Expected Outcomes

If you toss a fair coin $100$100 times, roughly how many heads should you get? If a spinner lands on red with probability $\dfrac{1}{4}$41​ and you spin it $80$80 times, how many reds would you predict? These are questions about expected outcomes — using a known probability to predict how often an event happens over many trials. This lesson develops the single most useful prediction formula in GCSE probability, $\text{expected frequency} = P \times n$expected frequency=P×n, and then applies it to comparing predictions with results and to reasoning about whether a game is fair.

This is AO1 (using the formula) and AO2/AO3 content (interpreting predictions and arguing about fairness). OCR phrases these questions with command words such as Work out, Calculate, Estimate and Show that, so showing the multiplication clearly is what earns the marks.

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Key Vocabulary

Term	Definition
Expected frequency	The number of times an event is predicted to occur: $P \times n$P×n.
Probability $P$P	The chance of the event happening on a single trial.
$n$n	The number of trials (e.g. spins, rolls or games).
Observed frequency	The number of times the event actually occurred.
Fair game	A game in which each player has an equal expected outcome (e.g. equal expected winnings).
Expected value	The average result per trial in the long run.

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The Expected Frequency Formula

If an event has probability $P$P and an experiment is repeated $n$n times, the expected frequency of the event is

$\text{expected frequency} = P \times n$expected frequency=P×n

This is a prediction of the long-run count. It need not be a whole number, and the actual result will vary around it. The reasoning behind the formula is simple proportion: if an event happens, on average, a fraction $P$P of the time, then in $n$n trials it should happen about $P$P of those $n$n times, which is $P \times n$P×n. For instance, a fair coin lands heads half the time, so in $200$200 tosses we expect about $\tfrac{1}{2} \times 200 = 100$21​×200=100 heads.

It is essential to read "expected" correctly. The expected frequency is a long-run average, not a guarantee. If you toss a coin $200$200 times you will rarely get exactly $100$100 heads — you might get $94$94 or $107$107 — but the expected value $100$100 is the value the results cluster around, and over many repeats the average outcome closes in on it. This is why the formula is a prediction tool rather than a promise: it tells you what to expect, not what will certainly happen. The same three-quantity relationship, $\text{expected} = P \times n$expected=P×n, can be rearranged: if you know the target count and the probability, you can find the number of trials needed by computing $n = \dfrac{\text{expected}}{P}$n=Pexpected​.

Worked Example 1

A fair coin is tossed $250$250 times. Work out the expected number of heads.

Solution: $P(\text{heads}) = \dfrac{1}{2}$P(heads)=21​.

$\text{expected heads} = \dfrac{1}{2} \times 250 = 125$expected heads=21​×250=125

Worked Example 2

A fair six-sided dice is rolled $120$120 times. Work out the expected number of times it lands on $3$3.

Solution: $P(3) = \dfrac{1}{6}$P(3)=61​.

$\dfrac{1}{6} \times 120 = 20$61​×120=20

Worked Example 3

A spinner lands on red with probability $0.35$0.35. It is spun $200$200 times. Calculate the expected number of reds.

Solution: $0.35 \times 200 = 70$0.35×200=70

Common error: Reading "$200$200 times" as the probability or adding it to the probability. The number of trials always multiplies the probability.

Worked Example 4

A biased coin has $P(\text{heads}) = 0.6$P(heads)=0.6. The coin is flipped $250$250 times. Work out the expected number of tails.

Solution: $P(\text{tails}) = 1 - 0.6 = 0.4$P(tails)=1−0.6=0.4.

$0.4 \times 250 = 100$0.4×250=100

Common error: Using $P(\text{heads})$P(heads) when the question asks about tails. Find the relevant probability first.

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Expected Outcomes from a Probability Table

When a probability distribution is given in a table, read off the probability you need and multiply by the number of trials.

Worked Example 5

A biased spinner has this distribution.

Colour	Red	Blue	Green	Yellow
Probability	0.3	0.25	0.15	0.3

The spinner is spun $400$400 times. Work out the expected number of greens.

Solution: $P(\text{green}) = 0.15$P(green)=0.15.

$0.15 \times 400 = 60$0.15×400=60

Worked Example 6

A four-sided dice has $P(1) = 0.4$P(1)=0.4, $P(2) = 0.1$P(2)=0.1, $P(3) = 0.2$P(3)=0.2 and $P(4) = 0.3$P(4)=0.3. It is rolled $500$500 times. Work out the expected number of times it lands on an odd number.

Solution: Odd outcomes are $1$1 and $3$3, so $P(\text{odd}) = 0.4 + 0.2 = 0.6$P(odd)=0.4+0.2=0.6.

$0.6 \times 500 = 300$0.6×500=300

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When the Probability Must Be Estimated First

Sometimes you are given an experiment, must estimate the probability as a relative frequency, then predict for a larger number of trials.

Worked Example 7

A spinner is spun $80$80 times. It lands on green $30$30 times. The spinner is spun another $500$500 times. Estimate the number of greens.

Solution: Estimate $P(\text{green}) = \dfrac{30}{80} = \dfrac{3}{8}$P(green)=8030​=83​.

$\dfrac{3}{8} \times 500 = 187.5$83​×500=187.5

So about $187$187 or $188$188 greens.

Common error: Giving a fractional count as the final answer without comment. Since you cannot have half a green, write "about $188$188" and note it is an estimate.

Worked Example 8

A football team plays $40$40 matches and wins $22$22. (a) Estimate the probability the team wins a match. (b) Next season the team plays $50$50 matches. Estimate the number of wins.

Solution: (a) Relative frequency $= \dfrac{22}{40} = \dfrac{11}{20} = 0.55$=4022​=2011​=0.55. (b) Expected wins $= 0.55 \times 50 = 27.5$=0.55×50=27.5, so about $27$27 or $28$28 wins.

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Fair-Game Reasoning

A game is fair if every player has the same expected outcome — for example, the same expected winnings, or an equal chance of winning overall. To decide, compare the relevant probabilities or expected values.

It is easy to be misled by the word "fair". A game played with a perfectly fair dice can still be an unfair game if the winning conditions favour one player — for instance, if you win on $1$1–$4$4 and your opponent only on $5$5–$6$6, the dice is fair but the game is not. So "fair game" is not about the equipment being unbiased; it is about each player having the same expected outcome. The reliable method is always the same: work out each player's probability of winning (or each player's expected winnings if money is involved) and compare. If the two are equal, the game is fair; if not, it is not, and the player with the larger value is favoured.

When money is involved, fairness is judged on expected winnings rather than just the chance of winning. A game can give you a small chance of a large prize and still be fair, or even favourable, if the expected value of your winnings matches or exceeds what you pay to play. This is exactly how organisers set prizes for a charity stall or a fairground game — they choose the numbers so that, on average, the stall makes money, which means the player's expected value is below the cost of playing.

Worked Example 9

In a game, a fair six-sided dice is rolled. Aisha wins if the score is $1$1, $2$2, $3$3 or $4$4; Ben wins if the score is $5$5 or $6$6. Is the game fair? Give a reason.

Solution: $P(\text{Aisha wins}) = \dfrac{4}{6} = \dfrac{2}{3}, \qquad P(\text{Ben wins}) = \dfrac{2}{6} = \dfrac{1}{3}$P(Aisha wins)=64​=32​,P(Ben wins)=62​=31​

The game is not fair because Aisha is twice as likely to win as Ben.

Worked Example 10

A spinner has sections numbered $1, 2, 3, 4$1,2,3,4. The probabilities are $P(1) = 0.4$P(1)=0.4, $P(2) = 0.2$P(2)=0.2, $P(3) = 0.2$P(3)=0.2, $P(4) = 0.2$P(4)=0.2. A game lets one player win on an even number and the other on an odd number. Show that the game is fair.

Solution:

• Even outcomes: $P(2) + P(4) = 0.2 + 0.2 = 0.4$P(2)+P(4)=0.2+0.2=0.4.
• Odd outcomes: $P(1) + P(3) = 0.4 + 0.2 = 0.6$P(1)+P(3)=0.4+0.2=0.6.

This gives $0.4$0.4 versus $0.6$0.6, so the game is not fair. (Worked through to show the comparison; a fair game would need both totals equal to $0.5$0.5.) To make it fair, the win conditions would need to split the probability $0.5$0.5 each — for example, one player wins on a $1$1 and the other on $2, 3$2,3 or $4$4.

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Extended Worked Examples

Worked Example 11