Probability describes how likely an event is under a stated model. Fairness asks whether the rules give the participants the opportunity that the game claims to give them. Neither question can be answered just by counting the names of possible results. First identify what can happen, how the result is generated and whether the individual possibilities are equally likely.
A bag contains three red counters and one blue counter. There are two colour names, but that does not make the colours equally likely. If each of the four counters is equally likely to be selected, red has three opportunities and blue has one. The probability of red is three-quarters, not one-half. The difference comes from the objects being selected, not the number of labels used afterwards.
This guide is Primary Mathematics enrichment. It builds on fractions, comparison and systematic listing; the later two-stage examples provide a bridge towards subsequent probability study. It is not a statement that formal probability, conditional probability or every notation below is required in the Singapore Primary syllabus. Use the MOE Primary curriculum page and the learner’s actual school programme to separate required work from extension.
Start with colour choices and one-event questions. Move to two coins or two selections only after the learner can identify a complete list of outcomes. All games here are classroom activities using counters or points, with no betting or financial stakes. All experimental records are explicitly illustrative; none is presented as a real study or a measured student result.
Chance language · Equal likelihood · Single events · Two-stage outcomes · Replacement · Experiments · Fairness · 24 questions · Worked answers · Teaching
1. Say what the event is before judging its chance
An outcome is one complete result of the activity. For one ordinary six-sided die, the recorded outcomes are one, two, three, four, five and six. An event groups the outcomes that satisfy a condition. “A number greater than four” contains five and six. “An even number” contains two, four and six. The event is not the same thing as one particular outcome.
The activity also needs a defined stopping point. Rolling once is different from rolling twice and recording the sum. Drawing one counter is different from drawing until a red counter appears. A calculation for the first activity cannot automatically answer the second, even when the same equipment is used.
For the finite equally likely models in this guide, an event with no allowed outcomes is impossible and has probability zero. An event containing every allowed outcome is certain and has probability one. Other probabilities lie between these endpoints. A chance of one-half means the event and its complement have equal probability; it does not mean that the next two trials must contain one of each.
These terms and the equally likely counting rule can also be checked in OpenStax’s probability terminology. Our worked examples below use their own stated objects and conditions. The arithmetic is justified by the lists and models shown, not by a claim that every real coin, die or handmade spinner is physically perfect.
“Possible” is weaker than “likely”
Drawing blue from a bag with nine red counters and one blue is possible. It is not the more likely colour. Conversely, drawing red is highly favoured in this model but not guaranteed. One blue result does not contradict a probability of nine-tenths for red. An unlikely allowed event can still happen.
When a child says, “Red will happen because there are more red counters,” ask whether a blue counter could be selected. The better conclusion is “Red is more likely.” Probability should sharpen the claim rather than turn a comparison into a promise.
State what “not” includes
For a die, the complement of “greater than four” is “four or less,” containing one through four. It is not “less than four,” which would omit four. Complement events partition the complete outcome list into the event and everything outside it. The two probabilities add to one because every outcome belongs to exactly one side.
This connects to Logic, Deduction and Truth Conditions. Carefully reading “greater than,” “at least,” “and” and “or” changes the event before any fraction is calculated. A correct division attached to the wrong event remains an incorrect answer.
2. Count equally likely possibilities, not convenient category names
For a finite collection of equally likely outcomes, the probability of an event is the number of favourable outcomes divided by the total number of outcomes. The equal-likelihood condition is essential. Without it, a list can still describe possibilities, but its length alone does not determine their probabilities.
Imagine six identical-size cards bearing A, A, A, B, B and C. Shuffle them so that each physical card is equally likely to be drawn. There are six individual selections, not three equally likely letter selections. A has probability three-sixths, B two-sixths and C one-sixth. Combining the cards into letter categories does not redistribute their chances equally.
Equal-sized spinner sectors
Consider a model spinner with eight equal sectors, equally likely stopping sectors and a rule to re-spin on a dividing line. Three sectors are marked blue and five yellow. Blue therefore has probability three-eighths and yellow five-eighths. The pointer does not select a colour name first; it selects a sector under the model.
If the sectors have unequal angles, counting sectors is insufficient. A red half-circle and two quarter-circle sectors, blue and green, give red half the ideal angular stopping space. Under a uniform-angle model, the probabilities are one-half, one-quarter and one-quarter. Three regions do not imply three equal chances.
A mechanism can invalidate an attractive-looking model
A person deliberately choosing their favourite card is not a uniform random selection. A counter distinguishable by touch may be selected more often than another. A handmade spinner can have friction or an uneven pivot. These possibilities do not make probability useless. They tell us to distinguish the mathematical assumption from evidence about the physical mechanism.
For a school calculation, write the assumption: each card equally likely; independent fair die rolls; uniform stopping angle. For an actual experiment, describe how the equipment and procedure approximate it. Do not claim that a few balanced results prove a device perfectly fair or that one unbalanced run proves dishonesty.
Comparing two bags
Bag A contains two red and three blue counters; Bag B contains five red and five blue. With equal selection of individual counters, Bag A gives red probability two-fifths and Bag B one-half. Compare fractions using a common denominator: four-tenths is less than five-tenths, so Bag B gives the larger red chance.
The raw red counts alone are not enough. A bag containing ten red counters and ninety blue offers a smaller red chance than one containing two red and three blue. The relevant comparison includes each bag’s whole collection. This is the same reference-whole discipline used in fraction work.
3. Work through single-event questions with an explicit outcome list
Example A: A die event
Roll a fair six-sided die numbered one to six. The event “at least four” contains four, five and six. Three favourable faces among six equally likely faces give probability three-sixths, or one-half. “More than four” would contain only five and six, giving one-third instead.
A useful check is to list the complement. For “at least four,” the complement is one, two or three, also one-half. Their probabilities sum to one. The wording, list, fraction and complement all describe the same event.
Example B: A bag with three colours
A bag contains four orange, three purple and five green counters, all equally likely to be selected. There are twelve counters. Purple has probability three-twelfths, or one-quarter. Not green has probability seven-twelfths because it includes both orange and purple.
Adding the orange and purple counts is valid because those colour categories do not overlap. A single counter cannot be both orange and purple in this model. If categories overlap, adding their counts without correcting the overlap can count the same outcome twice.
Example C: An overlapping “or” event
Select one of the numbers one through eight uniformly. Consider “even or greater than five,” with “or” including the possibility that both descriptions hold. The even numbers are two, four, six and eight. Numbers greater than five are six, seven and eight. Their combined set is two, four, six, seven and eight.
The probability is five-eighths. Adding four-eighths and three-eighths would double-count six and eight. Listing the union once is a transparent Primary-level solution; a formal union formula is unnecessary when the set is this small.
Example D: A missing colour count
A bag contains six red counters and some blue counters. Each counter is equally likely to be drawn, and red and blue must have equal probabilities. The two colours therefore need equal counts. Six blue counters are required in total. If two blue counters are already present, add four more.
Notice that the total changes when counters are added. The red probability moves from six-eighths to six-twelfths. It would be wrong to keep the old denominator eight after changing the collection.
4. Two-stage results need complete records
For two independent fair coin tosses, record the first toss and then the second. The complete outcomes are HH, HT, TH and TT. Each has probability one-quarter under the stated model. The order is part of the record: HT and TH describe different sequences even though both contain one head.
Exactly one head occurs in HT or TH, so its probability is two-quarters, or one-half. At least one head occurs in HH, HT or TH, giving three-quarters. Exactly two heads has only HH, giving one-quarter. These different events cannot be answered by merely saying that heads and tails are equally likely on each toss.
Count first, then combine
A common error is to list the possible numbers of heads as zero, one and two and assign each one-third. Those three categories are not equally likely. The middle category contains two elementary sequences, while the others contain one each. The complete equally likely list must come before the grouping.
The same issue appears in two dice. The possible sums run from two to twelve, but those eleven sums are not equally likely. A sum can be made in several ordered ways. Distinguishing a first die from a second gives thirty-six equally likely face pairs when the dice are fair and independent.
Example E: A small sum and a central sum
A sum of four can occur as (1,3), (2,2) or (3,1), giving three thirty-sixths, or one-twelfth. A sum of seven has six pairs: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). Its probability is six thirty-sixths, or one-sixth.
A sum of seven is twice as likely as a sum of four in this model. That does not mean every short sequence of rolls contains exactly twice as many sevens. It compares the probabilities assigned to one trial, not a required timetable of future outcomes.
Example F: Which die is larger?
For the first die to exceed the second, a first value of one has no favourable second values; two has one; three has two; four has three; five has four; six has five. The total is zero plus one plus two plus three plus four plus five, or fifteen favourable pairs.
Thus the probability is fifteen thirty-sixths, or five-twelfths. A symmetry check gives the same probability for the second die being larger. Six equal-face pairs are ties, probability one-sixth. Five-twelfths plus five-twelfths plus one-sixth equals one, accounting for all outcomes.
Independence is an assumption, not a synonym for randomness
Saying both coins are individually fair does not by itself establish independence. A second recorded result could be deliberately copied from the first. Each position might still show heads half the time across many trials, while HT and TH never occur. The familiar four-outcome calculation requires independent tosses or another justified mechanism giving those four sequences equal probability.
For the worked questions here, independence is stated where needed. A learner does not need advanced notation to ask the right practical question: does the first result change the mechanism generating the second?
5. Returning a counter changes the second draw’s whole
Take a bag with three red and two blue counters. Draw once, inspect the colour and return the counter, then mix before a second draw. Under independent uniform selections from the restored bag, the chance of red at each draw is three-fifths. Both red has probability three-fifths times three-fifths, or nine twenty-fifths.
A labelled counting explanation reaches the same result. Label the five counters temporarily. There are five choices for the first selection and five for the second, making twenty-five ordered pairs. Three red choices at each stage give nine favourable pairs. Replacement allows the same physical counter to be selected twice.
Without replacement
Now keep the first counter outside the bag. There are five choices initially and four remaining choices for the second selection, making twenty equally likely ordered pairs under uniform sequential drawing. Three red choices first leave two red choices second. Six favourable pairs give six-twentieths, or three-tenths.
The changing denominator records a changing collection. After a red is removed, two red and two blue remain; the chance of another red is one-half. After a blue is removed, three red and one blue remain; the chance of red is three-quarters. The first result matters because it changes the second bag.
“Given that” identifies the stage you know
Suppose the question says a red counter has already been removed and asks for the next red chance. Start with the remaining four counters, not the original five. The answer is two-fourths. There is no need to multiply by the chance of the event already supplied as known.
This is an introductory conditional-reasoning example, not a requirement to introduce a formal conditional-probability formula. Label the remaining collection and count it correctly. The mathematical issue is choosing the right whole for the question actually asked.
Replacement must be part of the written rules
The phrases “two counters are drawn” and “a counter is drawn twice” can be ambiguous. A well-formed exercise states whether the first is returned. Without that information, more than one model may be reasonable. Asking which model applies is better than pretending the difference cannot affect the answer.
6. An experiment records what happened, not what had to happen
An illustrative record of forty spins contains eighteen red outcomes. Its observed red proportion is eighteen-fortieths, or nine-twentieths, equal to 45%. This is a summary of the invented record. It is not automatically the spinner’s exact probability, and it is not evidence from an actual classroom trial.
Compare this record with a model predicting one-half red. A forty-trial experiment does not have to produce exactly twenty red results. Chance variation is part of the model. Further well-controlled trials may provide more information about the mechanism, but no finite run proves perfect fairness merely by looking balanced.
Keep predictions and observations in separate columns
Before an activity, write the model prediction. After each trial, record the observed result without changing the prediction to fit it. At the end, calculate the observed fraction and describe any difference. This keeps “what the model assigns” separate from “what this run produced.”
A short optional investigation can use a bag of four labelled cards, two marked star and two marked circle. Select uniformly with replacement and mixing. Predict one-half stars, conduct a manageable number of trials, then compare different runs. Do not alter or discard inconvenient outcomes to make a record look fair.
A streak does not create a debt
After five heads in independent fair tosses, the next toss still has probability one-half for heads. The coin is not obliged to produce tails to restore balance. Independence means the earlier sequence does not alter the probability model for the next toss.
The probability of six heads specified in advance is a different question: it concerns the whole six-toss sequence. Once five heads are already known, the next-toss question concerns only one future event. Confusing these questions makes a correct statement about one sound like a contradiction of the other.
What would justify questioning a model?
Repeated observations very unlike a prediction can motivate inspecting the mechanism and collecting more evidence. Perhaps cards were not mixed, labels could be felt or results were recorded incorrectly. This guide does not give a formal significance test or a sample-size rule for declaring a device biased.
The appropriate learning outcome is more modest and more useful: record honestly, distinguish an assumption from an observation and avoid turning one surprising result into a certain explanation.
7. Define fairness before redesigning a game
For a simple one-round classroom game with equal rewards, no choices and exactly one winner, a natural fairness criterion is equal winning probabilities. On a fair die, giving Player A faces one, two and three and Player B faces four, five and six meets that criterion. Each receives three equally likely faces.
Giving A only six and B every other face does not. A’s chance is one-sixth and B’s five-sixths, even though each player has one named winning event. The event sizes differ. Equal numbers of event names are not equal chances.
Example G: Redesign an unequal spinner game
A four-sector equally likely spinner has one red and three blue sectors. A wins on red and B on blue. Their probabilities are one-quarter and three-quarters. Relabelling one blue sector red produces two sectors for each and equal one-half winning chances, provided the spinner mechanism is unchanged.
Another redesign can assign alternating numbered sectors to the players without changing the colours. Explain which rule has changed and recount the assigned sectors. Do not claim fairness merely because both players agree to play; agreement and equal probability are different descriptions.
Equal scores in one session do not prove equal opportunity
An unfair game can finish tied by chance, and a fair game can produce a large score difference in a short session. Judge the stated probability mechanism separately from the observed scoreboard. Both are worth discussing, but they answer different questions.
If a game includes player choices, starting-position advantages or unequal rewards, equal single-event probabilities may no longer settle fairness. The Mathematical Games guide examines deterministic strategies, where winning depends on the state and responses rather than a random draw. Do not mix those two kinds of argument.
A fair discussion makes the assumptions visible
Ask which outcomes each player receives, whether those outcomes are equally likely and whether the rewards and number of turns match. For this article, equal win chances is a deliberately limited criterion. It does not claim to define fairness in every social or educational setting.
The NRICH teacher article Progression in Primary Probability offers additional context for discussing chance language and fairness through games. Our proposed lesson choices are adaptable suggestions, not evidence that a particular number of rounds guarantees learning.
8. Practice: 24 questions
Keep the worked answers covered. Every die below is a fair six-sided die numbered one to six. Coin tosses are independent and fair where stated. Uniform draws mean each individual remaining object is equally likely. Report probabilities as exact simplified fractions unless another form is requested.
Questions 1–8: Outcomes, events and the correct whole
1. A spinner has four equally likely sectors labelled A, B, C and D. List the outcomes and find the probability of C.
2. Roll one fair die. Find the probability of a number greater than four.
3. For the same die, find the probability of a number four or less. How does it check Question 2?
4. A bag contains three red, two blue and five green counters. Find the probability of red on a uniform single draw.
5. For the bag in Question 4, find the probability of not green.
6. An eight-sector equally likely spinner has three A sectors and five B sectors. Find the probability of B.
7. Bag A has two red and three blue counters. Bag B has five red and five blue. Which gives the larger red probability on a uniform single draw?
8. An ideal uniform-angle spinner is half red, one-quarter blue and one-quarter green. Are the three colours equally likely? Give their probabilities.
Questions 9–16: Complete two-stage records
9. List all complete outcomes of two independent fair coin tosses in order.
10. In Question 9, find the probability of exactly one head.
11. In Question 9, find the probability of at least one head.
12. Roll two independent fair dice, distinguished as first and second. Find the probability of a sum of four.
13. With the same dice, find the probability of a sum of seven and compare it with Question 12.
14. With the same dice, find the probability that the first die shows a larger number than the second.
15. A bag has three red and two blue counters. Draw uniformly, replace and mix, then draw independently again. Find the probability that both draws are red.
16. Repeat Question 15 without replacement, selecting uniformly from the remaining counters. Find the probability that both draws are red.
Questions 17–24: Evidence and fairness
17. In the bag from Question 16, a red counter is known to have been removed already. Find the red probability on the next uniform draw.
18. A four-sector equally likely spinner has one red and three blue sectors. A wins on red and B on blue, with identical rewards. Is the one-round game fair by equal winning probability?
19. On one fair die, A wins on odd faces and B on even faces. Are their winning probabilities equal?
20. An illustrative forty-trial record contains eighteen red results. Find the observed red proportion. Does this record alone prove that the red probability is exactly 45%?
21. Five independent fair coin tosses have all shown heads. What is the probability of heads on the next independent fair toss?
22. A model gives green probability one-quarter on each independent trial. Must forty trials contain exactly ten green outcomes? Explain.
23. A bag contains six red and two blue counters. How many blue counters should be added to make the colours equally likely on a uniform single draw?
24. Cards numbered one, two, three and four are drawn uniformly twice without replacement. A wins if the sum is odd; B wins if it is even. Find both winning probabilities and judge equality of chance.
9. Worked answers and model checks
Answers 1–8
1. A, B, C, D; probability 1/4. Exactly one of the four equally likely sectors is marked C. The count refers to sectors, and every sector has the same chance by the question’s condition. An extra sector or a change in sector size would require recounting or a different model.
2. 1/3. The favourable faces are five and six, two among six equally likely outcomes. Thus the fraction is 2/6, simplified to 1/3. Four is excluded because the event says greater than four, not at least four. Writing the face list protects that distinction.
3. 2/3. Faces one, two, three and four satisfy the condition, so the probability is 4/6. This event is the complement of Question 2. Their probabilities, 2/3 and 1/3, sum to one, and their disjoint face lists cover all six outcomes.
4. 3/10. There are ten individual counters and three are red. Uniform selection makes each counter equally likely, not each colour. The three colour names would be an inappropriate denominator. The total count is the reference whole for the probability fraction.
5. 1/2. Not green includes all three red and both blue counters, giving five of ten. Alternatively, green has probability 5/10, so its complement is 1 − 5/10. The two methods agree because every counter is either green or not green.
6. 5/8. Five of the eight equally likely sectors produce B. A has probability 3/8, and the two add to one. There are only two labels, but that does not produce two equal probabilities because the labels occupy different numbers of equal sectors.
7. Bag B. Red probabilities are 2/5 for A and 5/10 = 1/2 for B. Express 2/5 as 4/10 to compare with 5/10. The second bag gives the larger proportion of red opportunities. Comparing only two red counters with five would not be enough without the totals.
8. No: red 1/2, blue 1/4, green 1/4. The uniform-angle assumption makes sector angle proportional to chance. The red region occupies twice the angular space of either other colour. Equal numbers of named colours do not establish equally likely colour outcomes.
Answers 9–16
9. HH, HT, TH, TT. The first letter records the first toss. Each first-toss outcome has both possible second outcomes, so the list is complete. Under independent fair tossing each sequence has probability 1/4. The categories zero, one and two heads would combine different numbers of these sequences.
10. 1/2. Exactly one head occurs in HT and TH. These are two of the four equally likely ordered outcomes, giving 2/4. HH has at least one head but not exactly one, so it is excluded. The qualifying word changes the event.
11. 3/4. HH, HT and TH contain at least one head. A complementary check excludes only TT, whose probability is 1/4. Therefore the required probability is 1 − 1/4. Counting HH twice because it contains two heads would count occurrences rather than qualifying complete outcomes.
12. 1/12. The three favourable pairs are (1,3), (2,2) and (3,1). There are 6 × 6 = 36 equally likely ordered pairs. The probability is 3/36. The different sums are not equally likely categories, so one out of eleven sums is not the correct calculation.
13. 1/6, twice Question 12. Six pairs give seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Hence 6/36 = 1/6, which is twice 1/12. This is a comparison of model probabilities, not a promise about exact frequencies in a short run.
14. 5/12. For first-die values one through six, the numbers of smaller second-die values are 0, 1, 2, 3, 4 and 5. Their sum is fifteen, so the probability is 15/36. The reverse inequality also has fifteen pairs, leaving six ties; these counts total thirty-six.
15. 9/25. Replacement restores five counters before the second independent draw. There are twenty-five equally likely ordered selections of labelled counters, with 3 × 3 = 9 red-red selections. Equivalently, multiply 3/5 by 3/5. The returned counter may be selected again.
16. 3/10. Without replacement there are 5 × 4 = 20 ordered selections. Three red choices first leave two red choices second, giving six favourable selections. Thus 6/20 = 3/10. The second denominator and favourable count both change after the first red is removed.
Answers 17–24
17. 1/2. The known removal leaves two red and two blue counters, four altogether. The next probability is 2/4. Do not multiply by the chance that the known first removal happened: the question starts after that information has already been supplied.
18. No. A has probability 1/4 and B 3/4. By the specified equal-win-chance criterion and identical rewards, B has an advantage. Relabelling one blue sector red would give two equally likely sectors to each colour and one-half winning chance each under the same spinner model.
19. Yes; each is 1/2. A receives one, three and five, while B receives two, four and six. Each event contains three of the six equally likely faces. A short session can still end with unequal scores; the equal probabilities describe opportunity per round, not guaranteed balance afterwards.
20. 9/20, or 45%; no exact probability is proved. Eighteen divided by forty describes the illustrative record. A finite observed proportion need not equal the underlying model probability. The record should be reported honestly without being relabelled as a proven physical probability or an actual experiment conducted here.
21. 1/2. The next toss is stated to be independent and fair, so the earlier heads do not change its two equally likely outcomes. Tails is not owed. Asking about six heads in advance would be a different question about an entire sequence rather than the next toss after five known results.
22. No. Ten is the model’s expected count, obtained from forty times one-quarter, but it is not a required outcome count. Different sequences can contain more or fewer greens. For instance, a sequence with eleven green results is allowed under the model and disproves the claim of an exact requirement.
23. Four blue counters. Equal colour chances require equal colour counts when individual counters are uniformly selected. Red has six, so blue must rise from two to six. The new total is twelve and each colour has probability six-twelfths. Retaining the old eight-counter denominator would ignore the addition.
24. A: 2/3; B: 1/3; not equal. An odd sum uses one odd and one even card. There are two of each parity: 2 × 2 selections odd-first and even-second, and four in the reverse order, eight favourable ordered pairs. Twelve pairs are possible without replacement. The other four pairs have even sums.
10. Turn an answer into a useful next lesson
Ask the learner to identify the event, the whole outcome set and the reason for equal likelihood before calculating a fraction. When an answer is wrong, those three points help distinguish a reading problem from an incomplete list or an incorrect mechanism assumption. More fraction arithmetic will not fix a missing outcome.
For a child counting colour names rather than counters, begin with a deliberately unequal bag and label the individual counters. For a child treating the three head-count categories as equally likely, write all four ordered coin sequences. For a child keeping the same denominator without replacement, physically remove the first counter and recount what remains.
A short explanation test
Ask, “What exactly does the denominator count?” A strong response names equally likely individual outcomes under the model. “It is the bigger number” or “there are two colours” does not yet identify the relevant whole. Accept a clear pointing explanation before insisting on formal vocabulary.
Then change one rule while preserving the numbers. Return the counter in one version and keep it outside in another. Change “greater than” to “at least.” Change equal spinner sectors to unequal ones. A learner who notices which part of the model changes is developing transferable reasoning rather than memorising the answer to one picture.
Separate chance from certainty
An outcome can be likely without being forced. A strategy can force an outcome without involving chance. These are different mathematical jobs. The Games and Winning Positions guide supplies deterministic counter games, while Remainder Cycles and Clock Arithmetic explains repeated structures that have no random selection at all.
Use the Primary Mathematics Learning Hub for the earlier fraction, listing and data routes. Treat uncertainty honestly: name the assumptions, calculate within them and state what the answer does not guarantee.
Sources and scope
The linked MOE page is the official school-scope reference. OpenStax supplies a terminology reference for finite probability and equally likely outcomes. NRICH supplies further teacher reading on Primary probability discussion. The worked examples, hypothetical records, 24 questions, answer explanations and proposed teaching sequence in this guide are original. No endorsement by those organisations, verified student outcome or live experiment is claimed.