Expected frequency and fair games

Expected frequency as number of trials × probability, expected payout and expected gain for games of chance and raffles, fair games, and using expected values to judge whether a game or stall makes money.

A probability on its own is a bit abstract. Turn it into "how many times out of how many" and it becomes useful. A school fete needs to know how many prizes to stock. A club running a raffle wants to know whether it will make money. A player wants to know whether a game is worth it. All of these come down to one idea: multiply the probability by the number of times you'll play.

Expected frequency

$$\text{expected frequency} = \text{number of trials} \times P(\text{event})$$

Why multiply? A probability of $\frac{1}{6}$ means "one time in every six, in the long run". In 300 trials there are 50 groups of six, so you expect about 50 successes. Multiplying by $\frac{1}{6}$ is the same as dividing into groups of six.

Example 1●

Two dice are rolled 180 times. How many times would you expect a total of 7?

$P(\text{total } 7) = \frac{6}{36} = \frac{1}{6}$

Six of the 36 cells in the two-dice grid have a total of 7.

$180 \times \frac{1}{6} = 30$

Multiply the probability by the number of trials.

About $30$ times.

Expected doesn't mean guaranteed. It is the long-run average, so the actual count will usually be near 30, not exactly 30.

Expected payout of a game

When a game has several prizes, each with its own chance, you weight every prize by its probability and add. The result is the expected payout: the average amount the game pays out per play in the long run.

Expected payout and expected gain

$$\text{expected payout} = \text{add up } (\text{prize} \times \text{probability}) \qquad \text{expected gain} = \text{expected payout} - \text{cost to play}$$

If the expected gain is zero, the game is fair: in the long run, neither the player nor the organiser comes out ahead. A negative expected gain means the player loses on average, and the organiser makes money. That is how every casino and charity raffle works.

Example 2●●

A fete wheel has 10 equal sectors: one pays $\$20$, two pay $\$5$, three pay $\$2$ and four pay nothing. Each spin costs $\$3$. Find the expected payout and the expected gain for the player. Who wins in the long run?

$P(\$20) = \frac{1}{10},\ P(\$5) = \frac{2}{10},\ P(\$2) = \frac{3}{10},\ P(\$0) = \frac{4}{10}$

Each of the 10 equal sectors has probability 1/10, so each prize's probability is its number of sectors over 10.

$\text{expected payout} = 20 \times \frac{1}{10} + 5 \times \frac{2}{10} + 2 \times \frac{3}{10} = 2 + 1 + 0.6 = \$3.60$

Weight each prize by its probability and add. The empty sectors add nothing.

$\text{expected gain} = \$3.60 - \$3 = \$0.60$

Subtract the cost of a spin.

The player wins on average. The stall expects to lose $200 \times \$0.60 = \$120$ over 200 spins.

A positive expected gain favours the player. Over 200 spins the stall expects to pay out 60 cents more than it takes, each spin.

That stall needs fixing. Charging $\$3.60$ would make the game fair. Charging $\$4$ gives the player an expected gain of $\$3.60 - \$4 = -\$0.40$, so over 200 spins the stall expects to make $200 \times \$0.40 = \$80$.

InteractiveRun the fete wheel

Try this

Change the number of each prize sector and the cost per spin. The working beside the wheel updates as you go.

Find a cost that makes the game exactly fair. Then keep the cost at $\$3$ and change the prizes so the stall expects to make about $\$100$ from 200 players. Press Run a night a few times. Why doesn't the actual profit match the expected profit every time?

Raffles and other decisions

Example 3●●

A footy club sells 500 raffle tickets at $\$5$ each. The prizes are worth $\$1{,}000$, $\$200$ and $\$100$. Find the expected gain for someone who buys one ticket, and the club's profit if all tickets sell.

$\text{expected payout} = \frac{1{,}000 + 200 + 100}{500} = \frac{1{,}300}{500} = \$2.60$

Each ticket has a 1 in 500 chance of each prize.

$\text{expected gain} = \$2.60 - \$5 = -\$2.40$

Subtract the ticket price.

$\text{profit} = 500 \times \$5 - \$1{,}300 = \$1{,}200$

The club collects 500 tickets at 5 dollars and gives away 1300 dollars of prizes. This matches 500 × 2.40.

Expected value doesn't say you shouldn't buy a raffle ticket. People happily pay $\$2.40$ on average to support the club and enjoy the chance of winning. It just tells you, honestly, what it costs.

Your turn

A game costs $\$2$. You roll one die and win $\$6$ if it shows a 6, and nothing otherwise. Find the expected gain. What price would make the game fair?

Show the answer

$\text{expected payout} = 6 \times \frac{1}{6} = \$1$. $\text{expected gain} = \$1 - \$2 = -\$1$, so you lose $\$1$ a game on average. The game is fair when the cost equals the expected payout, which is $\$1$.

Flashcards

Expected frequency

Number of trials × probability.

Expected payout

Add up each prize × its probability.

Fair game

Expected gain is zero: cost to play = expected payout.

Practice

A die is rolled 300 times. How many 3s would you expect?
The probability that a baby born in a NSW hospital is a boy is about $0.512$. How many boys would you expect among 2000 births?
Two dice are rolled 360 times. How many times would you expect a total of 8?
Jordan buys 4 of the 500 tickets in a raffle with one prize. What is the probability that Jordan wins?
A spinner game pays $\$10$ with probability $\frac{1}{5}$ and $\$5$ with probability $\frac{1}{5}$, and nothing otherwise. Find the expected payout in dollars.
The game in the previous question costs $\$4$ to play. Find the expected gain for the player, in dollars.
A game is fair when:
A stall expects to make $\$0.50$ per game. About how much should it expect to make from 240 games?
Exam style: A raffle sells 500 tickets at $\$5$ each, with prizes worth $\$1{,}000$, $\$200$ and $\$100$. Find the expected payout per ticket, in dollars.
Exam style: A charity game pays $\$30$ with probability $0.1$ and $\$15$ with probability $0.1$. What should a single play cost, in dollars, for the game to be fair?