SPI Mathematics · Probability

Two Dice and Calendar Pairs

Two dice give 36 equally likely ordered pairs; count the pairs that fit, or multiply the chances for each die.

Why this matters

Thirteen questions in the bank, one of them from the 2022 paper, and the chapter's MODERATE questions are mostly here. Sums, products, doublets and 'at least one six' all come from the same 6 by 6 grid, and the leap-year questions are the same idea with the two extra days as the pair.

Concept 1 of 3: The 36 outcomes of two dice

Write the first die down the side and the second across the top: 36 cells, each equally likely. (2, 5) and (5, 2) are different cells, which is why a sum of 7 has six ways and a sum of 2 has only one.

Definition

Two dice give $6 \times 6 = 36$ ordered outcomes.
A doublet shows the same number on both dice: (1, 1) to (6, 6), six outcomes.
Ways to make each sum: 2 and 12 have 1 way, 3 and 11 have 2, 4 and 10 have 3, 5 and 9 have 4, 6 and 8 have 5, and 7 has 6.
Sums run from 2 to 12, so a sum of 1 or 13 is impossible.
For a product, list the factor pairs and count each order: $12 = 2 \times 6 = 6 \times 2 = 3 \times 4 = 4 \times 3$ , four ways.

Ways to make a sum s with two dice

\text{ways}(s) = 6 - |s - 7|, \quad 2 \le s \le 12

Worked example

Two dice are thrown. Find the probability that the sum is 10 or more.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Probability · Two Dice and Calendar Pairs

In a single throw of a pair of dice, the probability of getting the sum as a perfect square is:

Counting (2, 5) and (5, 2) once

The dice are different, so every unequal pair appears twice in the 36. Only a doublet such as (3, 3) appears once.

Counting a sum twice in 'divisible by 4 or 6'

12 is divisible by both 4 and 6. List the sums that qualify (4, 6, 8, 12) and count each one once:

3 + 5 + 5 + 1 = 14

ways.

'Between' leaves out the end values

A product between 8 and 13 means 9, 10 or 12. Counting 8 as well adds two ways and gives a printed wrong option.

Drill 6 more on the 36 outcomes of two dice

Concept 2 of 3: Both dice at once: multiply, or use the complement

When the question is about each die separately, the two dice do not affect each other, so the chance that both things happen is the product of the two chances. 'At least one' is easiest from the other side: work out 'none' and take it from 1.

Definition

$P(\text{first does A and second does B}) = P(A) \times P(B)$ for two dice.
At least one six $= 1 - P(\text{no six on either}) = 1 - \frac{5}{6} \times \frac{5}{6}$ .
A product is odd only when both numbers are odd, so P(even product) $= 1 - \frac{1}{2} \times \frac{1}{2}$ .
Any multiplied answer can be checked by counting cells in the 36.

Two independent dice

P(A \text{ and } B) = P(A) \times P(B) \qquad P(\text{at least one}) = 1 - P(\text{none})

Worked example

Two dice are thrown. Find the probability that both show a number greater than 4.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Probability · Two Dice and Calendar Pairs

Two dice are rolled together. The probability that at least one of them shows a six, is:

Adding instead of using the complement

At least one six is not

\frac{1}{6} + \frac{1}{6} = \frac{12}{36}

: that counts (6, 6) twice. The right count is

1 - \frac{25}{36} = \frac{11}{36}

An even product needs only one even number

The product of two dice is even unless both are odd. Counting only 'both even' gives

\frac{1}{4}

instead of

\frac{3}{4}

Drill 2 more on both dice at once: multiply, or use the complement

Concept 3 of 3: Pairs of days: 53 Sundays and two visitors

A leap year is 52 full weeks plus 2 extra days, so 52 of every weekday are certain. The question is only about the 2 extra days, which form one of 7 equally likely consecutive pairs.

Definition

Leap year: 366 days $= 52$ weeks $+ 2$ days. The extra pair is one of (Sun, Mon), (Mon, Tue), ..., (Sat, Sun): 7 pairs.
53 of a given weekday: that day must be in the pair, and it is in exactly 2 of the 7, so $\frac{2}{7}$ .
Ordinary year: 365 days $= 52$ weeks $+ 1$ day, so 53 of a given weekday has probability $\frac{1}{7}$ .
Two people each choosing one of $d$ days: $d \times d$ equally likely pairs, of which $d$ are the same day.

Two people choosing from d days

P(\text{same day}) = \frac{d}{d^2} = \frac{1}{d} \qquad P(\text{different days}) = 1 - \frac{1}{d}

Worked example

An ordinary year (not a leap year) is chosen at random. Find the probability that it has 53 Mondays.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Probability · Two Dice and Calendar Pairs

What is the probability that a leap year selected at random will contain 53 Sundays?

Answering 1/7 for a leap year

A leap year has two extra days, and a given weekday falls in two of the seven possible pairs, so the answer is

\frac{2}{7}

\frac{1}{7}

is the ordinary-year answer.

Drill 2 more on pairs of days: 53 sundays and two visitors

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The 36 outcomes of two dice
Ways to make a sum s with two dice
$\text{ways}(s) = 6 - |s - 7|, \quad 2 \le s \le 12$
Both dice at once: multiply, or use the complement
Two independent dice
$P(A \text{ and } B) = P(A) \times P(B) \qquad P(\text{at least one}) = 1 - P(\text{none})$
Pairs of days: 53 Sundays and two visitors
Two people choosing from d days
$P(\text{same day}) = \frac{d}{d^2} = \frac{1}{d} \qquad P(\text{different days}) = 1 - \frac{1}{d}$

Watch out for (6)

Counting (2, 5) and (5, 2) once
→ The 36 outcomes of two dice
Counting a sum twice in 'divisible by 4 or 6'
→ The 36 outcomes of two dice
'Between' leaves out the end values
→ The 36 outcomes of two dice
Adding instead of using the complement
→ Both dice at once: multiply, or use the complement
An even product needs only one even number
→ Both dice at once: multiply, or use the complement
Answering 1/7 for a leap year
→ Pairs of days: 53 Sundays and two visitors

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