It is currently 17 Nov 2017, 10:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Calculate the odds of getting a hand of exactly one pair

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
VP
Joined: 06 Jun 2004
Posts: 1050

Kudos [?]: 185 [0], given: 0

Location: CA
Calculate the odds of getting a hand of exactly one pair [#permalink]

### Show Tags

27 Nov 2005, 17:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Calculate the odds of getting a hand of exactly one
pair (any pair) in a five card stud poker, using one 52-card deck.

Kudos [?]: 185 [0], given: 0

Director
Joined: 14 Sep 2005
Posts: 984

Kudos [?]: 220 [0], given: 0

Location: South Korea

### Show Tags

27 Nov 2005, 18:58
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 !

Kudos [?]: 220 [0], given: 0

VP
Joined: 06 Jun 2004
Posts: 1050

Kudos [?]: 185 [0], given: 0

Location: CA

### Show Tags

27 Nov 2005, 19:09
Try again ....Hint: Use the combination formula for the 3 cards + 2 cards possibility

Kudos [?]: 185 [0], given: 0

Director
Joined: 14 Sep 2005
Posts: 984

Kudos [?]: 220 [0], given: 0

Location: South Korea

### Show Tags

27 Nov 2005, 19:14
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 !

Kudos [?]: 220 [0], given: 0

Intern
Joined: 22 Mar 2005
Posts: 39

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

Location: Houston

### Show Tags

29 Nov 2005, 17:32
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%.

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

Manager
Joined: 30 Aug 2005
Posts: 183

Kudos [?]: 15 [0], given: 0

### Show Tags

05 Dec 2005, 05:55
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

Kudos [?]: 15 [0], given: 0

05 Dec 2005, 05:55
Display posts from previous: Sort by

# Calculate the odds of getting a hand of exactly one pair

 post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.