Dice, Cards & Coins — notes, question types & practice · Permutation, Combination & Probability

01

Coin tosses

$n$ tosses give $2^n$ equally likely outcomes.

Two coins: HH, HT, TH, TT. Exactly one head: $\dfrac{2}{4} = \dfrac{1}{2}$ .

Four coins: $2^4 = 16$ outcomes, and at least one head is $1 - \dfrac{1}{16} = \dfrac{15}{16}$ .

Three coins, at least one tail: "none" means HHH, so $1 - \dfrac{1}{8} = \dfrac{7}{8}$ .

02

One die

A fair die has 6 outcomes. Favourable counts come straight from the words.

Even: 3 of 6, so $\dfrac{1}{2}$ . Prime (2, 3, 5): $\dfrac{1}{2}$ . More than 4 (5 and 6): $\dfrac{1}{3}$ .

03

Two dice

Two dice give $6 \times 6 = 36$ outcomes. Questions live on the sum.

The sum counts form a ladder: sums 2 and 12 have 1 way each, sums 3 and 11 have 2, rising to sum 7 with 6 ways.

Sum 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — 6 of 36, so $\dfrac{1}{6}$ .

Sum 8 has 5 ways, so $P = \dfrac{5}{36}$ .

'At least' sums use the ladder too. Sum more than 9: sums 10, 11, 12 give $3 + 2 + 1 = 6$ of 36, so $\dfrac{1}{6}$ .

Rule: Count ordered pairs. $(3,4)$ and $(4,3)$ are different outcomes.

04

Three dice

Three dice give $6^3 = 216$ outcomes. Direct counting is slow — use the complement.

All three same: 6 outcomes, so $\dfrac{6}{216} = \dfrac{1}{36}$ .

At least one six: none shows 6 in $5^3 = 125$ ways, so $1 - \dfrac{125}{216} = \dfrac{91}{216}$ .

Exactly one six: choose the die that shows it (3 ways) and fill the other two from 1 to 5 ( $25$ ways): $75$ of $216$ , so $\dfrac{25}{72}$ .

Tip: "At least one" with dice or coins almost always means $1 - P(\text{none})$ .

05

The standard deck

Learn the deck's structure once. It feeds every card question.

52 cards, 4 suits: spades and clubs are black; hearts and diamonds are red.
Each suit has 13 cards: A, 2 to 10, J, Q, K.
Face cards: J, Q, K — 12 in all. Aces: 4. Honour cards (A, K, Q, J): 16.
Red cards 26, black cards 26. Number cards (2 to 10): 36, nine per suit.

$P(\text{heart}) = \dfrac{13}{52} = \dfrac{1}{4}$ . $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$ .

Note: "King or heart" overlaps — the king of hearts sits in both groups, so subtract it once.

06

Cards drawn together

Two cards drawn together behave like two draws without replacement.

Both kings: $\dfrac{^{4}C_2}{^{52}C_2} = \dfrac{6}{1326} = \dfrac{1}{221}$ .

Both hearts: $\dfrac{^{13}C_2}{^{52}C_2} = \dfrac{78}{1326} = \dfrac{1}{17}$ .

One heart and one spade: $\dfrac{13 \times 13}{1326} = \dfrac{169}{1326} = \dfrac{13}{102}$ .

Both from the same suit: $4 \times \dfrac{78}{1326} = \dfrac{312}{1326} = \dfrac{4}{17}$ .

Remember the total: $^{52}C_2 = 1326$ . It appears in nearly every two-card question.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

One die or one coin

How to spot it:

A single die is rolled or a coin is tossed, and one simple event is asked.

Method

Write the sample space (6 or $2^n$ outcomes).
List or count the favourable outcomes.
Divide and reduce.

Why it works:

Small sample spaces can be listed directly, with no risk.

Try this

A die is rolled once. Find P(prime number).

Show solution

Primes: 2, 3, 5 — three of them.
$\dfrac{3}{6} = \dfrac{1}{2}$ .

Answer

1/2

Type 2very common2 practice Q

Two dice and the sum

How to spot it:

'Two dice are thrown. Find P(sum …)' — any condition on the total.

\text{total} = 36

Method

Fix the total outcomes at 36.
Count the ordered pairs making the sum (or use the sum ladder).
Divide by 36 and reduce.

Why it works:

Ordered pairs keep every outcome equally likely, which the classical definition needs.

Try this

Two dice are thrown. Find P(sum = 5).

Show solution

Pairs: $(1,4), (2,3), (3,2), (4,1)$ — 4 ways.
$\dfrac{4}{36} = \dfrac{1}{9}$ .

Answer

1/9

Type 3common2 practice Q

Three dice (use the complement)

How to spot it:

'Three dice are thrown' with 'at least one' in the question.

1 - \left(\frac{5}{6}\right)^3

Method

Set the total at $6^3 = 216$ .
Count the 'none' case: $5^3 = 125$ .
Answer $1 - \dfrac{125}{216} = \dfrac{91}{216}$ .

Why it works:

'None shows six' is one clean power, while 'at least one' is many cases.

Try this

Three dice are thrown. Find P(at least one six).

Show solution

None: 125 of 216.
$1 - \dfrac{125}{216} = \dfrac{91}{216}$ .

Answer

91/216

Type 4very common2 practice Q

One card from the deck

How to spot it:

'One card is drawn from a pack of 52' — suit, colour or rank is asked.

Method

Recall the deck facts: 13 per suit, 12 faces, 4 aces.
Count the favourable cards.
Divide by 52 and reduce.

Why it works:

The deck's fixed structure gives every count without listing cards.

Try this

One card is drawn from a deck. Find P(face card).

Show solution

Face cards: $3 \times 4 = 12$ .
$\dfrac{12}{52} = \dfrac{3}{13}$ .

Answer

3/13

Type 5common2 practice Q

Two cards drawn together

How to spot it:

'Two cards are drawn from a pack' — 'both' or 'one of each' is asked.

^{52}C_2 = 1326

Method

Set the total at $^{52}C_2 = 1326$ .
Count the favourable pairs with nCr.
Divide and reduce.

Why it works:

A simultaneous draw is one 2-card selection, so nCr counts it directly.

Try this

Two cards are drawn from a deck. Find P(both spades).

Show solution

$\dfrac{^{13}C_2}{^{52}C_2} = \dfrac{78}{1326}$ .
$= \dfrac{1}{17}$ .

Answer

1/17

Type 6common2 practice Q

Coin toss patterns

How to spot it:

'Two/three coins are tossed' with exactly / at least / at most in the question.

2^n \text{ outcomes}

Method

Fix the outcomes at $2^n$ .
For 2–3 coins, list the outcomes and count directly.
'At least one' flips to $1 - P(\text{none})$ .

Why it works:

Few outcomes make listing safe, and the complement kills the 'at least' cases.

Try this

Three coins are tossed. Find P(at least one tail).

Show solution

None: HHH only, $\dfrac{1}{8}$ .
$1 - \dfrac{1}{8} = \dfrac{7}{8}$ .

Answer

7/8

08

Formula sheet

Coin outcomes

2^n

$n$ fair coins tossed together.

Two dice outcomes

6 \times 6 = 36

Ordered pairs, not unordered.

Three dice outcomes

6^3 = 216

At least one six (three dice)

1 - \left(\frac{5}{6}\right)^3 = \frac{91}{216}

Two-card pairs

^{52}C_2 = 1326

Total for any two-card question.

09

Shortcuts that save time

⚡ The two-dice sum ladder

Memorise how many ordered pairs make each sum: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2 to 12.

Example

Two dice are thrown. Find P(sum = 9).

Show solution

Ladder: 4 ways for sum 9.
$\dfrac{4}{36} = \dfrac{1}{9}$ .

Answer

1/9

⚡ Deck facts at your fingertips

13 per suit, 4 suits, 12 face cards, 4 aces, 26 red, 26 black. Every card answer starts from one of these.

Example

One card is drawn from a deck. Find P(ace).

Show solution

4 aces of 52.
$\dfrac{4}{52} = \dfrac{1}{13}$ .

Answer

1/13

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Treating $(3,4)$ and $(4,3)$ as one outcome — ordered pairs keep all 36 outcomes equally likely.

Mistake 02

Counting three-dice outcomes as 36 — three dice give $6^3 = 216$ outcomes.

Mistake 03

Calling the aces face cards — face cards are J, Q, K: 12 in all; aces are a separate group of 4.

Mistake 04

Forgetting the overlap in 'spade or ace' — the ace of spades sits in both groups, so subtract it once.

Mistake 05

Using 52 as the total for two cards — two cards together draw from $^{52}C_2 = 1326$ pairs.

11

Quick revision

Read this the night before the exam.

Coins: $2^n$ outcomes.
One die: 6 outcomes.
Two dice: 36 outcomes; sum ladder 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1.
Three dice: 216; at least one six $= \dfrac{91}{216}$ .
Deck: 52 cards, 4 suits of 13; 12 faces; 4 aces; 26 red.
Two cards: $^{52}C_2 = 1326$ pairs.

12

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.