AFCAT Numerical Ability · Probability

Playing Cards: the Deck, Or, Given That, and Drawing Two

Know the 52-card deck by suit, colour and rank, count the cards that fit, take off any overlap in an or, and for two or three cards multiply, with or without replacement.

All Probability notes

Why this matters

Most card questions are one card and one count, so the deck itself is the formula: 4 suits of 13, 26 cards of each colour, 12 face cards. The marks go on an or with an overlap (a seven that is also a spade), on whether the Ace counts high or low, and on forgetting that the deck shrinks when a card is not put back.

Concept 1 of 6: The 52-card deck

A standard deck has 4 suits of 13 cards. Hearts and diamonds are red; spades and clubs are black. Each suit runs Ace, 2 to 10, Jack, Queen, King.

Definition

Count the cards that fit, then divide by 52. Every rank appears 4 times, once in each suit, and every suit has 13 cards. In these questions the face cards (also called court or picture cards) are the Jack, Queen and King only.

Group	Cards	Probability of one card
One colour (red, or black)	26	$\tfrac{1}{2}$
One suit	13	$\tfrac{1}{4}$
One rank (for example the kings)	4	$\tfrac{1}{13}$
Face cards: J, Q, K	12	$\tfrac{3}{13}$ The Ace is not a face card here.
Number cards: 2 to 10	36	$\tfrac{9}{13}$
One named card (for example the 7 of clubs)	1	$\tfrac{1}{52}$

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Probability · Playing Cards

What is the probability of drawing a card that is not a face card?

Twelve face cards, not sixteen

With the Jack, Queen and King as face cards there are 12. Adding the four Aces gives 16, which these questions do not use.

Number cards are 2 to 10

Each suit has 9 number cards, 2 to 10. The Ace is not one of them unless the question says it counts as 1.

Drill 8 more on the 52-card deck

Concept 2 of 6: Ranges of values, and two conditions on one card

A condition on value picks out some ranks, and each rank gives 4 cards. 'Value 9 or 10' is 2 ranks, so 8 cards:

\tfrac{8}{52} = \tfrac{2}{13}

Definition

Cards that fit $= 4 \times$ the ranks that fit, when every suit is allowed.
Read how the Ace is valued: as 1 (low) or as 14 (high). The question says which, and it changes the count.
Two conditions on one card ('a black number card', 'a spade that is not a face card'): count inside one group. Black number cards are 9 in spades and 9 in clubs, 18.
A condition every card meets has probability 1; one no card meets has probability 0.

Cards in some ranks

P = \dfrac{4 \times \text{ranks that fit}}{52}

Worked example

With the Ace counted as 14, find the probability that a card drawn has value greater than 11.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Probability · Playing Cards

What is the probability of drawing a card with value less than

5

(considering Ace as

1

Ace high or Ace low

With the Ace as 1, 'value less than 4' is Ace, 2, 3: 12 cards. With the Ace as 14 it is 2 and 3 only: 8 cards.

Inclusive ranges

'Between 3 and 7 inclusive' is 3, 4, 5, 6, 7: five ranks, 20 cards. Leaving out an end gives 16.

Drill 9 more on ranges of values, and two conditions on one card

Concept 3 of 6: Or: add, then take off the overlap

'A king or a spade' counts 4 kings and 13 spades, but the king of spades is in both groups. Take it off once: 16 cards.

Definition

$n(A \text{ or } B) = n(A) + n(B) - n(A \text{ and } B)$ , the rule from the first page.
If no card is in both groups (a king or a queen; a diamond or a black card), simply add.
Find the overlap by asking which cards are in both groups: one rank and one suit share exactly one card; one colour and one rank share two.

Or with an overlap

P(A \text{ or } B) = \dfrac{n(A) + n(B) - n(A \text{ and } B)}{52}

Worked example

A card is drawn. Find the probability that it is a king or a spade.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Probability · Playing Cards

A card is selected. What is the probability it is a court card (Jack, Queen, King) or a heart?

The overlap is counted twice

A queen or a club is

4 + 13 - 1 = 16

cards. Writing 17 counts the queen of clubs twice.

Groups that cannot overlap

A diamond or a black card has no card in both groups, so it is

13 + 26 = 39

cards with nothing taken off.

Drill 10 more on or: add, then take off the overlap

Concept 4 of 6: Given that, or a deck with cards removed

Once you are told the card is red, only the 26 red cards are possible. Count inside them. A deck with some cards taken out works the same way: count in what is left.

Definition

$P(A \mid B) = \dfrac{n(A \text{ and } B)}{n(B)}$ , the coins rule applied to cards.
'Given a red card' leaves 26 cards; 'given a spade' leaves 13.
If cards are removed first, the new total is what remains. Check whether any removed card belonged to the group you want.

Given that

P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}

Worked example

A card is drawn and it is black. Find the probability that it is a king.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Probability · Playing Cards

A card is drawn. Given it's a face card, what is the probability it's a Queen?

Removed cards change both counts

Taking out the 4 kings leaves 48 cards and 12 spades, so a spade is

\tfrac{12}{48}

. Keeping 13 spades forgets that the king of spades went out with the kings.

Drill 2 more on given that, or a deck with cards removed

Concept 5 of 6: Two draws with replacement

If the first card goes back and the deck is shuffled, the second draw starts from 52 again. The draws are independent, so multiply.

Definition

With replacement, P(first A, then B) $= P(A) \times P(B)$ , both out of 52.
'Exactly one' of two draws has two orders: add P(A, then not A) and P(not A, then A).
Three draws: multiply three times. All three are hearts with probability $\left(\tfrac14\right)^3$ .

With replacement

P(A \text{ then } B) = P(A) \times P(B)

Worked example

A card is drawn, replaced, and a second card drawn. Find the probability that the first is a heart and the second a face card.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 5 · Probability · Playing Cards

Two cards are drawn with replacement. What is the probability that the first is an Ace and the second is a King?

Replacement keeps the totals

With replacement both draws are out of 52. Writing 51 for the second draw answers the without-replacement question.

Drill 3 more on two draws with replacement

Concept 6 of 6: Two or three cards without replacement

If the first card stays out, the second is drawn from 51, and if the first was a heart only 12 hearts are left. Multiply the step-by-step probabilities.

Definition

P(first A, then B) $= P(A) \times P(B \text{ after } A \text{ has gone})$ .
Both of a kind: $\tfrac{36}{52} \times \tfrac{35}{51}$ for two number cards, $\tfrac{26}{52} \times \tfrac{25}{51}$ for two red cards.
When only the second card has to match the first (same suit, same rank), the first card can be anything, so only the second step counts.
At least one: $1 - P(\text{none})$ , with the none worked out step by step.

Without replacement

P(A \text{ then } B) = P(A) \times P(B \mid A)

Worked example

Two cards are drawn without replacement. Find the probability that both are red.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 6 · Probability · Playing Cards

Two cards are drawn without replacement. What is the probability that they form a pair (same rank)?

Same suit as the first card

For two cards of the same suit the first card can be any card, then 12 of the 51 left match:

\tfrac{12}{51} = \tfrac{4}{17}

. Multiplying by

\tfrac{13}{52}

as well answers 'both are hearts' instead.

Shrink both counts

After one face card is drawn and kept, 11 face cards remain in 51 cards. Keeping 12 or 52 for the second draw gives the with-replacement answer.

Drill 7 more on two or three cards without replacement

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 (5)

Ranges of values, and two conditions on one card
Cards in some ranks
$P = \dfrac{4 \times \text{ranks that fit}}{52}$
Or: add, then take off the overlap
Or with an overlap
$P(A \text{ or } B) = \dfrac{n(A) + n(B) - n(A \text{ and } B)}{52}$
Given that, or a deck with cards removed
Given that
$P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$
Two draws with replacement
With replacement
$P(A \text{ then } B) = P(A) \times P(B)$
Two or three cards without replacement
Without replacement
$P(A \text{ then } B) = P(A) \times P(B \mid A)$

Reference tables (1)

The 52-card deck6 rows

Group	Cards	Probability of one card
One colour (red, or black)	26	$\tfrac{1}{2}$
One suit	13	$\tfrac{1}{4}$
One rank (for example the kings)	4	$\tfrac{1}{13}$
Face cards: J, Q, K	12	$\tfrac{3}{13}$ The Ace is not a face card here.
Number cards: 2 to 10	36	$\tfrac{9}{13}$
One named card (for example the 7 of clubs)	1	$\tfrac{1}{52}$

Watch out for (10)

Twelve face cards, not sixteen
→ The 52-card deck
Number cards are 2 to 10
→ The 52-card deck
Ace high or Ace low
→ Ranges of values, and two conditions on one card
Inclusive ranges
→ Ranges of values, and two conditions on one card
The overlap is counted twice
→ Or: add, then take off the overlap
Groups that cannot overlap
→ Or: add, then take off the overlap
Removed cards change both counts
→ Given that, or a deck with cards removed
Replacement keeps the totals
→ Two draws with replacement
Same suit as the first card
→ Two or three cards without replacement
Shrink both counts
→ Two or three cards without replacement

Drill the questions on this topic

45 questions from the bank, with cart and Word export.

Drill all 45 questions