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Expected value

Probability · level 7, lesson 2 of 7

A question to start

An old fairground game called chuck-a-luck works like this. You bet 1 dollar on a number from 1 to 6. Three dice are rolled. If your number shows on one die, you get your dollar back plus 1 dollar; on two dice, your dollar back plus 2; on all three, your dollar back plus 3. If it shows on none, you lose your dollar.

The stallholder's patter goes: "Three dice, and each one has a 1 in 6 chance of your number. That's three chances in six, an even bet, and sometimes I pay double or triple!" It sounds generous. Is it? By the end of this lesson you will find the exact amount the game takes from a player per bet, in the long run: about 8 cents.

The idea

The long-run average

Take the simplest game. Pay 1 dollar and roll a die; if you roll a 6 you are paid 5 dollars. Let \(X\) be your profit: \(+4\) dollars on a 6 (paid 5, minus the 1 you spent), and \(-1\) dollar otherwise.

\(x\) \(+4\) \(-1\)
\(P(X = x)\) \(\tfrac16\) \(\tfrac56\)

Imagine playing \(N\) times, where \(N\) is large. About \(\tfrac16 N\) of the games give \(+4\) and about \(\tfrac56 N\) give \(-1\). The total profit is about

\[4 \cdot \tfrac16 N + (-1) \cdot \tfrac56 N = \left(4 \cdot \tfrac16 - 1 \cdot \tfrac56\right) N = -\tfrac16 N.\]

The average profit per game is the total divided by \(N\), and the \(N\) cancels: \(-\tfrac16\) of a dollar, about \(-17\) cents. What is left is "each value times its probability, added up". That is the definition.

The definition

The expected value (or mean) of a discrete random variable \(X\) is

\[E(X) = \sum x \, P(X = x),\]

the sum over every value \(x\) of the value times its probability. It is also written \(\mu\) (Greek "mu") or \(\mu_X\). It is a weighted average of the values: each value is weighted by how often it happens.

Three things to notice:

  • It is a long-run average, not a prediction for one go. In one game of the die stall you win 4 or lose 1; you never lose 17 cents.
  • It need not be a possible value. For one fair die, \(E(X) = \tfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\), though nobody rolls 3.5.
  • It is the balance point of the probability histogram. If the bars were blocks of wood on a weightless ruler, the ruler would balance on a pin placed at \(E(X)\). For a symmetric histogram, like the sum of two dice, the balance point is the centre, 7.

Why it is called "expected" is a historical accident; "long-run average" is the honest name. The link between the two, the fact that averages really do settle down near \(E(X)\), is the law of large numbers, proved in level 8.

Fair games, precisely

In early levels "fair" meant "everyone has the same chance". With money, a sharper definition is:

A game is fair when the expected profit of a player is 0.

Then nobody gains or loses in the long run. The die stall above is unfair to the player by about 17 cents a game. To make it fair, choose the payout \(w\) so that the expected amount received equals the cost:

\[w \cdot \tfrac16 = 1 \quad \Longrightarrow \quad w = 6 \text{ dollars}.\]

There is a neat way to see it: in six games you expect one 6, so a fair stall pays back the 6 dollars it took in those six games. Real stalls, lotteries and casinos are never fair: the negative expected value for the player is how they pay their bills.

Rules for expected value

Two rules make expected values much easier.

Rule 1: scaling and shifting. For numbers \(a\) and \(b\),

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

Reason: \(E(aX + b) = \sum (ax + b) P(x) = a \sum x P(x) + b \sum P(x) = a E(X) + b \cdot 1\), because the probabilities add to 1. Example: a die stall pays 2 dollars for every spot on the die and charges 8 dollars to play. Profit is \(2X - 8\), so \(E = 2(3.5) - 8 = -1\) dollar.

Rule 2: adding. For any two random variables \(X\) and \(Y\) on the same experiment,

\[E(X + Y) = E(X) + E(Y).\]

Reason: \(E(X + Y)\) adds \((x + y)\) times the probability of each outcome. Split it into the sum of the \(x\) parts and the sum of the \(y\) parts; the first is \(E(X)\) and the second is \(E(Y)\). Nothing in this argument needs \(X\) and \(Y\) to be independent, and that is what makes the rule so powerful.

Two dice: each has mean 3.5, so the expected sum is \(3.5 + 3.5 = 7\), with no table needed. (The long way agrees: the 36 sums add to 252, and \(252 \div 36 = 7\).)

Counting with indicators

Rule 2 gives a beautiful trick. To find the expected number of times something happens, write the count as a sum of indicators: random variables that are 1 if a particular thing happens and 0 if not. An indicator \(I\) with \(P(I = 1) = p\) has \(E(I) = 1 \cdot p + 0 \cdot (1 - p) = p\).

Example. Roll a die 10 times. Let \(S\) be the number of sixes. Then \(S = I_1 + I_2 + \dots + I_{10}\), where \(I_k\) is 1 if roll \(k\) is a six. Each \(E(I_k) = \tfrac16\), so \(E(S) = 10 \cdot \tfrac16 = \tfrac53 \approx 1.67\).

A surprising example. Ten people throw their hats in a pile and each takes one at random. What is the expected number who get their own hat back? Let \(I_k\) be 1 if person \(k\) gets their own hat; \(P(I_k = 1) = \tfrac{1}{10}\). So the expected number is \(10 \cdot \tfrac{1}{10} = 1\). With 100 people, it is still \(100 \cdot \tfrac{1}{100} = 1\). The indicators here are not independent (if nine people have their own hats, so does the tenth), but Rule 2 does not care.

When the average is infinite

Expected value is not always finite. In the St Petersburg game, a coin is tossed until the first head; if that happens on toss \(k\), you are paid \(2^k\) dollars. The chance of the first head on toss \(k\) is \(\left(\tfrac12\right)^k\), so

\[E = 2 \cdot \tfrac12 + 4 \cdot \tfrac14 + 8 \cdot \tfrac18 + \dots = 1 + 1 + 1 + \dots,\]

which grows without limit. Yet nobody would pay a million dollars to play: half the time you win only 2. This puzzle (posed in the 1700s) is one reason people think about risk as well as the average, which is the subject of the next lesson.

Worked examples

1. Chuck-a-luck. Let \(X\) be the player's profit on a 1-dollar bet. The number of dice showing your number is like three independent trials with chance \(\tfrac16\). Out of the \(6^3 = 216\) equally likely outcomes:

  • none of the three dice show your number: \(5^3 = 125\) outcomes, profit \(-1\);
  • exactly one: \(3 \times 5^2 = 75\) outcomes (choose which die, the other two miss), profit \(+1\);
  • exactly two: \(3 \times 5 = 15\) outcomes, profit \(+2\);
  • all three: 1 outcome, profit \(+3\).

Check: \(125 + 75 + 15 + 1 = 216\). ✓

\[E(X) = \frac{-125 + 75 + 30 + 3}{216} = \frac{-17}{216} \approx -0.079.\]

The player loses about 7.9 cents per dollar bet. The patter's "three chances in six" counts the outcomes with two or three matches several times over: the chance that your number shows at all is \(\tfrac{91}{216} \approx 0.42\), less than a half.

2. A raffle. 500 tickets are sold at 2 dollars each. There is one prize of 100 dollars and four prizes of 25 dollars. Expected prize money per ticket: \(100 \cdot \tfrac{1}{500} + 25 \cdot \tfrac{4}{500} = \tfrac{200}{500} = 0.40\) dollars. Expected profit: \(0.40 - 2 = -1.60\) dollars. The organisers collect 1000 dollars and pay out 200, raising 800, which is \(500 \times 1.60\).

3. Roulette. A European roulette wheel has 37 pockets: 0, and 1 to 36, of which 18 are red. A 1-dollar bet on red pays 1 dollar profit if red comes up and loses the dollar otherwise.

\[E = 1 \cdot \tfrac{18}{37} + (-1) \cdot \tfrac{19}{37} = -\tfrac{1}{37} \approx -0.027.\]

The single green zero is the whole of the casino's edge: about 2.7 cents per dollar, every bet, for ever.

4. Making a game fair. A game costs 2 dollars. You roll a die. On a 6 you receive \(w\) dollars, on a 4 or 5 you receive 2 dollars, otherwise nothing. Find \(w\) for a fair game.

Expected amount received: \(w \cdot \tfrac16 + 2 \cdot \tfrac26 = \tfrac{w + 4}{6}\). Fair means this equals the 2-dollar cost: \(\tfrac{w + 4}{6} = 2\), so \(w + 4 = 12\) and \(w = 8\) dollars.

Try it

  1. A spinner gives 0 points with probability \(\tfrac12\), 3 points with probability \(\tfrac13\) and 12 points with probability \(\tfrac16\). Find the expected score.
  2. A game costs 2 dollars. Toss two coins; you receive 5 dollars if both are heads. Find the expected profit. What price would make it fair?
  3. Using the distribution of the larger of two dice from lesson 1 (\(P(M = m) = \tfrac{2m - 1}{36}\)), find \(E(M)\).
  4. \(E(X) = 3.5\). Find \(E(3X + 1)\) and \(E(10 - X)\).
  5. A raffle sells 1000 tickets at 5 dollars. Prizes: one of 1000 dollars and two of 250 dollars. Find the expected profit per ticket.
  6. A phone costs 300 dollars and has a 5% chance of breaking in a year. Insurance costs 20 dollars a year. What is the expected cost of breakage without insurance? Why might someone still buy it?
  7. Shuffle a deck of 52 cards and turn them over one by one while counting "ace, two, three, ..., king, ace, two, ..." (four times through). A match is when the card turned shows the rank you say. Find the expected number of matches.
  8. A coin is tossed until the first head, at most 3 times. You win 2, 4 or 8 dollars if the first head is on toss 1, 2 or 3, and nothing if there is no head. Find the expected win. What would the fair price be?
Answers
  1. \(0 \cdot \tfrac12 + 3 \cdot \tfrac13 + 12 \cdot \tfrac16 = 0 + 1 + 2 = 3\) points.
  2. Expected amount received: \(5 \cdot \tfrac14 = 1.25\) dollars. Expected profit \(1.25 - 2 = -0.75\) dollars. A price of 1.25 dollars would make it fair.
  3. \(E(M) = \tfrac{1 \cdot 1 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 7 + 5 \cdot 9 + 6 \cdot 11}{36} = \tfrac{1 + 6 + 15 + 28 + 45 + 66}{36} = \tfrac{161}{36} \approx 4.47\).
  4. \(E(3X + 1) = 3(3.5) + 1 = 11.5\). \(E(10 - X) = 10 - 3.5 = 6.5\).
  5. Expected prize: \(\tfrac{1000 + 2 \times 250}{1000} = 1.50\) dollars. Expected profit: \(1.50 - 5 = -3.50\) dollars.
  6. \(0.05 \times 300 = 15\) dollars a year, less than the 20 dollar price, so on average the buyer loses 5 dollars a year. Someone might still buy it because a sudden 300 dollar bill could be much harder to pay than a steady 20 dollars: the expected value ignores how much a big loss hurts.
  7. Let \(I_k\) be 1 if card \(k\) matches the rank called. Whatever rank is called at position \(k\), 4 of the 52 cards have that rank, so \(P(I_k = 1) = \tfrac{4}{52} = \tfrac{1}{13}\). Expected matches: \(52 \cdot \tfrac{1}{13} = 4\).
  8. \(2 \cdot \tfrac12 + 4 \cdot \tfrac14 + 8 \cdot \tfrac18 + 0 \cdot \tfrac18 = 1 + 1 + 1 = 3\) dollars. A price of 3 dollars would make it fair.

Common mistake

Reading the expected value as "what will happen". An expected profit of \(-0.17\) dollars does not mean you lose 17 cents a game. It means the average over many games is close to \(-0.17\). In a handful of games anything can happen, which is exactly why games of chance tempt people.

Using profit and payout inconsistently. Decide first whether \(X\) is the amount you are paid or your profit (paid minus cost). Profit = payout \(-\) cost, and by Rule 1, \(E(\text{profit}) = E(\text{payout}) - \text{cost}\). Mixing them (adding the stake back in for some values but not others) is the most common error in game questions.

Where it leads