A pack of cards

What is in a pack

Board questions often say "one card is drawn from a well-shuffled pack of $52$ cards". Well shuffled means every card is equally likely to be drawn. To answer, you need to know what the pack holds.

• Four suits of $13$ cards each: spades and clubs are black, hearts and diamonds are red.
• So $26$ cards are red and $26$ are black.
• Each suit has an ace, the numbers $2$ to $10$, and a jack, a queen and a king.
• The face cards are the jacks, queens and kings: $3$ in each suit, so $12$ in all.
• There are $4$ cards of each kind: four aces, four sevens, four kings, and so on.

One card from a full pack

Divide the favourable cards by $52$, then simplify.

• A red card: $\dfrac{26}{52} = \dfrac{1}{2}$.
• A spade: $\dfrac{13}{52} = \dfrac{1}{4}$.
• An ace: $\dfrac{4}{52} = \dfrac{1}{13}$.
• A face card: $\dfrac{12}{52} = \dfrac{3}{13}$.
• A black face card: $\dfrac{6}{52} = \dfrac{3}{26}$.
• The queen of hearts: $\dfrac{1}{52}$, since there is only one.

"Or" questions: count each card once

What is the probability of drawing a king or a heart? There are $4$ kings and $13$ hearts. But the king of hearts is in both groups. Count it once:

so the probability is $\dfrac{16}{52} = \dfrac{4}{13}$.

The complement then answers a "neither" question. Neither a king nor a heart leaves $52 - 16 = 36$ cards, so the probability is $\dfrac{36}{52} = \dfrac{9}{13}$.

When cards are taken out

If some cards are removed, the pack is smaller. Both the top and the bottom of the fraction may change.

Suppose all four aces are removed. Then $48$ cards are left.

• $P(\text{a king}) = \dfrac{4}{48} = \dfrac{1}{12}$, since no king was removed.
• $P(\text{a red card}) = \dfrac{24}{48} = \dfrac{1}{2}$, since two red aces went.

A small pack. Five cards, the ten, jack, queen, king and ace of diamonds, are laid face down. Ishita picks one.

• $P(\text{the queen}) = \dfrac{1}{5}$.
• If she keeps the queen aside and picks again from the four left, $P(\text{the ace}) = \dfrac{1}{4}$ and $P(\text{the queen}) = 0$.

Practise with new numbers

Try it in Python

The program builds the pack as a list of (rank, suit) pairs, then counts some kinds of card.

Common mistakes

• Forgetting that cards were removed. Divide by the cards left, not by $52$.
• Counting a card twice. In "a king or a heart", the king of hearts belongs to both groups.
• Taking the ace as a face card. The face cards are only the jacks, queens and kings.
• Leaving a fraction unsimplified. $\dfrac{12}{52}$ should be written as $\dfrac{3}{13}$.

Remember these

• $52$ cards: $4$ suits of $13$, $26$ red and $26$ black.
• $12$ face cards, and $4$ cards of each kind.
• For "or", add the two groups and take away the cards in both.
• After removing cards, use the new total.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. One card is drawn from a full pack. What is the probability that it is a diamond?

2. What is the probability of drawing a red face card?

3. What is the probability of drawing a queen or a club?

4. All the hearts are removed from a pack. One card is drawn from the rest. What is the probability that it is red?

5. Kunal says the probability of an ace is $\dfrac{1}{52}$. What did he forget?