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Calculate the odds of getting a hand of exactly one pair

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VP
Joined: 06 Jun 2004
Posts: 1059
Location: CA
Followers: 2

Kudos [?]: 54 [0], given: 0

Calculate the odds of getting a hand of exactly one pair [#permalink]  27 Nov 2005, 16:40
Calculate the odds of getting a hand of exactly one
pair (any pair) in a five card stud poker, using one 52-card deck.
Director
Joined: 14 Sep 2005
Posts: 993
Location: South Korea
Followers: 2

Kudos [?]: 50 [0], given: 0

There are 13 kinds of cards from A to K.

The probability of getting WW XYZ (X, Y, and Z are three different kinds of cards)
= ( 1/52 * 1/51 * 13 ) * ( 1/12 * 1/11 * 1/10 * 12C3 )
= 1/1224

No. of cases of arranging WWXYZ = 4!/2! = 12

1/1224* 12 = 1/102

I'm not convinced, tho...[/b]
_________________

Auge um Auge, Zahn um Zahn !

VP
Joined: 06 Jun 2004
Posts: 1059
Location: CA
Followers: 2

Kudos [?]: 54 [0], given: 0

Try again ....Hint: Use the combination formula for the 3 cards + 2 cards possibility
Director
Joined: 14 Sep 2005
Posts: 993
Location: South Korea
Followers: 2

Kudos [?]: 50 [0], given: 0

gamjatang wrote:
There are 13 kinds of cards from A to K.

The probability of getting WW XYZ (X, Y, and Z are three different kinds of cards)
= ( 1/52 * 1/51 * 13 ) * ( 1/12 * 1/11 * 1/10 * 12C3 )
= 1/1224

No. of cases of arranging WWXYZ = 4!/2! = 12

1/1224* 12 = 1/102

I'm not convinced, tho...[/b]

The red part should be 5...

1/1224 * 60 = 5/102
_________________

Auge um Auge, Zahn um Zahn !

Intern
Joined: 22 Mar 2005
Posts: 39
Location: Houston
Followers: 0

Kudos [?]: 0 [0], given: 0

First choose the pair = 13*4c2 = 13*6 = 78

Now you have 50 cards left in the deck and you must pick one of 48 "good" ones b/c if you pick another of our pair cards you will get 3 of a kind.

Now you have 49 cards left in the deck and you must not match your pair cards or the card you just selected (two pair). So that leaves 44 "good" ones and 5 bad ones.

Finally, you have 48 cards left in the deck and you must pick a non pair card. There are still 2 left from your original pair that you don't want, and there are now 3 left from your last card that you picked and 3 left from the one before that. So you have 48-8=40 good ones left.

Therefore 48*44*40/6 = 14080 ways to get the other three cards without pairing or making three of a kind (don't forget to divide by 3! to eliminate duplicate arrangements).

So in closing you have 78*14080 = 1,098,240 ways to make exactly one pair. Divide this by 52c5 = 2,598,960 to get Prob = 42.2%.
Manager
Joined: 30 Aug 2005
Posts: 187
Followers: 1

Kudos [?]: 3 [0], given: 0

I only know the basics about cards, know nothing about poker thanks to my religious upbringing, no gambling allowed.
I will be screwed if I get a Q like this
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