eduKatePunggol · Secondary 4 Mathematics
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Two Dice: Possible Sums Are Not Equally Likely. Choose the route closest to your question.
Route 1 · Chapters 1–2
Route 2 · Chapters 3–4
Route 3 · Chapters 5–6
Route 4 · Chapters 7–8
Full chapter index · Independent practice · Secondary 4 Mathematics support
If your child counts the eleven possible sums from two to twelve and gives each probability 1/11, check whether those sums are equally likely. With two fair, independent six-sided dice, the 36 ordered face pairs are equally likely. Different sums collect different numbers of those pairs.
Secondary 4 Mathematics tuition in Punggol can make the distinction concrete: sum two occurs only as (1, 1), while sum seven occurs in six ordered ways. Their probabilities are 1/36 and 6/36 = 1/6. A Punggol Secondary 4 Maths tutor should identify the equally likely outcomes before choosing the denominator.
This guide assumes fair dice and independent rolls throughout its numerical examples. It is a Mathematics explanation, not gambling advice. Small-group 3-pax support should distinguish a possible result from an equally likely elementary outcome, then use a complete sample-space table to prevent missing or repeated cases.
Full chapter index
Build equally likely outcomes · Chapters 1–2
Count the required event · Chapters 3–4
Check overlap and assumptions · Chapters 5–6
Practise and continue · Chapters 7–8
Chapter 1 of 8 · Build equally likely outcomes
1. What does equally likely mean here?
For one fair die, each face has probability 1/6. Independence means that knowing the first die’s result does not change the second die’s face probabilities. Therefore each ordered pair has probability 1/6 × 1/6 = 1/36. There are six choices for each die and 36 pairs in total.
The phrase possible sums describes a different grouping of outcomes. The sums two through twelve are eleven labels, but equal numbers of labels do not imply equal probabilities. To use favourable-outcome count divided by total-outcome count directly, the outcomes being counted must have equal probability under the stated model.
Chapter 2 of 8 · Build equally likely outcomes
2. Why distinguish the first and second die?
Call the dice red and blue, or record the first roll and second roll. The pair (2, 5) then means red two and blue five, while (5, 2) means red five and blue two. These are different elementary outcomes, though they share the same sum.
The physical dice need not actually have different colours for this bookkeeping to work. We can label which die or which roll produced each result. But an unordered pair such as ‘two and five’ combines two ordered outcomes, whereas ‘two and two’ combines only one. Those unordered categories are not all equally likely.
The counts for sums 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 are respectively 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 and 1. They total 36. For sum seven, the pairs are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).
For sum twelve, only (6, 6) works. This is why central sums are more likely in the ideal model. The symmetry gives a useful check: sum two and sum twelve each have one pair, sum three and sum eleven each have two, and so on. Do not confuse this theoretical pattern with a guarantee for a short experiment.
| Sum | Number of ordered pairs |
|---|---|
| 2 | 1 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
| 6 | 5 |
| 7 | 6 |
| 8 | 5 |
| 9 | 4 |
| 10 | 3 |
| 11 | 2 |
| 12 | 1 |
For a sum greater than nine, include sums ten, eleven and twelve. Their pair counts are three, two and one, giving six favourable pairs. Probability is 6/36 = 1/6. Counting just the three sum labels over eleven would ignore their different weights.
Read the boundary word carefully. A sum at least nine also includes nine, which has four pairs, giving 10/36 = 5/18. ‘Greater than nine’ and ‘at least nine’ define different events. Write the allowed sum labels first, then count the pairs belonging to those labels.
Chapter 5 of 8 · Check overlap and assumptions
5. What if the event says at least one six?
There are six pairs with the first die six and six with the second die six. But (6, 6) appears in both groups, so the union contains 6 + 6 − 1 = 11 pairs. Probability is 11/36, not 12/36.
A complement check gives the same answer. Neither die six has probability (5/6)² = 25/36, so at least one six has probability 1 − 25/36 = 11/36. These calculations rely on the stated independence. ‘At least one’ includes the case of two sixes, but counts that case once.
Chapter 6 of 8 · Check overlap and assumptions
6. What changes if fairness or independence is not given?
If the dice are biased, their face probabilities may differ, so the 36 pairs need not all have probability 1/36. If the rolls are dependent, even fair individual face distributions do not by themselves guarantee a uniform distribution over pairs. The model must support the equally likely counting step.
The list of possible pairs can still be useful, but probabilities must then be assigned using the supplied information. Do not invent bias or dependence in a standard fair-independent exercise; equally, do not claim that every real pair of dice necessarily matches the ideal assumptions. State the conditions that justify the calculation.
Question 1: find the probability of sum twelve. Question 2: find the probability of sum six. Question 3: find the probability of an even sum. Question 4: find the probability of exactly one six. Use two fair, independent six-sided dice throughout.
Answers: 1) 1/36. 2) Five pairs, giving 5/36. 3) Both odd or both even gives 9 + 9 = 18 pairs, so 1/2. 4) Five pairs with only the first die six and five with only the second die six, giving 10/36 = 5/18. Exactly one excludes (6, 6), while at least one includes it.
Bring the original event wording and the denominator your child used. A diagnostic note such as ‘treats sum labels as equally likely’ identifies the model error. In a small-group 3-pax lesson, students can build a six-by-six table and compare one rare sum with one common sum before simplifying fractions.
For parents considering Secondary 4 Mathematics tuition in Punggol, the useful prompt is ‘What exactly are the equally likely outcomes?’ Once that is clear, check the event boundary and any overlapping cases. A complete sample space connects the probability fraction to actual outcomes rather than a count of convenient labels.
