Expected Value

Is a $5 raffle ticket a good deal? Should you play a carnival game? The expected value answers questions like these: it’s the average result you’d get per try if you repeated an experiment many, many times. Businesses use it to set prices, and it’s the key idea behind insurance, lotteries, and games of chance.

Key ideas

The formula

The expected value of a discrete random variable $X$ is

$$E(X) = \sum x \cdot P(X = x)$$

Multiply each value by its probability, then add up the products. You’ll also see it written $E(X) = x_1 P(x_1) + x_2 P(x_2) + \dots + x_n P(x_n)$.

What it means

$E(X)$ is the long-run average. If you roll a fair die thousands of times, the average of all your rolls will be very close to $E(X) = 3.5$.

The expected value does not have to be a value $X$ can actually take. You can never roll $3.5$; it’s an average, not a prediction for one roll.

Connection to the weighted mean

Suppose a class of $20$ students has $4$ students with no siblings, $10$ with one sibling, and $6$ with two. The mean number of siblings is a weighted mean:

$$\frac{0(4) + 1(10) + 2(6)}{20} = 0\left(\frac{4}{20}\right) + 1\left(\frac{10}{20}\right) + 2\left(\frac{6}{20}\right) = 1.1$$

The fractions $\tfrac{4}{20}$, $\tfrac{10}{20}$, $\tfrac{6}{20}$ act as weights. Expected value is the same calculation, with probabilities as the weights.

Games and fair games

For a game, let $X$ be your net gain: what you win minus what you paid to play. (A loss is a negative gain.)

• If $E(X) \gt 0$, the game favours you in the long run.
• If $E(X) \lt 0$, the game favours the organizer.
• If $E(X) = 0$, the game is fair: on average, nobody gains.

A shortcut: expected net gain $=$ expected winnings $-$ cost to play.

Worked examples

Example 1: One die

Find the expected value of the number rolled on a fair die.

Solution. Each value has probability $\tfrac{1}{6}$:

$$\begin{aligned} E(X) &= 1\left(\tfrac{1}{6}\right) + 2\left(\tfrac{1}{6}\right) + 3\left(\tfrac{1}{6}\right) + 4\left(\tfrac{1}{6}\right) + 5\left(\tfrac{1}{6}\right) + 6\left(\tfrac{1}{6}\right) \\ &= \frac{21}{6} = 3.5 \end{aligned}$$

In the long run, the average roll is $3.5$.

Example 2: From a distribution table

In a town, $X$ is the number of cars owned by a randomly chosen household:

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.10$	$0.35$	$0.40$	$0.15$

Find $E(X)$ and explain what it means.

Solution.

$$E(X) = 0(0.10) + 1(0.35) + 2(0.40) + 3(0.15) = 0 + 0.35 + 0.80 + 0.45 = 1.6$$

On average, households in this town own $1.6$ cars. For example, $1000$ households would have about $1600$ cars in total.

Example 3: A school raffle

A school sells $800$ raffle tickets at $5 each. The prizes are one $1000 gift card, two $250 gift cards, and five $50 gift cards. Find the expected net gain for someone who buys one ticket.

Solution. Find the expected winnings first. The total prize money is $1000 + 2(250) + 5(50) = 1750$ dollars, spread over $800$ tickets:

$$E(\text{winnings}) = 1000\left(\tfrac{1}{800}\right) + 250\left(\tfrac{2}{800}\right) + 50\left(\tfrac{5}{800}\right) = \frac{1750}{800} = 2.1875$$

Now subtract the cost of the ticket:

$$E(\text{net gain}) = 2.1875 - 5 = -2.8125$$

On average, each ticket loses about $2.81. That’s fine for a fundraiser: the school expects to raise $800(5) - 1750 = 2250$ dollars, which is $800 \times 2.8125$. ✓

Example 4: Is the carnival game fair?

A carnival game costs $2 to play. You roll two dice. A sum of $2$ or $12$ wins $20, a sum of $7$ wins $5, and anything else wins nothing. Is the game fair? What price would make it fair?

Solution. From the two-dice distribution, $P(2 \text{ or } 12) = \tfrac{2}{36}$ and $P(7) = \tfrac{6}{36}$.

$$E(\text{winnings}) = 20\left(\tfrac{2}{36}\right) + 5\left(\tfrac{6}{36}\right) + 0 = \frac{70}{36} \approx 1.944$$ $$E(\text{net gain}) = \frac{70}{36} - 2 \approx -0.056$$

The game isn’t fair: players lose about $6$ cents per game on average. It would be fair if the price equalled the expected winnings: $\tfrac{70}{36}$ dollars, or about $1.94.

Common mistakes

Forgetting to subtract the cost. Expected winnings and expected net gain are different. In Example 3, the ticket price must come off: $2.1875 - 5$, not just $2.1875$.

Counting the cost twice. If you use net values in the table (like $1000 - 5 = 995$ for the top prize and $-5$ for losing), don’t subtract the $5 again at the end. Use one method or the other.

Dividing by the number of values. $E(X)$ is not the plain average of the values unless they’re equally likely. Use the probabilities as weights.

Expecting E(X) to be a possible outcome. An expected value of $3.5$ or $1.6$ is a long-run average. It tells you nothing certain about a single trial.

Leaving out a value with zero payoff. It contributes $0$ to the sum, but its probability still matters: the probabilities must add to $1$. Check this before you compute.

Practice

1. (Warm-up) Find $E(X)$.

$x$	$1$	$2$	$3$
$P(X = x)$	$0.5$	$0.3$	$0.2$

Solution

$$E(X) = 1(0.5) + 2(0.3) + 3(0.2) = 0.5 + 0.6 + 0.6 = 1.7$$

2. (Warm-up) In a coin game, heads wins you $3 and tails loses you $1. Find your expected gain per game.

Solution

$$E(X) = 3\left(\tfrac{1}{2}\right) + (-1)\left(\tfrac{1}{2}\right) = 1.5 - 0.5 = 1$$

You expect to gain $1 per game on average, so the game favours you.

3. (Warm-up) A spinner has $8$ equal sections numbered $1$ to $8$. Find the expected value of the number spun.

Solution

$$E(X) = \frac{1 + 2 + \dots + 8}{8} = \frac{36}{8} = 4.5$$

4. (Core) A company sells a one-year phone protection plan for $60. From past data, $4\%$ of customers make a claim, and each claim costs the company $900. Find the company’s expected profit per plan.

Solution

Expected payout per plan: $900(0.04) + 0(0.96) = 36$ dollars.

Expected profit per plan: $60 - 36 = 24$ dollars. The company expects to make $24 per plan in the long run.

5. (Core) A scratch ticket costs $2. It has a $0.001$ chance of winning $500, a $0.01$ chance of winning $20, and otherwise wins nothing. Find the expected net gain per ticket.

Solution

$$E(\text{winnings}) = 500(0.001) + 20(0.01) = 0.5 + 0.2 = 0.7$$$$E(\text{net gain}) = 0.7 - 2 = -1.3$$

On average, a buyer loses $1.30 per ticket.

6. (Core) Three coins are flipped, and $X$ is the number of heads. Find $E(X)$.

Solution

The distribution is $P(0) = \tfrac{1}{8}$, $P(1) = \tfrac{3}{8}$, $P(2) = \tfrac{3}{8}$, $P(3) = \tfrac{1}{8}$.

$$E(X) = 0\left(\tfrac{1}{8}\right) + 1\left(\tfrac{3}{8}\right) + 2\left(\tfrac{3}{8}\right) + 3\left(\tfrac{1}{8}\right) = \frac{12}{8} = 1.5$$

7. (Core) This distribution has $E(X) = 1.5$. Find $a$ and $b$.

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.2$	$a$	$b$	$0.1$

Solution

The probabilities add to $1$: $0.2 + a + b + 0.1 = 1$, so $a + b = 0.7$.

The expected value: $0(0.2) + 1a + 2b + 3(0.1) = 1.5$, so $a + 2b = 1.2$.

Subtracting the first equation from the second: $b = 0.5$, so $a = 0.2$.

Check: $0.2 + 2(0.5) + 0.3 = 1.5$. ✓

8. (Challenge) You roll a die until you get a $6$, but stop after $3$ rolls no matter what. On average, how many rolls do you make? (The distribution is $P(1) = \tfrac{6}{36}$, $P(2) = \tfrac{5}{36}$, $P(3) = \tfrac{25}{36}$; see discrete random variables.)

Solution

$$E(X) = 1\left(\tfrac{6}{36}\right) + 2\left(\tfrac{5}{36}\right) + 3\left(\tfrac{25}{36}\right) = \frac{6 + 10 + 75}{36} = \frac{91}{36} \approx 2.53$$

On average, you make about $2.53$ rolls.

9. (Challenge) A multiple-choice test gives $+4$ points for a correct answer and $-1$ point for a wrong answer. Each question has $5$ choices.

• (a) Find the expected score from a random guess.
• (b) You can rule out one choice for certain and guess among the rest. Should you guess? Explain using expected value.

Solution

(a) $E = 4\left(\tfrac{1}{5}\right) + (-1)\left(\tfrac{4}{5}\right) = \tfrac{4}{5} - \tfrac{4}{5} = 0$. A blind guess is a fair game: on average it neither helps nor hurts.

(b) With $4$ choices left: $E = 4\left(\tfrac{1}{4}\right) + (-1)\left(\tfrac{3}{4}\right) = 1 - 0.75 = 0.25$. The expected score is positive, so guessing helps in the long run.