1The idea in one paragraph

Many trials produce a number: the score on a die, the number of heads in three tosses, the money you win in a game. A random variable is that number, named before you know its value. Its probability distribution lists every value it can take with the probability of each, in a table or as a formula, and because one of those values must happen, the probabilities add to 1. The expected value E(X) multiplies each value by its probability and adds the results: it is the average value of X over a very large number of trials, and it is where the distribution balances. When X is a player's gain in a game, E(X) = 0 means that, in the long run, neither the player nor the organiser comes out ahead: the game is fair.

2Random variables

A random variable is a quantity whose value is decided by the outcome of a trial. It is written with a capital letter, usually X. A particular value it might take is written with the lower-case letter, x. So P(X = x) means "the probability that the random variable X takes the value x", and P(X = 3) means the probability that it takes the value 3.

A random variable is discrete when its possible values can be listed separately, usually because they are counts: the number of sixes, the number of faulty items, a score. That matches the discrete data of 4.1. (Random variables that measure, such as a time or a mass, are continuous, and appear in 4.9 and, at HL, in 4.14.)

Here is the example used through this page. Two fair four-sided dice, each numbered 1 to 4, are rolled, and X is the larger of the two scores (if they are equal, X is that score). Figure 1 shows the 16 equally likely outcomes in a sample space diagram, each cell labelled with its value of X.

Sixteen outcomes collapse into four values, 1, 2, 3 and 4, and the values are not equally likely. Only (1, 1) gives X = 1. Five cells give X = 3: (3, 1), (3, 2), (3, 3), (1, 3), (2, 3). That is exactly the lesson of 4.5: count equally likely outcomes, then group them.

3Probability distributions

The probability distribution of a discrete random variable gives every possible value with its probability. From Figure 1:

Figure 2 draws it as a bar chart, one bar per value.

Every probability distribution obeys two rules, and questions test both.

0 ≤ P(X = x) ≤ 1 for every x, and Σ P(X = x) = 1. The probabilities add to 1 because X must take exactly one of its values.

Check: 1/16 + 3/16 + 5/16 + 7/16 = 16/16 = 1. A distribution that does not add to 1 has an error in it; find it before going on.

Reading probabilities off a distribution. Add the probabilities of the values that fit, and watch the inequality signs carefully, because for a discrete variable < and ≤ give different answers.

Conditional probability from 4.6 works on a distribution too. Given that X is at least 3, the chance that it is 4 is

The top is P(X = 4 and X ≥ 3), which is just P(X = 4), since X = 4 already satisfies X ≥ 3.

4Distributions with an unknown constant

The guide says distributions will be given as a table or as a formula. Either way, a common first part gives the probabilities in terms of an unknown and asks you to find it. There is only one fact to use: the probabilities add to 1.

From a formula. A random variable has P(X = x) = k(x² + 2) for x ∈ {0, 1, 2, 3}. Find k.

Write out each term, as here, rather than trying to add in your head: the method mark is for showing the substitution of every value. The distribution is then 2/22, 3/22, 6/22, 11/22, shown in Figure 3. It is fine to leave these unsimplified until the end; 6/22 and 3/11 are the same.

From a table. The same idea handles a table with an unknown in it.

Then 0.25 + 3c + 0.3 + c = 1, so 4c = 0.45 and c = 0.1125. Check every probability afterwards: each must lie between 0 and 1. If an unknown comes out making some probability negative or greater than 1, the value is impossible and something earlier is wrong. When the equation is quadratic, for example when some probabilities are written as k² and others as k, this check is how you reject one root.

5Expected value

Suppose you rolled the two four-sided dice 1600 times. You would expect X = 1 about 1600 × 1/16 = 100 times, X = 2 about 300 times, X = 3 about 500 times and X = 4 about 700 times, using the expected-number rule from 4.5. The mean of all 1600 values would then be

Each value is multiplied by the fraction of the time it occurs, which is its probability. The 1600 cancels, and that is the definition.

E(X) = Σ x P(X = x): multiply each value by its probability, and add. This is the expected value, or mean, of X.

Three facts about E(X) prevent most misunderstandings.

• It is a long-run average, not a prediction for one trial. In any single roll X is 1, 2, 3 or 4. Over many rolls the average of the values settles near 3.125.
• It need not be a value X can take. 3.125 is not a possible score, just as the mean number of children in a family is not a whole number.
• It is the balance point of the distribution. Imagine each bar in Figure 4 as a weight on a beam, placed at its value of x and as heavy as its probability. The beam balances at E(X). The bars lean towards 4, so the balance point sits above the middle value of 2.5.

For the formula distribution of section 4:

Using E(X) to find a second unknown. Two unknowns need two equations: the probabilities add to 1, and E(X) is given.

Given E(X) = 2.7, find a and b.

On Paper 1 solve the pair by hand as here. On Paper 2 the GDC's simultaneous equation solver is quicker, but write both equations first; they carry the method marks.

Variance is not on this page. The spread of a random variable, Var(X), is HL content (4.14). At SL you need only E(X).

6Applications: games, fairness and decisions

The guide's application is games of chance. Let X be the player's gain: what the player wins minus what they paid to play. Then E(X) is the average gain per game over many games.

A game is fair when E(X) = 0, where X is the player's gain. If E(X) < 0 the organiser profits in the long run; if E(X) > 0 the player does.

A worked game. At a school fair a player pays $4 and rolls one fair die. A 6 wins $12, a 4 or a 5 wins $4, and anything else wins nothing. Is the game fair?

Work in gains, not prizes, because the $4 stake is paid every time. A 6 gives 12 − 4 = $8; a 4 or 5 gives 4 − 4 = $0; a 1, 2 or 3 gives 0 − 4 = −$4.

The expected gain is −$0.67. The game is not fair: on average a player loses about 67 cents a game, and the fair raises about that much per game in the long run. Figure 5(a) shows the distribution with its balance point below zero.

Making it fair. What prize for a 6 would make the game fair? Call the prize p. The gain on a 6 becomes p − 4, and the other gains do not change.

A $16 prize makes the game fair, as Figure 5(b) shows: the gain on a 6 is now $12, and the beam balances at 0.

Alternatively keep the prizes and change the stake. The expected prize is 12 × 1/6 + 4 × 2/6 + 0 × 3/6 = 2 + 4/3 = 10/3, so a stake of $3.33 (3 s.f.) would make the game fair. A fair game charges exactly the expected prize. Real casinos and lotteries never do: every game they run has E(X) < 0 for the player, which is how they pay their costs and make a profit. Whether that is "fair" in the everyday sense is a good question for TOK; in the mathematical sense it is not.

A business application. The same calculation prices risk. An insurer sells phone cover for $60 a year. From its records (invented figures), in a year a policy leads to a $150 repair with probability 0.12, a $600 replacement with probability 0.04, and no claim otherwise. Let Y be the insurer's profit on one policy.

The insurer expects $18 profit per policy, so across 5000 policies about 5000 × 18 = $90,000, before its running costs. That is expected value and expected number of occurrences working together. The first probability, 0.84, came from 1 − 0.12 − 0.04: the probabilities must add to 1, even when the question only mentions two of them.