Number of hands containing two different pairs












0












$begingroup$


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).










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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
















0












$begingroup$


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).










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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).










share|cite|improve this question











$endgroup$




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






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edited Jan 14 at 18:57









N. F. Taussig

44.8k103358




44.8k103358










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


















  • $begingroup$
    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










1 Answer
1






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oldest

votes


















2












$begingroup$

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:
$$binom82timesbinom42timesbinom42times24=24192$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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:
$$binom82timesbinom42timesbinom42times24=24192$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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:
$$binom82timesbinom42timesbinom42times24=24192$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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:
$$binom82timesbinom42timesbinom42times24=24192$$






share|cite|improve this answer











$endgroup$



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:
$$binom82timesbinom42timesbinom42times24=24192$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 19:38

























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


















  • $begingroup$
    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


















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