Question 1165681: Suppose you are playing a Texas Hold’em game with three friends, using a standard 52-card deck. At the starting round each player is dealt 2 cards, and at each continuing round each player is dealt an additional card. Each player can be dealt no more than 5 cards in total. Assume the game is fair and no player fold till the final round.
If you have a pair at the initial round, what is the probability for you to get a Full House (three of a kind and a pair)?
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! That's a fun probability problem! To calculate the probability of getting a **Full House** (a three-of-a-kind and a pair) given you start with a specific pair in Texas Hold'em (simplified rules), we need to figure out the probabilities over the remaining three rounds where you are dealt one card each time.
Here are the simplified rules we're following:
* Total Players: 4 (You + 3 friends).
* Your Starting Hand: A specific pair (e.g., $A\heartsuit A\diamondsuit$).
* Total cards dealt: 5 cards to you in total (2 start + 3 subsequent rounds).
* Total cards remaining to be dealt in the hand: $4 \times 3 = 12$ to your opponents, and 3 to you, for a total of **15 cards**.
We only care about the **three cards** you are dealt in the subsequent rounds (let's call them $C_3, C_4, C_5$) that, combined with your starting pair, must form a Full House.
Your starting hand is $\{R, R\}$. For a Full House, the final 5 cards must be either:
1. **Three of a Kind of Rank R** (The two remaining $R$ cards) **AND** **a Pair of a New Rank $S$** (where $S \neq R$).
2. **A Pair of Rank R** (Your original pair) **AND** **Three of a Kind of a New Rank $S$** (where $S \neq R$).
However, since you only get **three** more cards, the possible combinations that make a Full House are:
### Favorable Outcome Analysis
The three cards you receive must satisfy **one** of two structures:
| Structure | Cards Needed | Example (Starting with $A, A$) |
| :---: | :---: | :---: |
| **I. R-Full House (R-RR + S-S)** | $\mathbf{R}$ and $\mathbf{S}$ and $\mathbf{S}$ | You need a third Ace, and a pair of another rank (e.g., $K, K$). |
| **II. S-Full House (R-R + S-SS)** | $\mathbf{S}$ and $\mathbf{S}$ and $\mathbf{S}$ | You need three cards of a new rank (e.g., $K, K, K$). |
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### Calculating the Probability
We will calculate the number of favorable 3-card combinations and divide by the total number of 3-card combinations.
#### 1. Total Possible Outcomes (Denominator)
You are being dealt 3 more cards, but your 3 opponents are also being dealt 3 more cards each. Before the 3 cards are dealt, 4 players have 2 cards each, so $4 \times 2 = 8$ cards are gone.
* Total cards remaining in the deck: $52 - 8 = 44$.
* Since the game is fair and cards are dealt sequentially, the order matters if we track the rounds. However, since the question asks for the probability *of getting* the final hand, it's easiest to treat your final three cards as one single draw of 3 cards from the available deck pool.
The cards already dealt are $4 \times 2 = 8$. The cards remaining are $52 - 8 = 44$.
Total ways to be dealt your next 3 cards from the deck (without considering opponents' cards *yet*):
$$\text{Total Outcomes} = \binom{44}{3} = \frac{44 \times 43 \times 42}{3 \times 2 \times 1} = \mathbf{13,244}$$
#### 2. Favorable Outcomes (Numerator)
Let $R$ be the rank you hold (e.g., Aces).
**Case I: R-Full House (R-RR + S-S)**
You need 1 more $R$ card and a pair of a new rank $S$.
* There are 2 cards of rank $R$ left in the deck. Choose $\mathbf{1}$: $\binom{2}{1} = 2$.
* There are 12 other ranks ($S \neq R$). Choose $\mathbf{1}$ rank: $\binom{12}{1} = 12$.
* For that rank $S$, there are 4 suits. Choose $\mathbf{2}$ suits to form the pair: $\binom{4}{2} = 6$.
* Total ways for Case I: $2 \times 12 \times 6 = \mathbf{144}$ ways.
**Case II: S-Full House (R-R + S-SS)**
You need three cards of a new rank $S$.
* There are 12 other ranks ($S \neq R$). Choose $\mathbf{1}$ rank: $\binom{12}{1} = 12$.
* For that rank $S$, there are 4 suits. Choose $\mathbf{3}$ suits to form the three-of-a-kind: $\binom{4}{3} = 4$.
* Total ways for Case II: $12 \times 4 = \mathbf{48}$ ways.
**Total Favorable Outcomes** = Case I + Case II $= 144 + 48 = \mathbf{192}$
#### 3. Final Probability
$$\text{P}(\text{Full House}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{192}{13,244}$$
$$\text{P}(\text{Full House}) \approx \mathbf{0.0145} \text{ or } \mathbf{1.45\%}$$
**The probability for you to get a Full House is approximately $\mathbf{0.0145}$.**
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