IB Analysis and Approaches HL · Discrete random variables and the binomial distribution · Section 2 of 4

Expected value, variance and fair games

3 worked examples · By Álvaro López

Key idea

The expected value $E (X)$ is the mean value that $X$ tends to if you repeat the experiment many times; the variance measures how far the values spread around that mean. With them you can decide whether a game is fair (expected gain equal to zero). In the IB exam they are calculated by hand in Paper 1 and with the GDC in Paper 2.

Expected value of a discrete variable

If you play many times, you do not expect the most frequent value but the weighted average: each value counts according to its probability.

The expected value does not have to be a value that $X$ can take. A normal die has $E (X) = 3.5$ , which never comes up in a single roll: it is the long-term average, not a concrete result.

If the expected value is a decimal, leave it as a decimal or as a simplified fraction; do not round it to an integer.

Variance and standard deviation

The formula $E (X^{2}) - [E (X)]^{2}$ is the most convenient by hand: first calculate $E (X)$ (the mean) and then $E (X^{2})$ by squaring the values of $x$ before multiplying by the probability. The probabilities are not squared.

Games and expected gain

A game of chance can be described by a random variable $G$ : the gain of the player (the prize minus what they pay to play, with a negative sign if they lose). The value $E (G)$ is what the player wins or loses on average per game if they play many times.

Be careful with the meaning of ‘gain’: if the question gives the gross prize, you must subtract what is paid to play to obtain the net gain. Always interpret the result: “on average, the player loses 0.80 euros per game”.

Also, for $n$ games the total expected gain is $n \cdot E (G)$ .

Worked examples

Example 1 Basic

Mean and standard deviation of a table

A student records the number of assignments handed in per week. The random variable $X$ has this distribution:

$x$	0	1	2	3	4
$P (X = x)$	$0.1$	$0.3$	$0.3$	$0.2$	$0.1$

a) Calculate $E (X)$ .

b) Calculate $Var (X)$ .

c) Calculate the standard deviation $σ$ .

Solution

a) $E (X) = 0 \cdot 0.1 + 1 \cdot 0.3 + 2 \cdot 0.3 + 3 \cdot 0.2 + 4 \cdot 0.1 = 0.3 + 0.6 + 0.6 + 0.4 = 1.9$

b) We calculate $E (X^{2})$ using the squares of the values: $E (X^{2}) = 0 \cdot 0.1 + 1 \cdot 0.3 + 4 \cdot 0.3 + 9 \cdot 0.2 + 16 \cdot 0.1 = 0.3 + 1.2 + 1.8 + 1.6 = 4.9$ $Var (X) = 4.9 - 1. 9^{2} = 4.9 - 3.61 = 1.29$

c) $σ = 1.29 \approx 1.14$

Answer: a) $1.9$ ; b) $1.29$ ; c) $1.29 \approx 1.14$ .

Example 2 Medium

Two unknown probabilities

The random variable $X$ has the distribution

$x$	1	2	3	4
$P (X = x)$	$a$	$0.3$	$b$	$0.1$

and it is known that $E (X) = 2.4$ .

a) Find the values of $a$ and $b$ .

b) Find $Var (X)$ .

c) Find $P (X > E (X))$ .

Solution

a) There are two unknowns, so two equations are needed. The probabilities sum to 1: $a + 0.3 + b + 0.1 = 1 \Rightarrow a + b = 0.6$ And the expected value is $2.4$ : $1 \cdot a + 2 \cdot 0.3 + 3 b + 4 \cdot 0.1 = 2.4 \Rightarrow a + 3 b = 1.4$ Subtracting the first equation from the second: $2 b = 0.8$ , so $b = 0.4$ and $a = 0.2$ . Both are between 0 and 1.

b) With the complete table: $E (X^{2}) = 1 \cdot 0.2 + 4 \cdot 0.3 + 9 \cdot 0.4 + 16 \cdot 0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6$ $Var (X) = 6.6 - 2. 4^{2} = 6.6 - 5.76 = 0.84$

c) As $E (X) = 2.4$ , the condition $X > 2.4$ is equivalent to $X \geq 3$ : $P (X \geq 3) = 0.4 + 0.1 = 0.5$

Answer: a) $a = 0.2$ , $b = 0.4$ ; b) $0.84$ ; c) $0.5$ .

Example 3 Hard

A raffle with two gold balls

At a fair, playing costs 6 euros. There is a bag with 6 balls, 2 gold and 4 grey. The player draws 2 balls at random, without replacement: if both are gold they win 30 euros, if only one is gold they win 6 euros and if none is gold they win nothing. (The 6 euros entry fee has already been paid.)

Let $G$ be the player's net gain in one game, that is, the prize minus 6 euros.

a) Write down the probability distribution of $G$ .

b) Find $E (G)$ and interpret the result.

c) The organiser changes the prize for a single gold ball to $k$ euros and leaves the others unchanged. Find $k$ so that the game is fair.

d) With the original game, what is the organiser's expected gain after 150 games?

Solution

a) There are $(2 6) = 15$ equally likely pairs.

Two gold: $(2 2) = 1$ pair, $P = \frac{1}{15}$ , and $G = 30 - 6 = 24$ .
One gold: $(1 2) (1 4) = 8$ pairs, $P = \frac{8}{15}$ , and $G = 6 - 6 = 0$ .
No gold: $(2 4) = 6$ pairs, $P = \frac{6}{15} = \frac{2}{5}$ , and $G = - 6$ .

$g$	$- 6$	$0$	$24$
$P (G = g)$	$\frac{2}{5}$	$\frac{8}{15}$	$\frac{1}{15}$

Check: $\frac{6}{15} + \frac{8}{15} + \frac{1}{15} = 1$ .

b) $E (G) = (- 6) \cdot \frac{6}{15} + 0 \cdot \frac{8}{15} + 24 \cdot \frac{1}{15} = \frac{- 36 + 24}{15} = - \frac{4}{5}$ That is, $- 0.80$ euros: on average, the player loses 0.80 euros per game. The game is not fair and favours the organiser.

c) With a prize of $k$ for one gold ball, the gain in that case is $k - 6$ : $E (G) = - 6 \cdot \frac{6}{15} + (k - 6) \cdot \frac{8}{15} + 24 \cdot \frac{1}{15} = \frac{- 36 + 8 k - 48 + 24}{15} = \frac{8 k - 60}{15}$ The game is fair if $E (G) = 0$ , that is, $8 k = 60$ and $k = \frac{15}{2} = 7.5$ euros.

d) The player's gain is the organiser's loss, so the organiser gains on average $0.80$ euros per game: $150 \cdot 0.80 = 120 euros$

Answer: b) $E (G) = - \frac{4}{5}$ euros; c) $k = 7.5$ euros; d) $120$ euros.

In the exam: define $G$ clearly (net or gross), write the table and the sum of products, and finish by interpreting the sign of $E (G)$ in the context of the game; without that sentence you lose the interpretation mark.

Common mistakes

Summary

$E (X) = \sum x P (X = x)$ : a weighted average by probability; it need not be a value that $X$ can take.
$Var (X) = E (X^{2}) - [E (X)]^{2}$ , with $E (X^{2}) = \sum x^{2} P (X = x)$ ; $σ = Var (X)$ .
In Paper 2 the GDC gives $E (X)$ and $σ$ when you enter values and probabilities as frequencies; still write down the set-up.
A game is fair if the expected net gain is zero; if it is positive it favours the player, if negative, the organiser.
The total expected gain after $n$ games is $n \cdot E (G)$ .
With two unknowns in a table: $\sum P = 1$ plus $E (X)$ or a cumulative probability give two equations.

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