Probability Without Replacement: Matching Letters

What You Will Learn

• How to count favorable outcomes when letters appear with different frequencies
• The difference between permutations and combinations in probability without replacement
• Why single-occurrence items cannot contribute to "matching" events
• How to systematically organize counting problems by categorizing possible outcomes
• The relationship between total sample space and favorable outcomes in discrete probability

Solution: Method 1 — Favorable Pair Counting

We'll count the number of ways to choose two cards that show the same letter, then divide by the total number of ways to choose any two cards.

Step 1 — Analyze the letter frequencies

First, let's count how often each letter appears in CALCULUS:

Total cards: 2 + 1 + 2 + 2 + 1 = 8

Step 2 — Calculate total possible pairs

When choosing 2 cards from 8 without replacement, order doesn't matter (we just want a pair). This is a combination problem:

Step 3 — Count favorable matching pairs

Only letters appearing at least twice can form matching pairs. Let's count each:

• C pairs: We have 2 C cards, so there's C(2,2) = 1 way to choose both
• L pairs: We have 2 L cards, so there's C(2,2) = 1 way to choose both
• U pairs: We have 2 U cards, so there's C(2,2) = 1 way to choose both

Step 4 — Calculate the probability

Using the fundamental probability formula:

This fraction is already in lowest terms since gcd(3, 28) = 1.

Solution: Method 2 — Sequential Choice Approach

Think of choosing the cards one after the other. The probability that the second card matches the first depends on what the first card was.

Step 1 — Identify the cases

After drawing the first card, we have different scenarios based on which letter we drew:

• If first card is C, L, or U: There's exactly 1 matching card left among 7 remaining
• If first card is A or S: There are 0 matching cards left among 7 remaining

Step 2 — Calculate conditional probabilities

Let's work through each case:

Step 3 — Apply the law of total probability

The probability of a matching pair is the sum of all successful paths:

Verification

Let's verify our answer by checking that it makes intuitive sense and by using an alternative counting method.

Reasonableness check

We have 8 cards total, and only 3 ways to get matching pairs out of 28 total pairs. This gives us 3/28 ≈ 0.107 or about 10.7%. This seems reasonable since most letters appear only once.

Direct enumeration verification

Let's label the cards as C₁, A, L₁, C₂, U₁, L₂, U₂, S. The matching pairs are:

• {C₁, C₂}
• {L₁, L₂}
• {U₁, U₂}

That's exactly 3 matching pairs out of C(8,2) = 28 total pairs. ✓

Method consistency check

Both methods gave us the same answer: 3/28. ✓

Common Pitfalls

The Pattern Behind This

This problem belongs to the family of "sampling without replacement" problems. The general pattern for finding matching pairs is:

Where:

• N is the total number of items
• nᵢ is the frequency of the i-th distinct item type
• The sum is over all item types where nᵢ ≥ 2

For CALCULUS: N = 8, and we have nC = 2, nL = 2, nU = 2 (the others have frequency 1, so they contribute 0 to the sum).

How to Spot This Problem Type

Watch for these key phrases that signal a matching-pairs probability problem:

• "Without replacement" — distinguishes from simpler replacement problems
• "Same [attribute]" — color, letter, number, etc.
• Items with repeated values — multiple cards/balls of the same type
• "Two [items] are chosen" — signals we're looking at pair outcomes

Real Applications

This type of calculation appears in several practical contexts:

• Quality control: Randomly sampling two products from a batch — what's the probability they have the same defect type?
• Genetics: Drawing two individuals from a population — what's the probability they share the same allele variant?
• Survey analysis: Randomly selecting two responses — what's the probability they give the same answer to a multiple-choice question?