two unknowns and a given $\mathrm{E}(X)$: set up and solve two equations;

• a distribution given as a function: find $k$, sometimes as an exact value;

• complete a distribution table from a sample space diagram or a tree diagram (A2 for a fully correct table, A1 for most entries correct);

• a game: find the expected gain, determine whether it is fair, find the cost or prize that makes it fair;

• a business decision combining expected value with probabilities from another distribution (Paper 2, the hardest parts).

Write the sum-to-1 equation and the substitution into $\mathrm{E}(X)$ every time; they carry the method marks. Give exact values (fractions) when asked, otherwise 3 significant figures, and money to the accuracy the question states.

In part (e) a common error is to set the expected prize equal to 0 instead of the expected gain. The equation for a fair game always uses the gain, or equivalently sets the expected prize equal to the cost of playing.

Questions that ask for an unknown probability may expect the answer to be rejected or accepted using the rule $0 \le \mathrm{P} \le 1$. When a quadratic gives two solutions, state why one is rejected; a negative probability or a probability above 1 is the reason markschemes accept.

Practice questions

1. A discrete random variable $X$ has the distribution in the table.

(a) Find $a$. [2] (b) Calculate $\mathrm{E}(X)$. [2] (c) Write down $\mathrm{P}(X > 4)$. [1]

2. The random variable $X$ has $\mathrm{P}(X = x) = \frac{x}{15}$ for $x \in \{1, 2, 3, 4, 5\}$. (a) Show that this is a valid probability distribution. [2] (b) Find $\mathrm{P}(X \ge 4)$. [2] (c) Find $\mathrm{E}(X)$, giving your answer as an exact value. [2]

3. Two fair dice are rolled and $S$ is the number of sixes shown. (a) Construct the probability distribution of $S$. [3] (b) Find $\mathrm{E}(S)$. [2]

4. A random variable takes the values 1, 2, 3 and 4 with probabilities $p$, $q$, 0.3 and 0.2 respectively. Given that $\mathrm{E}(X) = 2.6$, find $p$ and $q$. [4]

5. A spinner has four equal sectors marked $10, $2, $0 and $0. A player pays $3 to spin it once and receives the amount shown. Determine whether the game is fair. [3]

6. A charity sells 500 raffle tickets at $5 each. There is one prize of $1000 and two prizes of $250. Find the expected gain of a person who buys one ticket, and interpret your answer. [4]

7. The distribution of $X$ is given in the table, where $k > 0$.

(a) Find $k$. [3] (b) Find $\mathrm{E}(X)$. [2]

8. A fair eight-sided die has faces numbered 1 to 8, and $X$ is the score on one roll. (a) Calculate $\mathrm{E}(X)$. [2] (b) Explain what this value means, given that it cannot be scored on a single roll. [2]

9. An insurer sells bicycle theft cover for $60 a year. The probability that an insured bicycle is stolen in a year is 0.03, and the insurer then pays $1200. (a) Find the insurer's expected profit per policy. [3] (b) Find the number of policies the insurer must sell to expect a profit of at least $10,000. [2]

Summary

• A random variable $X$ assigns a number to each outcome of a random process; a discrete random variable takes separate, countable values.

• A probability distribution lists every value $x$ with $\mathrm{P}(X = x)$, as a table or a function; it can be built from a sample space diagram or a tree diagram, adding the probabilities of all outcomes that give the same value.

• Every distribution satisfies $0 \le \mathrm{P}(X = x) \le 1$ and $\sum \mathrm{P}(X = x) = 1$; the second rule finds one unknown, and a given $\mathrm{E}(X)$ gives a second equation.

• The expected value $\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$ is the long-run mean of $X$, the balance point of its distribution; it need not be a possible value of $X$.

• Over $n$ independent repetitions, the expected total is $n\,\mathrm{E}(X)$, and the expected number of times $X = x$ is $n\,\mathrm{P}(X = x)$

• A player's gain is prize minus cost; a game is fair when the expected gain is 0, equivalently when the cost equals the expected prize.

• The house edge is the organiser's expected profit per unit staked: 2.70% for a single-number bet on a European roulette wheel.

• Expected value guides repeated decisions (insurance premiums, business choices) but ignores risk, and it can mislead when huge prizes have tiny probabilities.

Self-check

• To do:

  I can say whether a random variable is discrete or continuous and list the values a discrete one can take.

• To do:

  I can use the notation $X$, $x$ and $\mathrm{P}(X = x)$ correctly.

• To do:

  I can build a probability distribution from a sample space diagram, adding the probabilities of outcomes with the same value.

• To do:

  I can build a probability distribution from a tree diagram, with or without replacement.

• To do:

  I can turn a distribution given as a function into a table, using only the allowed values.

• To do:

  I can find an unknown probability using $\sum \mathrm{P}(X = x) = 1$, and reject solutions that give impossible probabilities.

• To do:

  I can find probabilities such as $\mathrm{P}(X \ge 3)$ and $\mathrm{P}(X < 3)$ by listing the values included.

• To do:

  I can calculate $\mathrm{E}(X)$, showing the substitution, and check it with the GDC's one-variable statistics.

• To do:

  I can explain why $\mathrm{E}(X)$ can be a value that $X$ never takes.

• To do:

  I can find two unknown probabilities from the sum-to-1 rule and a given $\mathrm{E}(X)$.

• To do:

  I can define the gain in a game and calculate the expected gain, including the cost to play in every value.

• To do:

  I can decide whether a game is fair and find the cost or prize that makes it fair.

• To do:

  I can calculate a house edge and an insurer's or organiser's expected profit over many games or policies.

• To do:

  I can explain at least two limitations of expected value as a basis for decisions.

The outcomes of different turns are independent.

1. (a)

   Find
   (i) $\mathrm{E}(X)$;
   (ii) $\mathrm{Var}(X)$.

   [3]3 marks

   Markscheme (a)

   (i) $\mathrm{E}(X) = 1(0.45) + 3(0.35) + 6(0.2) = 2.7$ A1

   (ii) $\mathrm{E}(X^2) = 1(0.45) + 9(0.35) + 36(0.2) = 10.8$ OR use of GDC lists (M1)

   $\mathrm{Var}(X) = 10.8 - 2.7^2 = 3.51$ A1

   [3 marks]

2. (b)

   Zoltan takes 40 turns. Use the central limit theorem to find the probability that he moves more than 120 squares in total.

   [3]3 marks

   Markscheme (b)

   total $S \sim \mathrm{N}(40 \times 2.7, 40 \times 3.51)$ (M1)

   $= \mathrm{N}(108, 140.4)$

(b) $X < 3$ means $X = 1$ or $X = 2$, so $\mathrm{P}(X < 3) = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3}$.

Using the GDC's polynomial solver with coefficients 2, 1 and −0.72 gives $k = 0.4$ or $k = -0.9$.

A probability cannot be negative, so $k = -0.9$ is rejected (it would make $\mathrm{P}(X = 2) = -0.9$). With $k = 0.4$, the probabilities are 0.1, 0.4, 0.32 and 0.18, which all lie between 0 and 1 and add to 1.

So $k = 0.4$.

The answer lies between 1 and 4, and it is above the middle value 2.5 because the larger values are more likely. Both checks support it.

Subtracting the first equation from the second gives $b = 0.5$, and then $a = 0.2$. (The GDC's system solver with $a + b = 0.7$ and $a + 2b = 1.2$ gives the same.)

Check: $0 + 0.2 + 1.0 + 0.3 = 1.5$, and both values lie between 0 and 1. So $a = 0.2$ and $b = 0.5$.

(b) $\mathrm{P}(X = 0) = \frac{3}{8} \times \frac{5}{7} + \frac{5}{8} \times \frac{3}{7}$ (M1)

$= \frac{30}{56} = \frac{15}{28}$ A1

$\mathrm{P}(X = -2) = \frac{5}{8} \times \frac{4}{7}$ (M1)

$= \frac{20}{56} = \frac{10}{28}$ A1

(c) $\mathrm{E}(X) = 6 \times \frac{3}{28} + 0 \times \frac{15}{28} + (-2) \times \frac{10}{28}$ (M1)

$= -\frac{2}{28} = -\frac{1}{14}$ $(-0.0714)$ A1

(d) $350 \times \frac{1}{14}$ (M1)

$= \$25$ A1

Note: accept $-350 \times \mathrm{E}(X)$; FT from their (c).

(e) $(P - 2) \times \frac{3}{28} + 0 \times \frac{15}{28} - 2 \times \frac{10}{28} = 0$ M1A1

$P = \frac{26}{3} = \$8.67$ A1

Total [13]

$\mathrm{P}(S > 120) = 0.156$ $(0.155592\ldots)$ A1

Note: Accept the equivalent method using $\bar{X} \sim \mathrm{N}\left(2.7, \frac{3.51}{40}\right)$ and $\mathrm{P}(\bar{X} > 3)$. Follow through from part (a).

[3 marks]