https://mmh.mn/problem/one-pair-total

Calculating One-Pair Poker Hands Created with Pro May 5, 2026 at 12:35 PM Published May 6, 2026 at 01:40 PM

Solve

Here is the step-by-step solution to find the number of distinct five-card "one pair" hands.

To build a hand with exactly one pair, we need to make four independent choices:

1. Choose the rank for the pair There are 13 possible ranks in a standard deck (2 through Ace). We need to select 1 rank for our pair. (113)=13 ways

2. Choose the suits for the pair Once the rank is chosen, we must select 2 of the 4 available suits for that rank to form the pair. (24)=2×14×3=6 ways

3. Choose the ranks for the remaining three cards To ensure we only have one pair (and not a two-pair or a full house), the remaining three cards must be of different ranks from each other and different from the pair's rank. There are 12 remaining ranks, and we must choose 3. (312)=3×2×112×11×10=220 ways

4. Choose the suits for the remaining three cards For each of the 3 ranks chosen in the previous step, we can pick any of the 4 suits. (14)×(14)×(14)=43=64 ways Final Calculation

To find the total number of distinct one-pair hands, multiply the number of ways for each independent choice together: Total=13×6×220×64

13×6=78

78×220=17,160

17,160×64=1,098,240

There are 1,098,240 distinct five-card one-pair hands that can be dealt from a standard 52-card deck.

I did:

52/52 * 12/51

Then divide by total # of possible hands

It looks like you are trying to solve this by looking at the sequential probability of drawing cards one by one. While you can arrive at the right answer this way, there are a few mix-ups in the logic of your setup.

Here is a breakdown of why your approach didn't quite work, and how we can fix it to match your intuition!

1. Mixing Probability with Counting

The question asks "How many distinct... hands" (which is a count), rather than "What is the probability...". When you use fractions like 52/52, you are calculating the probability. If you divide that result by the total number of possible hands, you are essentially dividing a probability by a count, which will give you a remarkably tiny and incorrect number. 2. The wrong fraction for the pair

If you want to calculate the probability card-by-card, drawing 52/52 for the first card is correct (it can be anything). But for the second card to make a pair, it must match the rank of the first card. Since there are only 4 cards of each rank, and you already drew 1, there are only 3 matches left in the remaining 51 cards. So, your second fraction should be 3/51, not 12/51. 3. Forgetting the other three cards

A poker hand has 5 cards. After drawing the pair, you still need to draw 3 more cards, and they must be of different ranks to avoid making a two-pair or three-of-a-kind.

Card 3: Must be a different rank (48 safe cards left out of 50) → 48/50

Card 4: Must be a different rank again (44 safe cards left out of 49) → 44/49

Card 5: Must be a different rank again (40 safe cards left out of 48) → 40/48

4. Forgetting the Order

Drawing cards sequentially implies an order (e.g., getting the pair on cards 1 and 2). But in a hand of poker, you could get the pair on cards 1 and 4, or 3 and 5, etc. There are 10 different ways to position the 2 paired cards in a 5-card sequence (Calculated as 5 choose 2). You have to multiply your probability by 10. Putting your method together correctly:

If we correct your logic to find the probability of drawing a one-pair hand, it looks like this: P(One Pair)=(5252×513×5048×4944×4840)×10≈0.422569

Now, to find the total number of hands (the actual count), you multiply this probability by the total number of possible 5-card hands (2,598,960): 0.422569...×2,598,960=1,098,240

This gets you the exact same answer! However, because card-by-card probability requires you to remember to account for permutations (like multiplying by 10) and keeping track of shrinking decks, doing it via combinations (the method in my first response) is generally considered much easier and less prone to mistakes for these types of problems.

Discuss the scheme for picking each factor, I understand the prior ones of counting because it's much simpler. Help me to understand the finding of factors in counting

The shift from probability (dealing cards one by one) to combinatorial counting can feel a bit abstract at first.

To master the counting scheme, you have to change your perspective. Stop imagining a dealer dealing cards to you one at a time. Instead, imagine you are in a warehouse that contains a beautifully organized deck of cards, and you are building the hand from scratch using a specific blueprint.

Here is the blueprint for a "One Pair" hand, and the scheme for picking the factors to build it. The Scheme: Building the Hand Feature by Feature

Every factor in our equation represents an independent "decision" we have to make in the warehouse to guarantee we build a one-pair hand and absolutely nothing else.

Decision 1: What is the rank of our pair?

The Logic: You have 13 bins in your warehouse (Aces, 2s, 3s, up to Kings). You need to walk up and select exactly one bin to be your pair.

The Math: You are choosing 1 item out of 13 options.

The Factor: (113)=13

Decision 2: Which specific suits make up that pair?

The Logic: Let's say you chose the "Kings" bin. Inside that bin, there are 4 cards (Hearts, Diamonds, Clubs, Spades). A pair requires exactly 2 cards. You need to reach into the bin and grab 2 of those 4 cards.

The Math: You are choosing 2 items out of 4 options.

The Factor: (24)=6

Decision 3: What are the ranks of our three "kicker" cards?

The Logic: You still need 3 more cards for a 5-card hand. Crucially, to prevent making a "Three of a Kind" or "Two Pair," these 3 cards must have different ranks from your pair, and different ranks from each other. So, you look at the 12 bins you haven't touched yet. You need to select exactly 3 of those bins to represent your single cards.

The Math: You are choosing 3 items out of 12 options.

The Factor: (312)=220

Decision 4: Which specific suits make up those three kicker cards?

The Logic: Let's say you selected the 5s, 9s, and Jacks bins. You go to the 5s bin—there are 4 suits, and you need to pull exactly 1 card. Then you go to the 9s bin—4 suits, pull exactly 1. Then to the Jacks bin—4 suits, pull exactly 1.

The Math: You are choosing 1 out of 4, three separate times.

The Factor: (14)×(14)×(14)=43=64

Why we multiply them

In combinatorics, the Rule of Product states that if you have a sequence of independent choices to make, you multiply the number of options for each choice to find the total number of combinations.

Because every single "One Pair" hand ever dealt can be constructed by making Decision 1, then Decision 2, then Decision 3, then Decision 4, we multiply the factors: 13×6×220×64=1,098,240