Problem:
Suppose that we have a standard poker deck and we add $2$ cards to it. These two cards added are jokers and are considered wild cards. What is the probability of getting a full house?
Note: The best rules apply. That is, if a hand can be a full house or four of a kind then it is four of a kind and the hand is not a full house.
Answer:
Let $p$ be the probability we seek. Let
$p_0$ be the probability that we a full house without any wild cards. Let
$p_1$ be the probability that we get a full house with exactly $1$ wild card. Let
$p_2$ be the probability that we get a full house with exactly $2$ wild cards.
There are $54 \choose 5$ ways of selecting $5$ cards from this deck. We have:
\begin{align*}
p &= p_0 + p_1 + p_2 \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(50) }{5(4)(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(10) }{4(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(5) }{4(3) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(13)(51)(5) }{ 3 } \\
{{54} \choose 5} &= 54(53)(13)(17)(5) \\
p_0 &= \dfrac{ 13 { 4 \choose 3}(12){ 4 \choose 2 } }{ {{54} \choose 5} } \\
{ 4 \choose 3} &= 4 \\
{ 4 \choose 2} &= \dfrac{ 4(3) }{2} = 6 \\
p_0 &= \dfrac{ 13 (4)(12)(6) }{ {{54} \choose 5} } \\
\end{align*}
One way to get a full house is to have two pairs and a wild card. Another way is
three of a kind plus a wild card.
\begin{align*}
p_1 &= \dfrac{ {13 \choose 2} { 4 \choose 2}{ 4 \choose 2 } (48)(2) +
13{ 4 \choose 3} 48(46)(2) }
{ {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(6)(2)(3) (48)(2) + 13(4) (48)(46)(2) } { {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(48)( (6)(2)(3) (2) + (4) (46)(2)) } { {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(48)( 72+368) } { {{54} \choose 5} } \\
%
p_2 &= \dfrac{ 13 { 4 \choose 2} (50) }{ {{54} \choose 5} } \\
p_2 &= \dfrac{ 13 (6) (50) }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(48)( 72+368) + 13 (6) (50) }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(21120) + 13 (6) (50) }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(21120) + 13 (6) (50) }{ 54(53)(13)(17)(5) } \\
p &= \dfrac{ (4)(12)(6) + (21120) + (6) (50) }{ 54(53)(17)(5) } \\
p &= \dfrac{ (2)(12)(6) + 10560 + (6) (25) }{ 27(53)(17)(5) } \\
p &= \dfrac{ (24)(6) + 10560 + 150 }{ 27(53)(17)(5) } \\
p &= \dfrac{ 10854 }{ 27(53)(17)(5) } \\
p &= \dfrac{ 1206 }{ 3(53)(17)(5) } = \dfrac{ 402 }{ (53)(17)(5) } \\
p &= \dfrac{402}{4505}
\end{align*}
I believe the correct value is $\dfrac{104}{35139}$. Where did I go wrong?
I have updated my answer.
Answer:
Let $p$ be the probability we seek. Let
$p_0$ be the probability that we a full house without any wild cards. Let
$p_1$ be the probability that we get a full house with exactly $1$ wild card. Let
$p_2$ be the probability that we get a full house with exactly $2$ wild cards.
There are $54 \choose 5$ ways of selecting $5$ cards from this deck. We have:
\begin{align*}
p &= p_0 + p_1 + p_2 \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(50) }{5(4)(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(10) }{4(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(5) }{4(3) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(13)(51)(5) }{ 3 } \\
{{54} \choose 5} &= 54(53)(13)(17)(5) \\
p_0 &= \dfrac{ 13 { 4 \choose 3}(12){ 4 \choose 2 } }{ {{54} \choose 5} } \\
{ 4 \choose 3} &= 4 \\
{ 4 \choose 2} &= \dfrac{ 4(3) }{2} = 6 \\
p_0 &= \dfrac{ 13 (4)(12)(6) }{ {{54} \choose 5} } \\
\end{align*}
One way to get a full house is to have two pairs and a wild card. Another way is
three of a kind plus a wild card.
\begin{align*}
p_1 &= \dfrac{ {13 \choose 2} { 4 \choose 2}{ 4 \choose 2 } (48)(2) +
13{ 4 \choose 3} 48(46)(2) }
{ {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(6)(2)(3) (48)(2) + 13(4) (48)(46)(2) } { {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(48)( (6)(2)(3) (2) + (4) (46)(2)) } { {{54} \choose 5} } \\
p_1 &= \dfrac{ 13(48)( 72+368) } { {{54} \choose 5} } \\
%
p_2 &= \dfrac{ 0 }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(48)( 72+368) + 0 }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(21120) }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 13(21120) }{ 54(53)(13)(17)(5) } \\
p &= \dfrac{ (4)(12)(6) + 21120 }{ 54(53)(17)(5) } \\
p &= \dfrac{ (2)(12)(6) + 10560 }{ 27(53)(17)(5) } \\
p &= \dfrac{ (2)(4)(6) + 3520 }{ 9(53)(17)(5) } \\
p &= \dfrac{ 3568 }{ 9(53)(17)(5) } \\
p &= \dfrac{ 3568 }{ 40545 } \\
%
\end{align*}
Is my answer right now?
Here is an updated answer.
Let $p$ be the probability we seek. Let
$p_0$ be the probability that we a full house without any wild cards. Let $p_1$ be the probability that we get a full house with
exactly $1$ wild card. Let
$p_2$ be the probability that we get a full house with exactly $2$ wild cards.
There are $54 \choose 5$ ways of selecting $5$ cards from this deck. We have:
\begin{align*}
p &= p_0 + p_1 + p_2 \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(50) }{5(4)(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(10) }{4(3)(2) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(52)(51)(5) }{4(3) } \\
{{54} \choose 5} &= \dfrac{ 54(53)(13)(51)(5) }{ 3 } \\
{{54} \choose 5} &= 54(53)(13)(17)(5) \\
p_0 &= \dfrac{ 13 { 4 \choose 3}(12){ 4 \choose 2 } }{ {{54} \choose 5} } \\
{ 4 \choose 3} &= 4 \\
{ 4 \choose 2} &= \dfrac{ 4(3) }{2} = 6 \\
p_0 &= \dfrac{ 13 (4)(12)(6) }{ {{54} \choose 5} } \\
\end{align*}
One way to get a full house is to have two pairs and a wild card.
\begin{align*}
p_1 &= \dfrac{ {13 \choose 2} { 4 \choose 2}{ 4 \choose 2 }{2 \choose 1} }
{ {{54} \choose 5} } \\
{13 \choose 2} &= \dfrac{ 13(12) }{2} = 78 \\
{ 4 \choose 2 } &= \dfrac{4(3)}{2} = 6 \\
p_1 &= \dfrac{ 78(6)(6)(2) } { {{54} \choose 5} } \\
%
p_2 &= \dfrac{ 0 }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 78(6)(6)(2) }{ {{54} \choose 5} } \\
p &= \dfrac{ 13 (4)(12)(6) + 78(6)(6)(2) }{ 54(53)(13)(17)(5) } \\
p &= \dfrac{ 13 (2)(12)(6) + 78(6)(6) }{ 27(53)(13)(17)(5) } \\
p &= \dfrac{ 1872 + 78(6)(6) }{ 27(53)(13)(17)(5) } \\
p &= \dfrac{ 4680 }{ 27(53)(13)(17)(5) } \\
p &= \dfrac{ 520 }{ 3(53)(13)(17)(5) } \\
p &= \dfrac{ 104 }{ 3(53)(13)(17) } \\
p &= \dfrac{ 8 }{ 3(53)(17) } \\
p &= \dfrac{ 8 }{ 2703} \\
%
\end{align*}
Is it right now?
Best Answer
\begin{align} p_0 &= \frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}\binom{2}{0}}{\binom{54}{5}} \\ p_1 &= \frac{\binom{13}{2}\binom{4}{2}^2\binom{2}{1}}{\binom{54}{5}} \\ p_2 &= 0 \\ p&=p_0+p_1+p_2=\frac{3744+5616+0}{\binom{54}{5}}=\frac{9360}{\binom{54}{5}} \end{align} This turns out to be same as the probability of four of a kind, as shown in my answer here.