Number of hands containing two different pairs
$begingroup$
We simultaneously extract $5$ cards from a game set of $32$ cards ($8$ Hearts, $8$ Diamonds, $8$ Spades, $8$ Clubs). This set of $5$ cards is called a "hand". I want to know how to count the hands of $5$ cards containing two different pairs (for example 7788K, QQKK9).
probability combinatorics
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
$endgroup$
add a comment |
$begingroup$
We simultaneously extract $5$ cards from a game set of $32$ cards ($8$ Hearts, $8$ Diamonds, $8$ Spades, $8$ Clubs). This set of $5$ cards is called a "hand". I want to know how to count the hands of $5$ cards containing two different pairs (for example 7788K, QQKK9).
probability combinatorics
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
$endgroup$
$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21
add a comment |
$begingroup$
We simultaneously extract $5$ cards from a game set of $32$ cards ($8$ Hearts, $8$ Diamonds, $8$ Spades, $8$ Clubs). This set of $5$ cards is called a "hand". I want to know how to count the hands of $5$ cards containing two different pairs (for example 7788K, QQKK9).
probability combinatorics
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
$endgroup$
We simultaneously extract $5$ cards from a game set of $32$ cards ($8$ Hearts, $8$ Diamonds, $8$ Spades, $8$ Clubs). This set of $5$ cards is called a "hand". I want to know how to count the hands of $5$ cards containing two different pairs (for example 7788K, QQKK9).
probability combinatorics
probability combinatorics
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
edited Jan 14 at 18:57
N. F. Taussig
44.8k103358
44.8k103358
asked Jan 14 at 15:59
David LingardDavid Lingard
676
asked Jan 14 at 15:59
David LingardDavid Lingard
676
asked Jan 14 at 15:59
David LingardDavid Lingard
676
676
$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21
add a comment |
$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21
$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21
$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have $8$ ranks in each of $4$ suits. For two pairs, you need to choose $2$ ranks for the values of those pairs. There are $binom82=28$ ways to do this. Then, for each pair, you need to choose the suits. There are $binom42=6$ ways to do this for each pair. Finally, there are $6$ remaining ranks to choose for the unpaired card, and $4$ options for the suit gives $6times4=24$ options for the fifth card. The total is:
$$binom82timesbinom42timesbinom42times24=24192$$
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
You have $8$ ranks in each of $4$ suits. For two pairs, you need to choose $2$ ranks for the values of those pairs. There are $binom82=28$ ways to do this. Then, for each pair, you need to choose the suits. There are $binom42=6$ ways to do this for each pair. Finally, there are $6$ remaining ranks to choose for the unpaired card, and $4$ options for the suit gives $6times4=24$ options for the fifth card. The total is:
$$binom82timesbinom42timesbinom42times24=24192$$
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
add a comment |
$begingroup$
You have $8$ ranks in each of $4$ suits. For two pairs, you need to choose $2$ ranks for the values of those pairs. There are $binom82=28$ ways to do this. Then, for each pair, you need to choose the suits. There are $binom42=6$ ways to do this for each pair. Finally, there are $6$ remaining ranks to choose for the unpaired card, and $4$ options for the suit gives $6times4=24$ options for the fifth card. The total is:
$$binom82timesbinom42timesbinom42times24=24192$$
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
add a comment |
$begingroup$
You have $8$ ranks in each of $4$ suits. For two pairs, you need to choose $2$ ranks for the values of those pairs. There are $binom82=28$ ways to do this. Then, for each pair, you need to choose the suits. There are $binom42=6$ ways to do this for each pair. Finally, there are $6$ remaining ranks to choose for the unpaired card, and $4$ options for the suit gives $6times4=24$ options for the fifth card. The total is:
$$binom82timesbinom42timesbinom42times24=24192$$
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
You have $8$ ranks in each of $4$ suits. For two pairs, you need to choose $2$ ranks for the values of those pairs. There are $binom82=28$ ways to do this. Then, for each pair, you need to choose the suits. There are $binom42=6$ ways to do this for each pair. Finally, there are $6$ remaining ranks to choose for the unpaired card, and $4$ options for the suit gives $6times4=24$ options for the fifth card. The total is:
$$binom82timesbinom42timesbinom42times24=24192$$
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
edited Jan 14 at 19:38
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
answered Jan 14 at 16:44
Daniel MathiasDaniel Mathias
1,36018
1,36018
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
add a comment |
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
To choose the unpaired card, we only have to choose a different value, so 32-8=24. Why do you consider only 16 ways to do it.
$endgroup$
– David Lingard
Jan 14 at 18:41
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@DavidLingard Thanks for pointing that out. I guess I just had 16 on my brain.
$endgroup$
– Daniel Mathias
Jan 14 at 18:56
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
@N.F.Taussig David is correct, as there are $6$ ranks available for the fifth card, and $4$ suits gives $6times4=24$
$endgroup$
– Daniel Mathias
Jan 14 at 19:16
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
I am not sure why I am confusing ranks and suits today. However, you are correct.
$endgroup$
– N. F. Taussig
Jan 14 at 19:21
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
$begingroup$
thanks for the answer!!
$endgroup$
– David Lingard
Jan 15 at 10:01
add a comment |
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$begingroup$
For clarity, you should indicate that this $32$-card deck is a subset of the standard $52$-card poker deck, with (presumably) $2$ through $6$ excluded.
$endgroup$
– Daniel Mathias
Jan 14 at 19:21