Card & Combinatorics Problems

Easy3–4 hrsCardsCombinationsHypergeometric

Overview

Card problems are pure combinatorics: count favorables, divide by total. The mental model is a fraction: (ways to get what you want) / (total ways). For card hands, total is always C(52,k). For favorable, multiply choices for each independent requirement. The hypergeometric distribution formalizes 'sampling without replacement'.

How to Recognize

→Drawing cards from a standard deck without replacement
→Asked for probability of a specific hand (flush, pair, etc.)
→Problems with labeled/unlabeled items, 'choose k from n'

Step-by-Step Approach

1.Total outcomes = C(52,k) for drawing k cards
2.Favorable outcomes = product of combinations for each requirement
3.Use hypergeometric for sampling without replacement
4.For ordered arrangements: use permutations P(n,k) = n!/(n−k)!
5.Check: are order and repetition relevant?

Key Formulas

\binom{n}{k} = \frac{n!}{k!(n-k)!}

— combinations (order doesn't matter)

P(n,k) = \frac{n!}{(n-k)!}

— permutations (order matters)

P(\text{flush}) = \frac{4\binom{13}{5}}{\binom{52}{5}} \approx 0.198\%

04Hypergeometric:

P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

Worked Examples

0Quick Example

P(pair of aces in 5-card hand) = C(4,2)·C(48,3)/C(52,5) = 6·17296/2598960 ≈ 3.99%.

Problem

In 5-card poker, what is P(full house) — three of one rank, two of another?

Solution

Choose the rank for the triple: C(13,1) = 13 ways.

Choose 3 suits for the triple from 4 available: C(4,3) = 4 ways.

Choose the rank for the pair: C(12,1) = 12 ways (any of remaining 12 ranks).

Choose 2 suits for the pair: C(4,2) = 6 ways.

Favorable = 13·4·12·6 = 3,744. Total = C(52,5) = 2,598,960.

P = 3744/2598960 ≈ 0.144%.

Answer

$P(\text{full house}) = \frac{3744}{2598960} \approx 0.144\%$ . Framework: multiply independent choices for each component of the hand.

Common Mistakes

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Multiplying when you should add, or adding when you should multiply. The rule: AND (intersection of independent events) → multiply. OR (union of mutually exclusive events) → add.
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Forgetting to divide by repetitions when items are identical. C(n,k) already accounts for this.
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Using permutations when order doesn't matter (poker hands, for example).
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For 'at least one' — always use the complement (1 − P(none)) rather than trying to enumerate all cases.

Practice Problems

Click "Show Answer" to reveal

What is the probability of drawing exactly 2 red cards in a 4-card hand?

Hint: Hypergeometric: 26 red, 26 black, draw 4.

A bag has 5 red and 3 blue balls. Draw 3 without replacement. What is P(2 red, 1 blue)?

Hint: Hypergeometric: N=8, K=5 red, n=3 drawn.

How many ways can you arrange the letters in 'MISSISSIPPI'?

Hint: Multinomial coefficient: n! divided by factorials of repeated letters.

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