math.wiki
Probability › Random variables › Expected value
Open on the map
Statistics · S.MD.A.2 · S.MD.A.3

Expected value

A claw machine usually gives nothing, sometimes a small prize, and now and then a big one. What is one play worth on average, and what does that average actually promise?

You need
The law of large numbers and the gambler's fallacyMath 7Probability models: equally likely or notMath 7
Where you are 11 in Random variables
Probability4
Random variables11
▴ 1 aboveContinuous random variables and density curvesCumulative probability
Expected value
Linear transformations of a random variableNormal random variables▾ Show all 11 (5 more)

Why this exists

Vardan stands at an arcade claw machine (made-up numbers throughout this lesson). A play costs 2 dollars. The sign says the claw grabs a big plush worth 15 dollars with probability 0.08, a small one worth 5 dollars with probability 0.12, and otherwise nothing. Is a play worth 2 dollars?

Averaging the three prize values, \displaystyle \frac{15 + 5 + 0}{3} \approx 6.67 dollars, says yes, and is badly wrong: it treats the big plush as if it came up a third of the time, when nothing comes up 80\% of the time. A fair average must give each value a say in proportion to how often it happens. That average is the expected value. It turns a whole distribution into one number, says what a game pays per play in the long run, and lets a business compare choices whose outcomes are uncertain.

The lesson takes two days: the first builds the expected value and its long-run meaning, the second uses it on games and decisions, and marks where it stops being a good guide.

Definition

A random variable X attaches a number to each outcome of a chance process, and its probability distribution lists the possible values x_1, x_2, \ldots, x_n with their probabilities p_1, p_2, \ldots, p_n, which are not negative and add to 1.

The expected value (or mean) of X multiplies each value by its probability and adds the products:

\begin{aligned} E(X) = \mu_X &= x_1 p_1 + x_2 p_2 + \cdots + x_n p_n \\ &= \sum x_i p_i . \end{aligned}

It has the units of X. A game is a random variable for the player’s net gain (what comes back minus what was paid), and a game is fair when its expected net gain is 0.

Intuition

The expected value is the mean from a frequency table with the relative frequencies replaced by probabilities. Picture 100 plays of the claw machine that come out exactly in proportion: 8 big prizes, 12 small ones, 80 misses. The total prize value is 8 \cdot 15 + 12 \cdot 5 + 80 \cdot 0 = 180 dollars, or 1.80 dollars a play. Dividing by 100 first gives the same thing in the expected-value form: 15 \cdot 0.08 + 5 \cdot 0.12 + 0 \cdot 0.80 = 1.80.

0 5 10 15 0 0.2 0.4 0.6 0.8 0.80 0.12 0.08 mean 1.80 Probability Prize value of one play (dollars)
Figure 1The claw machine’s prize value per play (made-up probabilities). The triangle at 1.80 is the balance point of the bars; no single play lands there.

The figure shows the second picture: the expected value is the balance point of the distribution. Hang a weight of 0.80 at 0, 0.12 at 5 and 0.08 at 15 on a light rod, and it balances at 1.80. The heavy weight at 0 pulls the point far to the left, which is why the plain average of the values, 6.67, misses so badly.

Notice that 1.80 dollars is not a prize anyone can win. The expected value is an average, and an average need not be a possible value, any more than a family has 2.3 children.

Results

Weighted mean. E(X) = \sum x_i p_i. Values with larger probabilities pull it toward themselves.

Inside the range. The expected value lies between the smallest and largest values of X, and it equals the plain average of the values when all the values are equally likely.

Balance. The probability-weighted distances from the mean cancel: \sum (x_i - \mu_X)\, p_i = 0.

Long run. If the chance process is repeated independently under the same conditions, the average of the observed values settles near E(X) as the number of repetitions grows.

Fair game. A game is fair when its expected net gain is 0.

Why it is true

The long-run average. Repeat the process N times and suppose value x_i turns up N_i times. The average of the N results is

\begin{aligned} &\frac{x_1 N_1 + \cdots + x_n N_n}{N} \\ &\quad = x_1 \frac{N_1}{N} + \cdots + x_n \frac{N_n}{N}. \end{aligned}

Each fraction \displaystyle \frac{N_i}{N} is the relative frequency of x_i, and by the law of large numbers it settles near p_i as N grows. So the average settles near x_1 p_1 + \cdots + x_n p_n = E(X). This is why the weighted sum, and not the plain average of the values, is the right definition: it is the number the data average heads toward.

Inside the range. Let m be the smallest value and M the largest. Each x_i p_i lies between m p_i and M p_i because p_i \ge 0. Adding, and using \sum p_i = 1, gives m \le E(X) \le M. If all n values are equally likely, each \displaystyle p_i = \frac{1}{n} and the sum is the plain average.

Balance. Splitting the sum and using \sum p_i = 1,

\begin{aligned} \sum (x_i - \mu_X)\, p_i &= \sum x_i p_i - \mu_X \sum p_i \\ &= \mu_X - \mu_X \cdot 1 = 0 . \end{aligned}

Worked examples

Examples 1 to 3 belong to the first day (computing the expected value and reading it as a long-run average), Examples 4 to 7 to the second (games, fair games and decisions).

Example 1

The claw machine’s prize value

Let X be the prize value, in dollars, of one play of Vardan’s claw machine. Find E(X) and say what it means.

Solution

Lay out the distribution with a column for the products.

Prize value x (dollars)Probability P(x)x \cdot P(x)
00.800
50.120.60
150.081.20
Total11.80

The probabilities add to 1, so this is a valid distribution. Adding the last column, E(X) = 0 + 0.60 + 1.20 = 1.80 dollars. Over many plays, the claw machine gives out prizes worth about 1.80 dollars per play on average. No single play is worth 1.80 dollars: each one gives 0, 5 or 15.

Example 2

From a frequency table to a distribution

Zabel, who runs the arcade, records how many times each of 200 customers played the claw machine on one visit: 90 played once, 60 twice, 30 three times and 20 four times. Let X be the number of plays by a customer chosen at random from these 200. Find E(X) two ways.

Solution

As a mean from a frequency table, the total number of plays is 1 \cdot 90 + 2 \cdot 60 + 3 \cdot 30 + 4 \cdot 20 = 380, and \displaystyle \frac{380}{200} = 1.90.

As an expected value, choosing a customer at random gives each value its relative frequency as its probability: \displaystyle P(1) = \frac{90}{200} = 0.45, P(2) = 0.30, P(3) = 0.15, P(4) = 0.10. Then

\begin{aligned} E(X) &= 1(0.45) + 2(0.30) \\ &\quad + 3(0.15) + 4(0.10) \\ &= 1.90 . \end{aligned}

The two answers agree because dividing the total by 200 is the same as weighting each value by its share. A randomly chosen customer of these 200 played the machine 1.90 times on average, which again is not a number of plays anyone made.

Example 3

What the long run does and does not promise

Arpi plays the claw machine 50 times and wins 6 big prizes and 5 small ones. (a) Find her average prize value per play and compare it with E(X) = 1.80 dollars. (b) Zabel expects about 10{,}000 plays a year. How much money does she expect to keep after paying for prizes?

Solution

(a) Arpi’s prizes total 6 \cdot 15 + 5 \cdot 5 = 115 dollars, so her average is \displaystyle \frac{115}{50} = 2.30 dollars a play, 0.50 dollars above the expected value. That is no contradiction. The expected value describes the average over many plays; in 50 plays one extra big prize moves the average by \displaystyle \frac{15}{50} = 0.30 dollars, so short-run averages swing.

(b) The law of large numbers says the average prize value per play will settle near 1.80 dollars, so the prizes should come to about 10{,}000 \times 1.80 = 18{,}000 dollars. The plays bring in 10{,}000 \times 2 = 20{,}000 dollars, leaving Zabel about 2{,}000 dollars, or 0.20 dollars a play. The promise is about the average: the total kept will be close to 2{,}000 dollars in relative terms, not exactly 2{,}000.

Example 4

The expected value of a game

Let G be Vardan’s net gain from one play, the prize value minus the 2 dollars paid. Find E(G). Is the game fair?

Solution

The three outcomes give net gains of 0 - 2 = -2, 5 - 2 = 3 and 15 - 2 = 13 dollars, with the same probabilities as before.

Net gain g (dollars)P(g)g \cdot P(g)
-20.80-1.60
30.120.36
130.081.04

So E(G) = -1.60 + 0.36 + 1.04 = -0.20 dollars. The same answer comes from E(X) - 2 = 1.80 - 2 = -0.20, since every value dropped by 2 and the weights add to 1 (why this shortcut always works is Linear transformations of a random variable). The game is not fair: over many plays a player loses about 20 cents a play, which is exactly what Zabel keeps.

Example 5

Making the game fair

Find (a) the price per play, and (b) the value of the big plush at a price of 2 dollars, that would make the claw game fair.

Solution

(a) The game is fair when the expected prize value equals the price, so a price of 1.80 dollars makes it fair: E(G) = 1.80 - 1.80 = 0.

(b) Call the big plush’s value v. The expected prize value is 0.08v + 5(0.12) = 0.08v + 0.60, and fairness asks for this to equal 2:

\begin{aligned} 0.08v + 0.60 &= 2 \\ 0.08v &= 1.40 \\ v &= 17.50 . \end{aligned}

A big plush worth 17.50 dollars makes the game fair. Check: 0.08 \times 17.50 = 1.40, and 1.40 + 0.60 = 2.

Example 6

A repair plan for the claw machine

A service company offers Zabel a repair plan for the claw machine: 120 dollars a year, and every repair is covered. Without it, her yearly repair bill is 0 dollars with probability 0.60, 150 dollars with probability 0.30, and 600 dollars with probability 0.10. Should she buy the plan?

Solution

Let C be the yearly repair bill without the plan.

\begin{aligned} E(C) &= 0(0.60) + 150(0.30) + 600(0.10) \\ &= 45 + 60 = 105 \text{ dollars.} \end{aligned}

The plan costs 120 dollars for certain, 15 dollars a year more than the expected bill. By expected value alone, she should not buy it: over many years, and across several machines, paying repairs herself costs about 15 dollars a year less per machine. This is also why the company can offer the plan: it expects to take in 15 dollars a year more than it pays out on each machine.

Expected value is not the whole answer, though. Without the plan there is a 0.10 chance of a 600 dollar bill in a single year. If a bill that size would close the arcade, paying a steady 15 dollars a year above the expected cost to remove that risk can be the sensible choice. That trade is what every insurance policy sells.

Example 7

A one-time choice

Vardan wins the arcade’s high-score contest and may take 40 dollars in cash, or draw once from a box in which 1 card in 10 pays 350 dollars and the rest pay 10 dollars. Compare the choices.

Solution

Let D be the payout of the draw.

\begin{aligned} E(D) &= 350(0.1) + 10(0.9) \\ &= 35 + 9 = 44 \text{ dollars.} \end{aligned}

By expected value the draw is better, by 44 - 40 = 4 dollars. The cash would have to be 44 dollars for the two to be equal.

0 50 100 150 200 250 300 350 0 0.5 1 1 mean 40 Take the cash 0 50 100 150 200 250 300 350 0 0.5 1 0.9 0.1 mean 44 Take the draw Dollars Vardan leaves with
Figure 2Vardan’s one-time choice (Example 7). The means are close, 40 against 44 dollars, but the two pictures are not.

But the long-run meaning of 44 needs many draws, and Vardan gets one. With probability 0.9 he walks away with 10 dollars, 30 less than the sure cash; the draw beats the cash only one time in ten. Someone who would be happy with 40 dollars and disappointed with 10 has a sound reason to take the cash, and someone who wants the long shot at 350 has a sound reason to draw. The expected value reports the center of each distribution; for a one-time choice the whole distribution matters, and the spread is measured in the variance of a random variable.

Boundaries

  • The expected value needs a probability distribution: values with probabilities that are not negative and add to 1. If the probabilities are guesses, the expected value is only as good as the guesses.
  • Its long-run meaning needs repeats that are independent and under the same conditions. For a single event (one draw, one contest, one season) it is a summary of the distribution, not a forecast of the result.
  • It measures center only. Two choices with the same expected value can carry very different risk, as in Example 7.
  • With infinitely many possible values, the sum may not settle. A game that pays 2^k dollars with probability \displaystyle \frac{1}{2^k} for k = 1, 2, 3, \ldots adds 1 dollar to the sum for every k, so it has no finite expected value.

Where this goes next

The spread of a random variable around its expected value, weighted the same way, is its variance. Adding a constant or changing units moves the expected value in a predictable way (Linear transformations of a random variable), and the expected values of two random variables add, which combining random variables puts beside the rule for spreads. Famous distributions come with ready-made expected values: the binomial and the geometric. For a continuous random variable the balance point moves from bars to a density curve. A fair game links back to odds: the payout balances the odds.

Exercises

Exercises 1 to 8 go with the first day (Examples 1 to 3), Exercises 9 to 15 with the second (Examples 4 to 7), and the last group draws on both.

Part 1: computing and reading the expected value

  1. Let X be the number of times the claw machine jams in a day, with P(0) = 0.1, P(1) = 0.3, P(2) = 0.4 and P(3) = 0.2. Find E(X).
  2. A smaller claw machine gives a prize worth 8 dollars with probability 0.05, one worth 3 dollars with probability 0.25, and nothing otherwise. Find the expected prize value per play. Can a single play give exactly that value?
  3. A bonus button gives 0 tickets with probability 0.5, 10 tickets with probability 0.3, 25 tickets, or 100 tickets with probability 0.05. Find the probability of 25 tickets and the expected number of tickets.
  4. Over 40 visits, Vardan played the claw machine 2 times on 10 visits, 3 times on 14, 4 times on 12 and 5 times on 4. For a visit chosen at random, find the expected number of plays.
  5. Arpi plays the smaller machine of Exercise 2 300 times at 1.50 dollars a play. About how much prize value should she expect in total, and how far behind should she expect to be?
  6. A friend reads E(X) = 1.7 in Exercise 1 and says “the machine jams 1.7 times every day.” Correct the statement.
  7. A random variable takes the values 2, 6 and v with probabilities 0.5, 0.3 and 0.2, and E(X) = 5. Find v.
  8. After 20 plays on the main claw machine, Vardan’s average prize value is 2.75 dollars a play. Does this show the sign’s probabilities are wrong? Explain.

Part 2: games and decisions

  1. The arcade’s hoop-shooting game costs 3 dollars. It gives a prize worth 10 dollars with probability 0.2, a prize worth 4 dollars with probability 0.1, and nothing otherwise. Find the expected net gain per game. Is the game fair?
  2. What price would make the game of Exercise 9 fair, with the same prizes?
  3. At 3 dollars a game, what value of the top prize in Exercise 9 would make the game fair?
  4. You pay 1 dollar to roll a fair die. A six pays 5 dollars; anything else pays nothing. Find the expected net gain, and the payout for a six that would make the game fair.
  5. A three-year plan covering repairs to the arcade’s new air hockey table costs 200 dollars. Without it, the three-year repair bill is 0 dollars with probability 0.7, 250 dollars with probability 0.2 and 900 dollars with probability 0.1. What does expected value say? Give one reason to buy the plan anyway.
  6. Zabel can place one new machine by the door for the whole year. Machine A takes in 120 dollars on a day with probability 0.5 and 60 dollars otherwise; machine B takes in 200 dollars with probability 0.3 and 50 dollars otherwise. Which should she choose, and why is expected value a good guide here?
  7. The arcade sells all 400 tickets for a prize drawing at 5 dollars each, and the one prize is worth 1{,}200 dollars. Find the expected net gain for a ticket holder and the arcade’s profit.

Explaining

  1. Explain why the expected value of a random variable can never be larger than its largest value.
  2. Explain why the plain average of the possible values equals the expected value when every value is equally likely. Use the claw machine to show that otherwise the two can be far apart.
  3. Zabel and Vardan both look at E(G) = -0.20 dollars for the claw game. Explain why it is a better guide to Zabel’s choices than to Vardan’s choice to play once.
  4. Show that, for the distribution in Exercise 1, the probability-weighted distances (x - \mu_X)\,P(x) add to 0.
  5. A game has expected net gain 0. Does a player who plays it 10 times break even? Explain.
Answers
  1. E(X) = 1.7 jams. 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6.
  2. 1.15 dollars. 8(0.05) + 3(0.25) + 0(0.70) = 0.40 + 0.75. No: each play gives 0, 3 or 8 dollars; 1.15 is the long-run average.
  3. P(25) = 0.15 and the expected number is 11.75 tickets. The probabilities add to 1, so P(25) = 1 - 0.5 - 0.3 - 0.05 = 0.15; then 0 + 10(0.3) + 25(0.15) + 100(0.05) = 3 + 3.75 + 5.
  4. 3.25 plays. The probabilities are \displaystyle \frac{10}{40} = 0.25, 0.35, 0.30 and 0.10, and 2(0.25) + 3(0.35) + 4(0.30) + 5(0.10) = 0.5 + 1.05 + 1.2 + 0.5.
  5. About 345 dollars of prizes, from 300 \times 1.15, against 300 \times 1.50 = 450 dollars paid, so about 105 dollars behind. These are long-run approximations, not exact totals.
  6. The machine jams 0, 1, 2 or 3 times on a given day; over many days it averages about 1.7 jams a day.
  7. v = 11. 2(0.5) + 6(0.3) + 0.2v = 5 gives 2.8 + 0.2v = 5, so 0.2v = 2.2.
  8. No. Twenty plays is a short run: one extra big prize moves the average by \displaystyle \frac{15}{20} = 0.75 dollars. The law of large numbers speaks about many plays; only a long record well away from 1.80 dollars would cast doubt on the sign.
  9. E = -0.60 dollars; not fair. The expected prize value is 10(0.2) + 4(0.1) = 2.40 dollars, and 2.40 - 3 = -0.60.
  10. 2.40 dollars, the expected prize value.
  11. 13 dollars. 0.2v + 0.40 = 3 gives 0.2v = 2.60.
  12. \displaystyle E = -\frac{1}{6} \approx -0.17 dollars. The net gain is 4 with probability \displaystyle \frac{1}{6} and -1 with probability \displaystyle \frac{5}{6}, so \displaystyle E = \frac{4}{6} - \frac{5}{6}. The game is fair if a six pays 6 dollars: \displaystyle \frac{5}{6} - \frac{5}{6} = 0. The net payout of 5 to 1 matches the odds against a six.
  13. Expected bill 140 dollars, from 0.2(250) + 0.1(900) = 50 + 90, so by expected value she should skip the 200 dollar plan and save about 60 dollars on average. A reason to buy anyway: a 0.1 chance of a 900 dollar bill she may not be able to absorb.
  14. Machine B. E(A) = 120(0.5) + 60(0.5) = 90 dollars a day and E(B) = 200(0.3) + 50(0.7) = 60 + 35 = 95 dollars a day. Over hundreds of days the daily average settles near the expected value, so B should take in about 5 dollars a day more.
  15. -2 dollars per ticket: \displaystyle \frac{1200}{400} - 5 = 3 - 5. The arcade takes in 400 \times 5 = 2{,}000 dollars and pays 1{,}200, a profit of 800 dollars for certain, since every ticket was sold and the prize is awarded once.
  16. Each term x_i p_i is at most M p_i, where M is the largest value, because p_i \ge 0. Adding, E(X) \le M(p_1 + \cdots + p_n) = M.
  17. With n equally likely values each weight is \displaystyle \frac{1}{n}, so \displaystyle \sum x_i \cdot \frac{1}{n} = \frac{x_1 + \cdots + x_n}{n}, the plain average. For the claw machine the plain average is \displaystyle \frac{15 + 5 + 0}{3} \approx 6.67 dollars but E(X) = 1.80 dollars, because 0 carries weight 0.80.
  18. Zabel’s machine is played thousands of times, so the law of large numbers makes her average take close to 0.20 dollars a play. Vardan’s one play gives -2, 3 or 13 dollars, never -0.20; for one play the whole distribution matters more than its center.
  19. The distances from 1.7 are -1.7, -0.7, 0.3 and 1.3, and -1.7(0.1) - 0.7(0.3) + 0.3(0.4) + 1.3(0.2) = -0.17 - 0.21 + 0.12 + 0.26 = 0.
  20. Not necessarily. Expected value 0 says the average net gain per game settles near 0 over many games; in 10 games the player can easily be ahead or behind.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0