document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165681>Question 1165681</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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.\r<BR>\n" );
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document.write( "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)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852945 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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.\r<BR>\n" );
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document.write( "Here are the simplified rules we're following:<BR>\n" );
document.write( "* Total Players: 4 (You + 3 friends).<BR>\n" );
document.write( "* Your Starting Hand: A specific pair (e.g., $A\heartsuit A\diamondsuit$).<BR>\n" );
document.write( "* Total cards dealt: 5 cards to you in total (2 start + 3 subsequent rounds).<BR>\n" );
document.write( "* 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**.\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "Your starting hand is $\{R, R\}$. For a Full House, the final 5 cards must be either:<BR>\n" );
document.write( "1.  **Three of a Kind of Rank R** (The two remaining $R$ cards) **AND** **a Pair of a New Rank $S$** (where $S \neq R$).<BR>\n" );
document.write( "2.  **A Pair of Rank R** (Your original pair) **AND** **Three of a Kind of a New Rank $S$** (where $S \neq R$).\r<BR>\n" );
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document.write( "However, since you only get **three** more cards, the possible combinations that make a Full House are:\r<BR>\n" );
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document.write( "### Favorable Outcome Analysis\r<BR>\n" );
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document.write( "The three cards you receive must satisfy **one** of two structures:\r<BR>\n" );
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document.write( "| Structure | Cards Needed | Example (Starting with $A, A$) |<BR>\n" );
document.write( "| :---: | :---: | :---: |<BR>\n" );
document.write( "| **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., $K, K$). |<BR>\n" );
document.write( "| **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., $K, K, K$). |\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### Calculating the Probability\r<BR>\n" );
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document.write( "We will calculate the number of favorable 3-card combinations and divide by the total number of 3-card combinations.\r<BR>\n" );
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document.write( "#### 1. Total Possible Outcomes (Denominator)\r<BR>\n" );
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document.write( "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.<BR>\n" );
document.write( "* Total cards remaining in the deck: $52 - 8 = 44$.<BR>\n" );
document.write( "* 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.\r<BR>\n" );
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document.write( "The cards already dealt are $4 \times 2 = 8$. The cards remaining are $52 - 8 = 44$.<BR>\n" );
document.write( "Total ways to be dealt your next 3 cards from the deck (without considering opponents' cards *yet*):<BR>\n" );
document.write( "$$\text{Total Outcomes} = \binom{44}{3} = \frac{44 \times 43 \times 42}{3 \times 2 \times 1} = \mathbf{13,244}$$\r<BR>\n" );
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document.write( "#### 2. Favorable Outcomes (Numerator)\r<BR>\n" );
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document.write( "Let $R$ be the rank you hold (e.g., Aces).\r<BR>\n" );
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document.write( "**Case I: R-Full House (R-RR + S-S)**\r<BR>\n" );
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document.write( "You need 1 more $R$ card and a pair of a new rank $S$.<BR>\n" );
document.write( "* There are 2 cards of rank $R$ left in the deck. Choose $\mathbf{1}$: $\binom{2}{1} = 2$.<BR>\n" );
document.write( "* There are 12 other ranks ($S \neq R$). Choose $\mathbf{1}$ rank: $\binom{12}{1} = 12$.<BR>\n" );
document.write( "* For that rank $S$, there are 4 suits. Choose $\mathbf{2}$ suits to form the pair: $\binom{4}{2} = 6$.<BR>\n" );
document.write( "* Total ways for Case I: $2 \times 12 \times 6 = \mathbf{144}$ ways.\r<BR>\n" );
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document.write( "**Case II: S-Full House (R-R + S-SS)**\r<BR>\n" );
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document.write( "You need three cards of a new rank $S$.<BR>\n" );
document.write( "* There are 12 other ranks ($S \neq R$). Choose $\mathbf{1}$ rank: $\binom{12}{1} = 12$.<BR>\n" );
document.write( "* For that rank $S$, there are 4 suits. Choose $\mathbf{3}$ suits to form the three-of-a-kind: $\binom{4}{3} = 4$.<BR>\n" );
document.write( "* Total ways for Case II: $12 \times 4 = \mathbf{48}$ ways.\r<BR>\n" );
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document.write( "**Total Favorable Outcomes** = Case I + Case II $= 144 + 48 = \mathbf{192}$\r<BR>\n" );
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document.write( "#### 3. Final Probability\r<BR>\n" );
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document.write( "$$\text{P}(\text{Full House}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{192}{13,244}$$\r<BR>\n" );
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document.write( "$$\text{P}(\text{Full House}) \approx \mathbf{0.0145} \text{ or } \mathbf{1.45\%}$$\r<BR>\n" );
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document.write( "**The probability for you to get a Full House is approximately $\mathbf{0.0145}$.** </SPAN>\n" );
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