GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 15 Nov 2019, 03:35

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

What is the probability of a two pair in poker?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 114
What is the probability of a two pair in poker?  [#permalink]

Show Tags

New post Updated on: 25 Aug 2013, 22:36
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

HideShow timer Statistics

What is the probability of a two pair in poker?

So we have 52 cards. 13 pairs.
We choose our first two pairs from the 13 pairs. (13C2)
From the first pair we choose our pair from the 4 cards (4C2)
From the second pair we choose our pair from the 4 cards (4C2)

Here is my question. Since we already have picked 4 cards from our deck. Can I just say from the remaining 48 cards, I will pick 1?
So (48C1)?


Is this wrong because of the remaining 48 cards are also members of the 2 pairs we picked from?
So we have to make sure we pick that last car from a group of cards that will definitly not include
cards from the two other pairs?

So we have 11 pairs remaining. From the 11 we choose 1. (11C1) and from the 4 cards we pick 1.

So answer: (13C2)(4C2)(4C2)(11C1)(4C1)/(52C4)

Originally posted by alphabeta1234 on 25 Aug 2013, 17:08.
Last edited by Bunuel on 25 Aug 2013, 22:36, edited 1 time in total.
RENAMED THE TOPIC.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59071
Re: What is the probability of a two pair in poker?  [#permalink]

Show Tags

New post 25 Aug 2013, 22:48
alphabeta1234 wrote:
What is the probability of a two pair in poker?

So we have 52 cards. 13 pairs.
We choose our first two pairs from the 13 pairs. (13C2)
From the first pair we choose our pair from the 4 cards (4C2)
From the second pair we choose our pair from the 4 cards (4C2)

Here is my question. Since we already have picked 4 cards from our deck. Can I just say from the remaining 48 cards, I will pick 1?
So (48C1)?


Is this wrong because of the remaining 48 cards are also members of the 2 pairs we picked from?
So we have to make sure we pick that last car from a group of cards that will definitly not include
cards from the two other pairs?

So we have 11 pairs remaining. From the 11 we choose 1. (11C1) and from the 4 cards we pick 1.

So answer: (13C2)(4C2)(4C2)(11C1)(4C1)/(52C4)


\(C^1_{48}\) is wrong because if you choose the value you already have in pairs you'd have full house.

What is the probability of a two pair in poker?

Cards = 52.
Suits = 4.
Values = 13.

\(P(two pair)=\frac{C^2_{13}*C^2_4*C^2_4*C^1_{44}}{C^5_{52}}\)

\(C^2_{13}\) is the number of ways to choose which 2 values will be the pairs (for example, 2 kings, and 2 aces);
\(C^2_4\) is the number of ways to choose 2 suits for the first pair;
\(C^2_4\) is the number of ways to choose 2 suits for the second pair;
\(C^1_{44}\) is the number of ways to choose the remaining fifth card from 52-8=44 (exclude 4 cards we used for the first pair and 4 cards we used for the second pair). This can also be written as \(C^1_{11}*C^1_4\): choosing value for the fifth card and choosing suit for the fifth card.

For more check: let-s-play-poker-84956.html

Hope it helps.

P.S. Please name topics properly.
_________________
Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 114
Re: What is the probability of a two pair in poker?  [#permalink]

Show Tags

New post 30 Aug 2013, 14:10
I see. So the problem is that of the remaining 48 cards, some might be coming from the two pairs we initially picked from. If that is the case why cant we subtract 6? Why 8? We already chose 2 cards from the 2 pairs (8 cards) that leaves 6 cards from the two pairs in the 48 cards.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59071
Re: What is the probability of a two pair in poker?  [#permalink]

Show Tags

New post 31 Aug 2013, 05:48
alphabeta1234 wrote:
I see. So the problem is that of the remaining 48 cards, some might be coming from the two pairs we initially picked from. If that is the case why cant we subtract 6? Why 8? We already chose 2 cards from the 2 pairs (8 cards) that leaves 6 cards from the two pairs in the 48 cards.


Say the two pairs are aces and kings. The fifth card should be any but ace or king, so in order to get that we should choose from 52 - 4 aces and 4 kings.
_________________
GMAT Club Bot
Re: What is the probability of a two pair in poker?   [#permalink] 31 Aug 2013, 05:48
Display posts from previous: Sort by

What is the probability of a two pair in poker?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne