---
title: "Probability and Random Variables"
book: "High School Mathematics"
subject: math
language: en
chapter: 18
exercises: 11
source: https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables
---

# Chapter 18 — Probability and Random Variables

[Probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) models experiments whose outcome is uncertain: a die roll, a card drawn, a game played. This chapter sets up the vocabulary on finite sets, then introduces the central character of all of [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) theory: the *[random variable](#def-g11-prob-rv)* and its *[expectation](#def-g11-prob-expectation)*. These notions are reviewed and deepened in [Chapter 33](https://one-course.com/books/math/2/en/chapter/33-random-variables-and-the-binomial-distribution#ch-g12-randvar), and conditional probabilities follow in [Chapter 32](https://one-course.com/books/math/2/en/chapter/32-conditional-probability-and-independence#ch-g12-condprob).

## 18.1 Probability on a finite set

**Definition 18.1 (Probability model).**

An experiment with finitely many outcomes is modelled by its *sample space* $\Omega = \{\omega_1, \dots, \omega_n\}$ (the set of outcomes) together with a *[probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution)* $\P$ assigning to each outcome $\omega_i$ a number $p_i \geq 0$, with $p_1 + \dots + p_n = 1$. An *event* is a subset $A \subseteq \Omega$, and

$$
\P(A) = \sum_{\omega_i \in A} p_i .
$$

When all outcomes are equally likely ($p_i = \frac1n$: the *uniform model*),

$$
\P(A) = \frac{\text{number of outcomes in } A}
{\text{number of outcomes in } \Omega}.
$$

**Example 18.2.**

Roll two distinguishable dice: $\Omega$ is the set of the $36$ ordered pairs $(i, j)$ with $1 \leq i, j \leq 6$, all equally likely. The [event](#def-g11-prob-model) $S$ = “the sum is $7$” contains the $6$ pairs $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, $(6,1)$, so $\P(S) = \frac{6}{36} = \frac16$. The [event](#def-g11-prob-model) “the sum is $12$” contains only $(6,6)$: [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac{1}{36}$. Choosing the *right* [sample space](#def-g11-prob-model) — ordered pairs, not unordered — is what makes the [uniform model](#def-g11-prob-model) applicable.

**Proposition 18.3 (Rules of probability).**

For [events](#def-g11-prob-model) $A, B \subseteq \Omega$:

$$
\P(\emptyset) = 0, \qquad \P(\Omega) = 1, \qquad
\P(\bar A) = 1 - \P(A),
$$

$$
\P(A \cup B) = \P(A) + \P(B) - \P(A \cap B),
$$

where $\bar A$ is the *[complement](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations)* of $A$ (all outcomes not in $A$). In particular, $\P(A \cup B) = \P(A) + \P(B)$ when $A$ and $B$ are *[incompatible](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations)* ($A \cap B = \emptyset$).

**Proof.** The empty [event](#def-g11-prob-model) contains no outcome (empty sum), and $\Omega$ contains them all (total sum $1$). Every outcome is in exactly one of $A$, $\bar A$, so $\P(A) + \P(\bar A) = 1$. For the [union](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion): summing the $p_i$ over $A$ and then over $B$ counts the outcomes of $A \cap B$ twice; subtracting $\P(A \cap B)$ corrects the double count. ∎

**Example 18.4.**

In a class, $60\%$ study music ($M$), $45\%$ play a sport ($S$), and $25\%$ do both. The proportion doing at least one is $\P(M \cup S) = 0.60 + 0.45 - 0.25 = 0.80$, so $20\%$ do neither ([complement](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-operations)).

## 18.2 Random variables

**Definition 18.5 (Random variable, distribution).**

A *random variable* on $\Omega$ is a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $X$ assigning a number to each outcome. Its *distribution* is the table of its possible values $x_1, \dots, x_k$ together with their probabilities

$$
\P(X = x_i) = \P\bigl(\{\omega : X(\omega) = x_i\}\bigr),
\qquad
\sum_{i=1}^k \P(X = x_i) = 1 .
$$

**Example 18.6.**

A game: pay $2$ euros, roll one fair die, receive the value shown if it is at least $5$, nothing otherwise. Let $G$ be the *gain* (received minus paid). The outcomes $1, 2, 3, 4$ give $G = -2$; the outcome $5$ gives $G = 3$; the outcome $6$ gives $G = 4$. The [distribution](#def-g11-prob-rv) of $G$:

| $g$ | $-2$ | $3$ | $4$ |
| --- | --- | --- | --- |
| $\P(G = g)$ | $\dfrac46$ | $\dfrac16$ | $\dfrac16$ |

**Method 18.7 (Building a distribution table).**

List the possible values of $X$; for each value, collect the outcomes producing it and add their probabilities; check that the row of probabilities sums to $1$.

## 18.3 Expectation

**Definition 18.8 (Expectation).**

The *expectation* (or [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean)) of $X$ is the average of its values weighted by their probabilities:

$$
\E(X) = \sum_{i=1}^{k} \P(X = x_i)\; x_i .
$$

**Remark 18.9.**

$\E(X)$ is what the [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) of a huge number of independent repetitions of the experiment would be — the precise version of this statement is the law of large numbers, proved in [Chapter 34](https://one-course.com/books/math/2/en/chapter/34-sums-of-random-variables-and-the-law-of-large-numbers#ch-g12-sums). Note the formal analogy with the [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean) of a dataset with frequencies ([Definition 17.1](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean)): probabilities play the role of the relative frequencies $\frac{n_i}{n}$.

**Example 18.10.**

For the game of [Example 18.6](#ex-g11-prob-game):

$$
\E(G) = \frac46 \times (-2) + \frac16 \times 3 + \frac16 \times 4
= \frac{-8 + 3 + 4}{6} = -\frac16 \approx -0.17 .
$$

On average the player loses about $17$ cents per game: the game is unfavorable. A game is called *fair* when the expected gain is $0$.

**Proposition 18.11 (Linearity).**

For all reals $a, b$:

$$
\E(aX + b) = a\,\E(X) + b .
$$

**Proof.** The variable $aX + b$ takes the value $a x_i + b$ with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\P(X = x_i)$, so

$$
\E(aX + b) = \sum_i \P(X = x_i)(a x_i + b)
= a \sum_i \P(X = x_i)\,x_i + b \underbrace{\sum_i \P(X =
x_i)}_{=\,1} = a\,\E(X) + b .
$$

∎

## 18.4 Variance and standard deviation

**Definition 18.12 (Variance).**

The *variance* of $X$ measures the spread of its values around the [expectation](#def-g11-prob-expectation) $m = \E(X)$:

$$
\V(X) = \E\bigl((X - m)^2\bigr)
= \sum_{i=1}^{k} \P(X = x_i)\,(x_i - m)^2,
\qquad
\sigma(X) = \sqrt{\V(X)} .
$$

**Proposition 18.13 (Shortcut formula).**

$$
\V(X) = \E(X^2) - \E(X)^2 .
$$

**Proof.** Expand $(x_i - m)^2 = x_i^2 - 2m\,x_i + m^2$ inside the sum:

$$
\V(X) = \sum_i \P(X = x_i)\,x_i^2
- 2m \sum_i \P(X = x_i)\,x_i + m^2\sum_i \P(X = x_i)
= \E(X^2) - 2m^2 + m^2 ,
$$

which is $\E(X^2) - m^2$. This is the random-variable version of the identity proved for datasets in [Proposition 17.7](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#prop-g11-stat-shortcut). ∎

**Proposition 18.14 (Affine transformation).**

$\V(aX + b) = a^2\,\V(X)$ and $\sigma(aX + b) = \abs a\,\sigma(X)$.

**Proof.** By linearity, $aX + b$ deviates from its [expectation](#def-g11-prob-expectation) $a m + b$ by $(aX + b) - (am + b) = a(X - m)$: every deviation is scaled by $a$, hence every squared deviation by $a^2$, and so is their weighted [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean). The shift $b$ has disappeared — shifting a variable does not change its spread. ∎

**Example 18.15.**

For the game of [Example 18.6](#ex-g11-prob-game): $\E(G^2) = \frac46 \times 4 + \frac16 \times 9 + \frac16 \times 16
= \frac{16 + 9 + 16}{6} = \frac{41}{6}$, so

$$
\V(G) = \frac{41}{6} - \left(-\frac16\right)^2
= \frac{41}{6} - \frac{1}{36} = \frac{245}{36} \approx 6.8,
\qquad
\sigma(G) \approx 2.6 .
$$

The loss of $0.17$ per game on average comes with swings of typical size $2.6$ euros — [expectation](#def-g11-prob-expectation) without [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) tells only half the story.

![The distribution of the gain G of , and its expectation (G) = -1/6 (dashed): the balance point of the probability masses.](https://one-course.com/images/onecourse/chapters/math-2/g11-prob/fig-10cbaf11787d.svg)

*The [distribution](#def-g11-prob-rv) of the gain $G$ of [Example 18.6](#ex-g11-prob-game), and its [expectation](#def-g11-prob-expectation) $\E(G) = -\frac16$ (dashed): the balance point of the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) masses.*

## 18.5 Exercises

**Exercise 18.1 ★.**

A card is drawn from a standard $52$-card deck. Compute the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of drawing: a heart; a king; a heart or a king; neither a heart nor a king.

**Solution of Exercise 18.1.**

[Uniform model](#def-g11-prob-model) on $52$ cards. Hearts: $\frac{13}{52} = \frac14$. Kings: $\frac{4}{52} = \frac{1}{13}$. Heart or king: $\frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13}$ (the king of hearts is counted once). Neither: $1 - \frac{4}{13} = \frac{9}{13}$.

**Exercise 18.2 ★.**

Two fair dice are rolled and their sum $S$ is recorded. Using the $36$-outcome model of [Example 18.2](#ex-g11-prob-twodice), compute $\P(S = 6)$, $\P(S \geq 10)$ and $\P(S \text{ is even})$.

**Solution of Exercise 18.2.**

$S = 6$: the pairs $(1,5), (2,4), (3,3), (4,2), (5,1)$, so $\P(S = 6) = \frac{5}{36}$.

$S \geq 10$: three pairs for $10$, two for $11$, one for $12$: $\P(S \geq 10) = \frac{6}{36} = \frac16$.

$S$ even: sums $2, 4, 6, 8, 10, 12$ have $1 + 3 + 5 + 5 + 3 + 1 = 18$ pairs: [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac{18}{36} = \frac12$.

**Exercise 18.3 ★.**

A [random variable](#def-g11-prob-rv) has [distribution](#def-g11-prob-rv)

| $x$ | $-1$ | $0$ | $2$ | $5$ |
| --- | --- | --- | --- | --- |
| $\P(X=x)$ | $0.3$ | $0.2$ | $0.4$ | $p$ |

Find $p$, then compute $\E(X)$.

**Solution of Exercise 18.3.**

The probabilities sum to $1$: $p = 1 - 0.3 - 0.2 - 0.4 = 0.1$. Then

$$
\E(X) = 0.3 \times (-1) + 0.2 \times 0 + 0.4 \times 2 + 0.1 \times 5
= -0.3 + 0.8 + 0.5 = 1 .
$$

**Exercise 18.4 ★.**

$X$ satisfies $\E(X) = 3$ and $\V(X) = 4$. Give the [expectation](#def-g11-prob-expectation), [variance](#def-g11-prob-variance) and [standard deviation](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-variance) of $Y = 2X - 5$.

**Solution of Exercise 18.4.**

$\E(Y) = 2 \times 3 - 5 = 1$; $\V(Y) = 2^2 \times 4 = 16$; $\sigma(Y) = 4$.

**Exercise 18.5 ★★.**

A wheel of fortune has $10$ equal sectors: $5$ show $0$ euros, $3$ show $2$ euros, $2$ show $5$ euros. Playing costs $2$ euros. Give the [distribution](#def-g11-prob-rv) of the gain $G$ (winnings minus stake), compute $\E(G)$, and decide whether the game is favorable.

**Solution of Exercise 18.5.**

The winnings are $0$, $2$, $5$ euros with probabilities $0.5$, $0.3$, $0.2$; subtracting the $2$-euro stake, $G$ takes the values $-2, 0, 3$:

| $g$ | $-2$ | $0$ | $3$ |
| --- | --- | --- | --- |
| $\P(G=g)$ | $0.5$ | $0.3$ | $0.2$ |

$\E(G) = -1 + 0 + 0.6 = -0.4$: the player loses $40$ cents per game on average — unfavorable.

**Exercise 18.6 ★★.**

An urn contains $3$ red and $2$ blue balls; two balls are drawn *without* replacement. Let $X$ be the number of red balls drawn. Build the [distribution](#def-g11-prob-rv) of $X$ (enumerate the $20$ ordered draws, or use a [tree](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#met-g10-proba-tree)), and compute $\E(X)$.

**Solution of Exercise 18.6.**

Drawing in order: $\P(X = 2) = \frac35 \times \frac24 = \frac{3}{10}$ (red then red); $\P(X = 0) = \frac25 \times \frac14 = \frac{1}{10}$ (blue then blue); hence $\P(X = 1) = 1 - \frac{3}{10} - \frac{1}{10} = \frac{6}{10}$. Then

$$
\E(X) = 0 \times \frac1{10} + 1 \times \frac6{10} + 2 \times \frac3{10}
= \frac{12}{10} = 1.2 ,
$$

consistent with intuition: $\frac35$ of the two draws are red on average.

**Exercise 18.7 ★★.**

For the wheel of [Exercise 18.5](#exo-g11-prob-5), compute $\V(G)$ and $\sigma(G)$.

**Solution of Exercise 18.7.**

$\E(G^2) = 0.5 \times 4 + 0.3 \times 0 + 0.2 \times 9 = 3.8$, so by the shortcut formula $\V(G) = 3.8 - (-0.4)^2 = 3.8 - 0.16 = 3.64$ and $\sigma(G) \approx 1.9$ euros.

**Exercise 18.8 ★★.**

An insurance company sells a policy for $80$ euros. With [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.05$ the client claims $1000$ euros, with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.01$ the client claims $5000$ euros, otherwise nothing. Let $B$ be the company’s benefit on one policy. Give the [distribution](#def-g11-prob-rv) of $B$, compute $\E(B)$, and interpret.

**Solution of Exercise 18.8.**

The benefit is $80$ minus the claim:

| $b$ | $80$ | $-920$ | $-4920$ |
| --- | --- | --- | --- |
| $\P(B=b)$ | $0.94$ | $0.05$ | $0.01$ |

$\E(B) = 75.2 - 46 - 49.2 = -20$: at this price the company *loses* $20$ euros per policy on average. The premium is too low for the risks covered (it would need to exceed $100$ euros for a positive expected benefit).

**Exercise 18.9 ★★.**

A multiple-choice question has $4$ answers, one correct. A correct answer scores $+1$; to discourage guessing, a wrong answer scores $x < 0$. A student answers at random. For which value of $x$ is the student’s expected score zero?

**Solution of Exercise 18.9.**

The random score is $+1$ with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac14$ and $x$ with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac34$:

$$
\E = \frac14 + \frac{3x}{4} = 0 \iff x = -\frac13 .
$$

With the penalty $-\frac13$, random guessing brings nothing on average.

**Exercise 18.10 ★★.**

Two fair coins are tossed. Ann proposes: “if both coins match, I pay you $1$ euro; if they differ, you pay me $1$ euro.” Compute the expected gain of each player. Is the game fair? Same question with three coins, where Ann pays $2$ euros if all three match.

**Solution of Exercise 18.10.**

Two coins: they match (HH or TT) with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac24 \times 2 =
\frac12$. The player’s gain is $+1$ or $-1$ with equal probabilities: [expectation](#def-g11-prob-expectation) $0$ for both players — the game is fair.

Three coins: all match (HHH or TTT) with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac{2}{8} =
\frac14$. The player receives $2$ with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac14$ and pays $1$ with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac34$: $\E = \frac24 - \frac34 = -\frac14$. The game favours Ann by $25$ cents per round; it would be fair if she paid $3$ euros for a triple match.

**Exercise 18.11 ★★★.**

A lottery sells $1000$ tickets at $5$ euros each. One ticket wins $2000$ euros, five tickets win $200$ euros, twenty tickets win $25$ euros.

1. Give the [distribution](#def-g11-prob-rv) of the organizer’s total revenue minus prizes, and of a single player’s gain $G$ .
2. Compute $\E(G)$ and the total expected profit of the organizer. Check that the two answers are consistent.

**Solution of Exercise 18.11.**

*1.* The organizer receives $1000 \times 5 = 5000$ euros and pays out $2000 + 5 \times 200 + 20 \times 25 = 3500$ euros in all: the profit is the *constant* $1500$ (no randomness once all tickets are sold). A single player’s gain $G$ (prize minus the $5$-euro ticket):

| $g$ | $1995$ | $195$ | $20$ | $-5$ |
| --- | --- | --- | --- | --- |
| $\P(G=g)$ | $\dfrac{1}{1000}$ | $\dfrac{5}{1000}$ | $\dfrac{20}{1000}$ | $\dfrac{974}{1000}$ |

*2.*

$$
\E(G) = \frac{1995 + 5 \times 195 + 20 \times 20 - 974 \times 5}{1000}
= \frac{1995 + 975 + 400 - 4870}{1000} = -1.5 .
$$

Each player expects to lose $1.50$ euros; over $1000$ players that is $1500$ euros — exactly the organizer’s profit. The books balance: what the players lose on average is precisely what the organizer collects.

## 18.6 Problem: The unfinished game

**Problem 18.1.**

Weekend problem — Pascal, Fermat and the fair split of an interrupted bet: the birth of expectation, the magic of linearity, and what expectation cannot see

In 1654 the Chevalier de Méré — the same gambler whose dice bets opened the Middle School volume’s [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) problem — asked Blaise Pascal a deeper question: two equally skilled players must stop their series at the score of $5$ to $3$, six wins taking the $64$-pistole pot. *How should the pot be split fairly?* The letters between Pascal and Fermat answering it founded the concept at the heart of this chapter: [expectation](#def-g11-prob-expectation) ([Definition 18.8](#def-g11-prob-expectation)). You will re-derive their answer both ways, then meet linearity’s party trick and [expectation](#def-g11-prob-expectation)’s one blind spot.

**Part I — [Expectation](#def-g11-prob-expectation) fluency.**

1. Roll a fair die; your gain is (face $- 3$ ) euros. Build the [distribution](#def-g11-prob-rv) table of the gain $G$ ( [Method 18.7](#met-g11-prob-table) ) and compute $\E(G)$ .
2. Compute $V(G)$ with the shortcut formula ( [Proposition 18.13](#prop-g11-prob-konig) ) and $\sigma(G)$ (two decimals).
3. Without any new table, give $\E(2G - 1)$ and $V(2G - 1)$ ( [Proposition 18.14](#prop-g11-prob-affine) ).
4. A stall charges $m$ euros to roll two dice and pockets you the sum in euros. What is the *fair* price $m$ ? The stall charges $8$ : what is your expected loss per game?
5. A laptop worth $800$ euros has a $5\,\%$ chance of destruction each year. Compute the fair insurance premium, and the insurer’s expected profit per policy if it charges $60$ euros.

**Part II — Splitting the pot.** Score $5$–$3$, first to $6$ wins the $64$ pistoles, rounds are fair coin flips.

6. Two tempting wrong answers: split $5 : 3$ (proportionally to rounds won), or give everything to the leader. Say in one sentence each what is unfair about them.
7. Fermat’s method: player A needs $1$ more win, B needs $3$ ; imagine playing exactly $3$ more rounds regardless. List the $8$ equally likely outcomes, mark who wins the series in each, and deduce the fair split of the $64$ pistoles.
8. The subtle point Fermat had to defend: the real game might stop after one round — why is it nevertheless legitimate to reason on all $3$ imaginary rounds?
9. Pascal’s method, backwards from the end: compute A’s fair share at the (hypothetical) scores $5$ – $5$ , then $5$ – $4$ , then $5$ – $3$ , each time averaging the two equally likely futures. Compare with question 7.
10. Same question at the score $4$ – $3$ : how many imaginary rounds, which outcomes give B the series, and what is the fair split?
11. State the principle the two methods share — the fair value of an uncertain future is its *[expectation](#def-g11-prob-expectation)* — and name two modern industries running entirely on it (question 5 shows one).
12. A sanity check for every score: explain, with the linearity of [expectation](#def-g11-prob-expectation) ( [Proposition 18.11](#prop-g11-prob-linearity) ), why the two players’ fair shares must always add up to exactly the $64$ -pistole pot.

**Part III — Linearity’s party trick.**

13. Recover $\E(\text{sum of two dice}) = 7$ in one line with linearity ( [Proposition 18.11](#prop-g11-prob-linearity) ), no $36$ -cell table. For the *[variance](#def-g11-prob-variance)* of the sum, the rule $V(X + Y) = V(X) + V(Y)$ needs the dice to be independent (admitted): compute it.
14. The hat-check party: $n$ guests throw their hats in a pile and each takes one back at random. Let $X$ count the guests who recover their own hat. For one fixed guest, what is the [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) of recovering their own hat? Summing these $n$ chances (linearity!), compute $\E(X)$ . The answer should surprise you: it does not depend on $n$ .
15. Verify $\E(X) = 1$ by brute force for $n = 2$ and $n = 3$ (list the $2$ and $6$ equally likely hat-assignments and average the counts).
16. The guests’ recoveries are far from independent (if $n - 1$ guests got their own hats back, the last one must too). Where exactly did question 14’s computation *not* require independence? State linearity’s superpower.

**Part IV — What [expectation](#def-g11-prob-expectation) cannot see.**

17. Two $5$ -euro scratch cards: card A wins $10$ euros with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac12$ ; card B wins $1\,000$ euros with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $0.005$ . Check that both gains have the same [expectation](#def-g11-prob-expectation) , then compute both [variances](#def-g11-prob-variance) . Which card do real people buy, and what are they buying?
18. The St Petersburg game: flip until the first tails; if the first tails occurs at flip $n$ you receive $2^n$ euros. Compute the contribution of each $n$ to the [expectation](#def-g11-prob-expectation) and the [expectation](#def-g11-prob-expectation) itself. Would you pay $1\,000$ euros to play once? What does the clash teach about [expectation](#def-g11-prob-expectation) ?
19. Insurance from the customer’s side: pay $20$ euros, or face a $1\,\%$ risk of losing $1\,000$ ? Compare the two [expectations](#def-g11-prob-expectation) , and explain why buying the insurance can still be rational (what can a single family not “average over”?).
20. Finale — [expectation](#def-g11-prob-expectation) ’s biography, one sentence per chapter: born in 1654 to split a pot; its superpower, linearity, which adds [expectations](#def-g11-prob-expectation) without asking independence (the hats); its blind spot, risk, which [variance](#def-g11-prob-variance) measures (cards, St Petersburg, insurance); and its coronation next year, when the law of large numbers explains exactly why the house always wins.

**Solution of Problem 18.1.**

**1.** $G$ takes $-2, -1, 0, 1, 2, 3$, each with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac16$: $\E(G) = \frac{-2 - 1 + 0 + 1 + 2 +
3}{6} = \frac12$ euro.

**2.** $\E(G^2) = \frac{4 + 1 + 0 + 1 + 4 + 9}{6} =
\frac{19}{6}$; $V(G) = \frac{19}{6} - \frac14 = \frac{35}{12}
\approx 2.92$; $\sigma(G) \approx 1.71$.

**3.** $\E(2G - 1) = 2 \times \frac12 - 1 = 0$; $V(2G - 1) = 4\,V(G) = \frac{35}{3}$.

**4.** $\E(\text{sum}) = 7$: fair price $7$ euros. At $8$, the expected loss is $1$ euro per game.

**5.** Fair premium: $0.05 \times 800 = 40$ euros. Charging $60$: expected profit $20$ euros per policy.

**6.** Splitting $5:3$ rewards the *past* — but the pot belongs to whoever wins the *future*, and B might still win. Giving all to A treats a probable win as certain — B’s small chance is worth something.

**7.** Rounds to play: $3$; outcomes (A-win/B-win per round): AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. Player B takes the series only by winning all three: BBB alone. A wins the series in $7$ of $8$ cases: fair split $64 \times \frac78 = 56$ pistoles for A, $8$ for B.

**8.** Playing the extra, pointless rounds changes nobody’s victory: once A has her sixth win, later rounds are ceremony. But fixing the horizon at exactly $3$ rounds makes all $8$ outcomes equally likely — honest counting needs same-length stories, and padding them out costs nothing.

**9.** At $5$–$5$: one deciding round, $32$ each. At $5$–$4$: A gets $\frac{64 + 32}{2} = 48$. At $5$–$3$: the next round leads to $64$ (A wins it) or to the $5$–$4$ position worth $48$: A’s share $\frac{64 + 48}{2} = 56$ — Fermat’s $56 : 8$, re-derived by walking backwards.

**10.** A needs $2$, B needs $3$: play $4$ imaginary rounds ($16$ outcomes). B takes the series with $3$ or $4$ B-wins: $4 + 1 = 5$ outcomes. Split: $B$: $64 \times \frac{5}{16} = 20$; $A$: $44$.

**11.** The fair present value of an uncertain future payoff is its [expectation](#def-g11-prob-expectation) — probability-weighted average of what may come. Insurance (question 5) and finance (pricing uncertain payoffs) are this sentence turned into industries; so are casinos, from the other side of the table.

**12.** The pot always ends up somewhere: the two shares are [random variables](#def-g11-prob-rv) summing to the constant $64$, so by linearity $\E_A + \E_B = 64$ — fair shares exhaust the pot, at every score.

**13.** Linearity: $\E(X + Y) = 3.5 + 3.5 = 7$, table-free — true even for glued-together dice. [Variance](#def-g11-prob-variance) of one die: $\E(X^2) - 3.5^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}$; for independent dice $V(X + Y) = \frac{35}{12} + \frac{35}{12} = \frac{35}{6}$.

**14.** Each fixed guest recovers their own hat with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac1n$ (their hat is one among $n$). Summing over the $n$ guests: $\E(X) = n \times \frac1n = 1$ — one lucky guest on average, whether the party has two hats or two thousand.

**15.** $n = 2$: assignments: both right ($X = 2$) or swapped ($X = 0$): $\E = 1$. $n = 3$: the six assignments give $X = 3, 1, 1, 1, 0, 0$: $\E = \frac{6}{6} = 1$.

**16.** The computation added the [expectations](#def-g11-prob-expectation) of the $n$ individual indicators (“guest $i$ got their hat”), and linearity adds [expectations](#def-g11-prob-expectation) *unconditionally* — dependence would wreck a product or a [variance](#def-g11-prob-variance), but never a sum of [expectations](#def-g11-prob-expectation). That is linearity’s superpower: no independence questions asked.

**17.** Both gains: $\E = 0.5 \times 10 - 5 = 0$ and $0.005 \times 1000 - 5 = 0$. [Variances](#def-g11-prob-variance) (of the gain): A: $\E(G^2) = 0.5 \times 25 + 0.5 \times 25 = 25$; B: $0.005 \times 995^2 + 0.995 \times 25 \approx 4\,975$: $\sigma_A = 5$, $\sigma_B \approx 70.5$. People buy B: what they purchase is not [expectation](#def-g11-prob-expectation) (zero in both) but the *spread* — a small chance of a life-changing right tail.

**18.** First tails at flip $n$ has [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $2^{-n}$ and pays $2^n$: each $n$ contributes $2^{-n} \times 2^n = 1$ to the [expectation](#def-g11-prob-expectation), which is therefore $1 + 1 + 1 + \dots$: infinite. Yet no sane person pays $1\,000$ to play once: with [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) $\frac{15}{16}$ the payout is at most $16$. [Expectation](#def-g11-prob-expectation) summarizes the long run of *repeatable* bets; a once-in-a-lifetime game with a wild tail is not priced by its [mean](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#def-g11-stat-mean).

**19.** [Expectations](#def-g11-prob-expectation): insure $-20$; go bare $-0.01 \times 1000 = -10$. The naked strategy wins on average — for someone who can absorb the loss and repeat the bet many times. A family that cannot survive the $-1\,000$ is not in the averaging business: it pays $10$ extra to delete the [variance](#def-g11-prob-variance). Insurers average; households buy certainty.

**20.** Born splitting a pot between two impatient gamblers; empowered by linearity, which sums hopes without asking permission of dependence; blind to risk, which [variance](#def-g11-prob-variance) — the lotteries, the Petersburg game, the family’s policy — must measure instead; and crowned next year by the law of large numbers, which explains why the averages always, eventually, collect.