Poker Hands Probabilities with Combinatorics

TL;DR

Use $(r n) = \frac{n !}{r ! ( n - r )!}$ to count rank selections and multiply by suit choices when suits are independent.

Briefing Cornell Notes

Briefing

Combinatorics can turn Texas Hold’em hand rankings into exact counts—and those counts translate directly into probabilities that explain why “good” hands are rare. Using combinations and suit/rank choices from a 52-card deck, the transcript derives how many 5-card poker hands fall into key categories (straight flush, four of a kind, full house, flush, straight, three of a kind, two pair, one pair, and high card) and then divides each by the total number of 5-card selections, $(5 52)$ .

A straight flush requires five consecutive ranks in the same suit. There are 10 possible sequences of ranks (including A-2-3-4-5 as the “low” straight and 10-J-Q-K-A as the “high” straight). Each sequence can occur in any of 4 suits, giving $10 \times 4 = 40$ straight flushes. Four of a kind comes from choosing one rank out of 13, taking all 4 suits for that rank, and then choosing the remaining fifth card from the other 12 ranks (4 suit options each). That yields $13 \times 1 \times 12 \times 4 = 624$ hands, about 0.02%.

Full houses combine a three-of-a-kind plus a pair. The count starts by choosing the rank for the triple (13 choices) and selecting 3 of its 4 suits ( $(3 4)$ ). Then it chooses the rank for the pair from the remaining 12 ranks and selects 2 of its 4 suits ( $(2 4)$ ). The result is 3,744 full houses, about 0.14%—still uncommon.

Flushes are five cards of the same suit with no repeated ranks, but they must exclude straight flushes to avoid double-counting. The transcript counts flushes by choosing 5 distinct ranks from 13 and assigning a suit (4 choices), then subtracts the 40 straight flushes already counted. Straights are counted separately as 10 rank sequences, with each of the 5 cards able to take any suit ( $4^{5}$ ), and then subtracting the 40 straight flushes again.

Three of a kind is built by choosing the rank for the triple (13 choices), selecting 3 of its 4 suits ( $(3 4)$ ), then choosing two additional distinct ranks from the remaining 12 and assigning suits independently ( $4$ choices each). The transcript notes that having three of the same rank prevents the hand from being a straight flush, so no subtraction is needed. Two pair uses $(2 13)$ ways to pick the pair ranks, then $(2 4)$ suit choices for each pair, and finally a fifth card from the remaining 11 ranks with 4 suit choices. One pair chooses a rank for the pair (13), selects 2 of its 4 suits, then chooses 3 other distinct ranks from the remaining 12 and assigns suits.

The most frequent outcomes are one pair and high card. The transcript gives a one-pair count around 1,098,240 (about 42%), and a high-card count around 1,302,540 (about 50%), after subtracting straights, flushes, and straight flushes from the raw “five distinct ranks and suits” total. The takeaway is strategic: roughly 92% of hands are one pair or worse, so poker success depends heavily on playing—often bluffing and extracting value—when the starting hand is not “strong.”

Cornell Notes

Combinatorics turns poker hand categories into exact counts by selecting ranks and suits from a 52-card deck and then dividing by $(5 52)$ . The transcript derives key examples: straight flushes total $10 \times 4 = 40$ , four of a kind totals 624, and full houses total 3,744. Flushes and straights require careful subtraction to avoid double-counting straight flushes. The most common results are one pair (about 42%) and high card (about 50%), meaning “better than a pair” hands are rare (about 7.6%). That rarity drives strategy: most hands are weak, so learning to play bad hands matters.

How many straight flushes exist, and why does the count equal $10 \times 4$ ?

A straight flush needs five consecutive ranks in the same suit. There are 10 distinct rank sequences in poker: 1-2-3-4-5 through 10-J-Q-K-A (with Ace acting as both low and high to create two different sequences). Each sequence can be in any of 4 suits, so the total is

10 \times 4 = 40

What is the logic behind the 624 hands for four of a kind?

Pick the rank for the four cards: 13 choices. The four suits are forced because all four suits must appear for that rank. Then choose the fifth card’s rank from the remaining 12 ranks and choose its suit from 4 options, giving

13 \times 12 \times 4 = 624

Why do flush and straight counts subtract 40 straight flushes?

Flushes count hands with five cards of the same suit and distinct ranks, but some of those hands are also straights—specifically straight flushes. If straight flushes are counted inside both categories, they get double-counted. The transcript fixes this by subtracting the 40 straight flushes from both the flush and straight totals.

How is a full house counted from rank and suit choices?

Choose the rank for the triple (13 choices) and select 3 of its 4 suits (

(3 4)

). Then choose the rank for the pair from the remaining 12 ranks and select 2 of its 4 suits (

(2 4)

). Multiplying these choices yields 3,744 full houses (about 0.14%).

Why is one pair so common compared with four of a kind?

One pair only requires matching two cards by rank; the other three cards can be any three distinct ranks different from the pair rank, with suits chosen freely. Four of a kind is much more restrictive because it forces all four suits of a single rank plus only one extra card. The transcript’s probabilities reflect this: one pair is about 42%, while four of a kind is about 0.02%.

Review Questions

Which categories require explicit subtraction terms to prevent double-counting, and what overlap causes it?
Re-derive the straight flush count by listing the 10 possible rank sequences and multiplying by suit choices.
Using the transcript’s method, set up (but don’t fully compute) the combinatorial expression for the number of two pair hands.

Key Points

1
Use $(r n) = \frac{n !}{r ! ( n - r )!}$ to count rank selections and multiply by suit choices when suits are independent.
2
Straight flushes come from 10 consecutive rank sequences (Ace-low through Ace-high) times 4 suits, totaling 40.
3
Four of a kind is counted by choosing the rank (13), taking all 4 suits, then choosing the kicker from 12 remaining ranks with 4 suits each, totaling 624.
4
Full houses require a triple and a pair: choose the triple rank (13), choose 3 suits ( $(3 4)$ ), then choose the pair rank (12) and 2 suits ( $(2 4)$ ), totaling 3,744.
5
Flushes and straights must subtract the 40 straight flushes to avoid double-counting hands that satisfy both patterns.
6
Three of a kind and two pair can be counted without subtraction in the transcript’s approach because the structure prevents straight flush overlap.
7
Strategically, one pair (~42%) and high card (~50%) dominate, so most starting hands are weak; success depends on playing bad hands effectively, including bluffing.

Highlights

Straight flushes total just 40 hands because there are only 10 rank sequences and 4 suits.

Four of a kind is extremely rare: 624 hands out of

(5 52)

, about 0.02%.

Full houses land at 3,744 hands, roughly 0.14%, still far less common than pairs.

Flush and straight calculations require subtracting straight flushes to prevent double-counting.

With about 92% of hands at one pair or worse, poker strategy must be built around weak-hand play.

Topics

Combinatorics
Poker Probabilities
Texas Hold’em
Hand Counting
Double-Counting