Poker Hand Probabilities.md

Poker Hand Probabilities/674b227d-11c8-8012-ac6d-bb04699bc30f

In a standard game of Texas Hold'em or Five-Card Draw Poker, the deck consists of 52 cards, and probabilities for specific hands depend on the combinations of cards in a dealt hand.

Below are the probabilities of each poker hand in a five-card hand:

Hand	Frequency	Probability	Odds Against
Royal Flush	4	$$0.000154%$$	649,739 to 1
Straight Flush	36	$$0.00139%$$	72,192 to 1
Four of a Kind	624	$$0.024%$$	4,164 to 1
Full House	3,744	$$0.144%$$	693 to 1
Flush	5,108	$$0.197%$$	508 to 1
Straight	10,200	$$0.39%$$	254 to 1
Three of a Kind	54,912	$$2.1128%$$	46.3 to 1
Two Pair	123,552	$$4.7539%$$	20.0 to 1
One Pair	1,098,240	$$42.2569%$$	1.37 to 1
High Card	1,302,540	$$50.1177%$$	0.995 to 1

---

Key Concepts

• Royal Flush: Ace, King, Queen, Jack, and 10 of the same suit.
• Straight Flush: Five consecutive cards of the same suit (not Royal).
• Four of a Kind: Four cards of the same rank (e.g., four Kings).
• Full House: Three cards of one rank and two cards of another (e.g., three Aces and two Kings).
• Flush: Five cards of the same suit (not in sequence).
• Straight: Five consecutive cards of mixed suits.
• Three of a Kind: Three cards of the same rank.
• Two Pair: Two different pairs of cards.
• One Pair: Two cards of the same rank.
• High Card: None of the above combinations; the hand's value is determined by the highest card.

---

Calculation Insights

These probabilities are derived from combinatorics:

• The total number of possible five-card hands:

$$ \binom{52}{5} = \frac{52!}{5!(52-5)!} = 2,598,960 $$

Each specific hand type is calculated based on the ways to select cards to form that hand divided by the total number of hands.

For example:

• Royal Flush:

$$ \binom{4}{1} = 4 $$

(Only 4 Royal Flushes are possible, one per suit.)

• Four of a Kind:

$$ \binom{13}{1} \times \binom{12}{1} \times \binom{4}{4} \times \binom{48}{1} = 624 $$

You can adjust calculations for other poker variants or include probabilities for community cards in Texas Hold'em. Let me know if you'd like to explore this further!

Let's break down the calculation for Four of a Kind to understand why $\binom{13}{1} \times \binom{12}{1} \times \binom{4}{4} \times \binom{48}{1} = 624$:

---

What is Four of a Kind?

A Four of a Kind is a poker hand consisting of:

• Four cards of the same rank (e.g., four Kings: $\heartsuit K, \clubsuit K, \spadesuit K, \diamondsuit K$),
• One additional card of any rank (not matching the four of a kind).

---

Breaking Down the Calculation

1. Choose the Rank for the Four of a Kind: $\binom{13}{1}$
   • There are 13 possible ranks ($\text{A, 2, 3, ..., K}$).
   • We need to choose one rank for our Four of a Kind.

$$ \binom{13}{1} = 13 $$

2. Choose the Rank for the Additional Card: $\binom{12}{1}$
   • After selecting the rank for the Four of a Kind, 12 ranks remain for the fifth card.
   • We need to choose one of these remaining ranks.

$$ \binom{12}{1} = 12 $$

3. Choose All Four Suits for the Four of a Kind: $\binom{4}{4}$
   • For the Four of a Kind, we need all four suits (e.g., $\heartsuit, \clubsuit, \spadesuit, \diamondsuit$).
   • There is only one way to choose all four suits:

$$ \binom{4}{4} = 1 $$

4. Choose One Card from the Remaining Deck: $\binom{48}{1}$
   • After selecting the four cards of the same rank, 48 cards remain in the deck.
   • We need to choose one card to complete the hand.

$$ \binom{48}{1} = 48 $$

---

Combining These Choices

The total number of ways to form a Four of a Kind hand is:

$$ \binom{13}{1} \times \binom{12}{1} \times \binom{4}{4} \times \binom{48}{1} $$

Substitute the values:

$$ 13 \times 12 \times 1 \times 48 = 624 $$

---

Why Does This Work?

This formula counts all possible ways to form a valid Four of a Kind hand:

1. First, choose the rank for the Four of a Kind ($13$ choices).
2. Then, choose the rank for the kicker (the additional card, $12$ choices).
3. Next, select all four suits for the Four of a Kind ($1$ way).
4. Finally, pick one card from the remaining deck ($48$ ways).

This ensures all valid combinations are counted without overcounting or missing any possibilities.

Let me know if you'd like more assistance!