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Let's play poker.
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Updated on: 07 Oct 2009, 16:43
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Playing poker helps to understand the combination and viseversa. So, find below problems to master in either of them: In the poker game FiveCard Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K). A. What is the probability that we are dealt a fourofakind? B. What is the probability that we are dealt full house? (A full house is a hand with both a threeofakind and a twoofakind.) C. What is the probability that we are dealt two pairs? (Fifth is different) D. What is the probability that we are dealt three of a kind? (Others are different) E. What is the probability that we are dealt one pair? (Others are different) F. What is the probability that we are dealt all five of different ranks? G. What is the probability that we are dealt hands with every suit?
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Originally posted by Bunuel on 07 Oct 2009, 13:45.
Last edited by Bunuel on 07 Oct 2009, 16:43, edited 1 time in total.




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07 Oct 2009, 15:54
I'll give it a shot (although I'm going to cheat and use a calculator lol).
A. What is the probability that we are dealt a fourofakind?
13C1 ways to select the value of the 4 of a kind (2,3,...,K,A) 48C1 ways to select the fifth card
\(P = \frac{13C1 * 48C1}{52C5}\)
\(P = \frac{13*48}{2598960}\)
\(P = \frac{1}{4165}\)
B. What is the probability that we are dealt full house? (A full house is a hand with both a threeofakind and a twoofakind.)
13C1 ways to select the value of the 3 of a kind (2,3,...,K,A) 4C3 ways to select the suits of the 3 of a kind 12C1 ways to select the value of the pair 4C2 ways to select the suits of the pair
\(P = \frac{13C1 * 4C3 * 12C1 * 4C2}{52C5}\)
\(P = \frac{13*4*12*6}{2598960}\)
\(P = \frac{3744}{2598960}\)
\(P = \frac{6}{4165}\)
C. What is the probability that we are dealt two pairs? (Fifth is different)
13C2 ways to select the value of the two pairs 4C2 ways to select the suits of the first pair 4C2 ways to select the suits of the second pair 44C1 ways to select the fifth card
\(P = \frac{13C2 * 4C2 * 4C2 * 44C1}{52C5}\)
\(P = \frac{78 * 6 * 6 * 44}{2598960}\)
\(P = \frac{123552}{2598960}\)
\(P = \frac{198}{4165}\)
D. What is the probability that we are dealt three of a kind? (Others are different)
13C1 ways to select the value of the three of a kind 4C3 ways to select the suits of the three of a kind 12C2 ways to select the values of the fourth and fifth 4C1 ways to select the suit of the fourth card 4C1 ways to select the suit of the fifth card
\(P = \frac{13C1 * 4C3 * 12C2 * 4C1 * 4C1}{52C5}\)
\(P = \frac{13 * 4 * 66 * 4 * 4}{2598960}\)
\(P = \frac{54912}{2598960}\)
\(P = \frac{88}{4165}\)
E. What is the probability that we are dealt one pair? (Others are different)
13C1 ways to select the value of the pair 4C2 ways to select the suits of the pair 12C3 ways to select the value of the single cards 4C1 * 4C1 * 4C1 ways to select the suits of those card
\(P = \frac{13C1 * 4C2 * 12C3 * 4C1 * 4C1 * 4C1}{52C5}\)
\(P = \frac{13 * 6 * 220 * 4 * 4 * 4}{2598960}\)
\(P = \frac{1098240}{2598960}\)
\(P = \frac{352}{833}\)
F. What is the probability that we are dealt four different ranks?
I'm not sure I understand this question... isn't this essentially asking what's the probability of obtaining a pair?
G. What is the probability that we are dealt hands with every suit?
4C1 ways to select the suit which we are dealt two of 13C2 ways to select the cards of the suit we are dealt two of 13C1 * 13C1 * 13C1 ways to select the cards of the suit we are dealt one of
\(P = \frac{4C1 * 13C2 * 13C1 * 13C1 * 13C1}{52C5}\)
\(P = \frac{4 * 78 * 13 * 13 * 13}{2598960}\)
\(P = \frac{685464}{2598960}\)
\(P = \frac{2197}{8330}\)
Alright that took a bit of time! From my knowledge of poker the probabilities seem in the ball park of what they should be. If anyone wants to double check, please do! Good post, kudos.




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07 Oct 2009, 16:51
Great job AKProdigy87!!! I've just looked through the answers and seems that all are correct. +1 to you. As for F: It should state: "What is the probability that we are dealt all five of different rank?" Typo. Edited. But no doubt that you will answer this one as well.
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08 Oct 2009, 16:35
Bunuel wrote: Great job AKProdigy87!!!
I've just looked through the answers and seems that all are correct. +1 to you.
As for F: It should state: "What is the probability that we are dealt all five of different rank?" Typo. Edited. But no doubt that you will answer this one as well. Yeah I figured as much lol. It's just (13C5 * 4^5) / 52C5 I believe. Don't have a calculator handy to get a nice fractional probability lol. Great questions again!



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08 Oct 2009, 16:49
AKProdigy87 wrote: Bunuel wrote: Great job AKProdigy87!!!
I've just looked through the answers and seems that all are correct. +1 to you.
As for F: It should state: "What is the probability that we are dealt all five of different rank?" Typo. Edited. But no doubt that you will answer this one as well. Yeah I figured as much lol. It's just (13C5 * 4^5) / 52C5 I believe. Don't have a calculator handy to get a nice fractional probability lol. Great questions again! No need for calculator: it's right.
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Re: Let's play poker.
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07 Apr 2010, 08:05
I had a go with permutation for the sake of experimentation:
Some of these can also be done slightly quicker with permutations:
B. What is the probability that we are dealt full house? (A full house is a hand with both a threeofakind and a twoofakind.)
13P2 ways to select the value of the 3 of a kind and 2 of a kind (2,3,...,K,A) [Order is important] 4C3 ways to select the suits of the 3 of a kind 4C2 ways to select the suits of the pair
(13P2*4C3*4C2) / 52C5
However for hands where the order of the suit is not always important permutations do not make the process any quicker:
E. What is the probability that we are dealt one pair? (Others are different)
13P4 ways to select the four values (including one for the pair) 3! discount the order of the cards which are not in the pair 4C2 ways to select the suit of the pair 4C1 * 4C1 * 4C1 ways to select the suits of the remaining cards
( (13P4*4C2*4C1*4C1*4C1) / 3! ) / 52C5
And, if anybody feels like continuing the theme:
G. What is the probability that we are dealt a flush? (all the same suit)
H. What is the probability that we are dealt a straight? (5 cards in sequence, not all the same suit)
I. What is the probability that we are dealt a straight flush? (5 cards in sequence, all the same suit)
J. What is the probability that we are dealt a royal flush? (A,K,Q,J,10, all the same suit)



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23 Dec 2010, 14:52
AKProdigy87 wrote: C. What is the probability that we are dealt two pairs? (Fifth is different)
D. What is the probability that we are dealt three of a kind? (Others are different)
13C1 ways to select the value of the three of a kind 4C3 ways to select the suits of the three of a kind 12C2 ways to select the values of the fourth and fifth 4C1 ways to select the suit of the fourth card 4C1 ways to select the suit of the fifth card
\(P = \frac{13C1 * 4C3 * 12C2 * 4C1 * 4C1}{52C5}\)
\(P = \frac{13 * 4 * 66 * 4 * 4}{2598960}\)
\(P = \frac{54912}{2598960}\)
\(P = \frac{88}{4165}\)
What would be a way to solve it without using the combinations formula ? Thank you in advance!



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25 Dec 2010, 07:24
G. What is the probability that we are dealt a flush? (all the same suit) 4C1 for 4 different suits 13C5 for the combination of 5 different ranks out of 13 possible
\(P=\frac{4C1*13C5}{52C5}\approx \frac{1}{505}\) this is including the 40 different cases when we can have a straight flush
H. What is the probability that we are dealt a straight? (5 cards in sequence, not all the same suit) We can have 10 different cases for straights. Suits are of no importance, therefore each of the 5 cards could have 4 different suits: 4C1
\(P=\frac{10*4C1*4C1*4C1*4C1*4C1}{52C5}\approx \frac{1}{254}\) this is again including the 40 different cases when we can have a straight flush
I. What is the probability that we are dealt a straight flush? (5 cards in sequence, all the same suit) There are 10 different cases when we can have a straight and 4 different colours, therefore there are 4*10 cases for a straight flush.
\(P=\frac{40}{52C5}\approx \frac{1}{64974}\)
J. What is the probability that we are dealt a royal flush? (A,K,Q,J,10, all the same suit) We can have a royal flush in 4 different suits, therefore there are 4 different cases:
\(P=\frac{4}{52C5}\approx \frac{1}{649740}\)



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07 Jun 2012, 01:39
Bunuel wrote: Playing poker helps to understand the combination and viseversa.
So, find below problems to master in either of them:
In the poker game FiveCard Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K).
A. What is the probability that we are dealt a fourofakind?
B. What is the probability that we are dealt full house? (A full house is a hand with both a threeofakind and a twoofakind.)
C. What is the probability that we are dealt two pairs? (Fifth is different)
D. What is the probability that we are dealt three of a kind? (Others are different)
E. What is the probability that we are dealt one pair? (Others are different)
F. What is the probability that we are dealt all five of different ranks?
G. What is the probability that we are dealt hands with every suit? Bunuel if you could please show this sum by individual probability way it would be very helpful , combinotrics seems a little complex . I tried the first one : 1. p( four of a kind ): I guess this means exactly four of a kind , i e fifth one cannot be the same kind . so (13/52)*(12/51)*(11/50)*(10/49) * 4* (39/48) there are 13 ways to choose the same kind of suit ( spades, hearts , clubs , diamonds) and there are four different kinds . after we choose the four there are 48 cards left . But the fifth card cannot be of the same suit as the first 4 ( if the first 4 is heart then the fifth one cannot be a heart) so from 48 we have to remove the remaining hearts i .e 13  4 = 9 hence 48  9 = 39 so fifth card probability is 39/48 , I would like to stop here and request you proceed . so do we have to multiply (39/48) with 3 as there are 3 suits remaining : if anyone could show all these poker problems by individual probability method it would be very helpful . You could be eligible for a lot of Kidos!!



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28 Jun 2012, 22:39
Responding to a pm: Actually Prakash, fourofakind is generally interpreted as four cards with the same value i.e. all four are either 2s or 3s etc. In case we were talking about suits (spades, hearts etc), we would have said, four of the same suit. In poker, four of the same suit and the fifth different is nothing so it is kind of understood that four of a kind means four cards of the same value. So first step is to select a value (A/2/3/4/5...) out of the 13 different values in 13 ways. The moment you select a value, you have your 4 cards because every value has only 4 cards. All you have to do now is select the fifth card from the leftover 48 cards in 48 ways. Therefore, probability = 13*48/52C5
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03 Jul 2012, 05:01
Bunuel wrote: Playing poker helps to understand the combination and viseversa.
So, find below problems to master in either of them:
In the poker game FiveCard Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K).
A. What is the probability that we are dealt a fourofakind?
B. What is the probability that we are dealt full house? (A full house is a hand with both a threeofakind and a twoofakind.)
C. What is the probability that we are dealt two pairs? (Fifth is different)
D. What is the probability that we are dealt three of a kind? (Others are different)
E. What is the probability that we are dealt one pair? (Others are different)
F. What is the probability that we are dealt all five of different ranks?
G. What is the probability that we are dealt hands with every suit? F. What is the probability that we are dealt all five of different ranks? This is how i went about this one: No. Of ways first card can be selected = 13C1 X 4C1 No. Of ways second card can be selected = 12C1 X 4C1 No. Of ways third card can be selected = 11C1 X 4C1 No. Of ways fourth card can be selected = 10C1 X 4C1 No. Of ways fifth card can be selected = 9C1 X 4C1 So total No. Of ways = 13C1 X 4C1 X 12C1 X 4C1 X 11C1 X 4C1 X 10C1 X 4C1 X 9C1 X 4C1 Please explain why and how this is wrong???
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17 Jul 2012, 19:08
D. Probability of 3 of a kind is previously solved as
(13C1*4C3*12C2*4C1*4C1)/52C5.
Why is this approach wrong:
13C1 ways to select the value of the three of a kind 4C3 ways to select the suits of the three of a kind 48C2 ways to select 2 out of the remaining 48 cards
P = (13C1*4C3*48C2)/52C5
Thanks.



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17 Jul 2012, 20:47
manwiththeharmonica wrote: D. Probability of 3 of a kind is previously solved as
(13C1*4C3*12C2*4C1*4C1)/52C5.
Why is this approach wrong:
13C1 ways to select the value of the three of a kind 4C3 ways to select the suits of the three of a kind 48C2 ways to select 2 out of the remaining 48 cards
P = (13C1*4C3*48C2)/52C5
Thanks. Because 'three of a kind' implies that the other two cards must be different i.e. there shouldn't be another pair (otherwise it becomes a full house  poker terminology). In the question, it is mentioned that others are different. So you have to ensure that the other two cards are not the same. So you select 2 different values using 12C2 and then a suit of each value using 4C1. When you do 48C2, it includes the cases where both the cards have the same value.
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07 Nov 2012, 07:03
I also have another question. Why do we have to select two values at once when calculating the probability of getting a pair, i.e. why do I use 13C2 and can't calculate the prob. of getting two pairs with 13C1*4C2*12C1*4C2*44C1/52C5? What am I calculating with that last approach, do I take into account the order of the drawing with 13C1*4C2*12C1*4C2*44C1/52C5?



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07 Nov 2012, 20:33
BN1989 wrote: I also have another question. Why do we have to select two values at once when calculating the probability of getting a pair, i.e. why do I use 13C2 and can't calculate the prob. of getting two pairs with 13C1*4C2*12C1*4C2*44C1/52C5? What am I calculating with that last approach, do I take into account the order of the drawing with 13C1*4C2*12C1*4C2*44C1/52C5? Exactly! You are selecting two pairs in order when you do 13C1 and 12C1. You select the first pair with 13C1 and the second with 12C1. Say you select Kings as the first pair and Queens as the second pair. In another draw, you will select Queens as the first pair and Kings as the second pair. With 13C2, you count these 2 draws only once and hence you get only one set of 4 cards. With 13C1 and then 12C1, you count these two draws as two different set of 4 cards which is not correct. As expected 13C2 = 13*12/2 13C1*12C1 = 13*12 The difference is the division by 2 in the first case which is required since the ordering of the cards is immaterial. You still have the same hand.
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08 Nov 2012, 17:29
[quote="Bunuel"]Playing poker helps to understand the combination and viseversa.
So, find below problems to master in either of them:
In the poker game FiveCard Draw, each player is dealt a hand consisting of 5 cards from a deck of 52 cards. Each card in the deck has a suit (clubs, hearts, diamonds, or spades) and a value (A, 2,..., 10,J,Q,K).
D. What is the probability that we are dealt three of a kind? (Others are different)
I need help. I can't get where I'm wrong
1) 13*4 is the total number of ways to select 3 cards of the same value from 52 cards 2) after we have chosen 3 cards, we a left with 491 cards to choose from for card #4 and #5 (we should subtruct one because it is of the same value as the 3 chosen). So we have 48 cards to select from for card #5 and #4. 3) as we can choose any card from the 48 left ones, there are 48 ways to choose card #4 4) after we selected card #4, there are 47 cards left. But in order for card #5 to be different in suit and value from card #4, we should subtruct 3 cards of the same value and 12 cards of the same suit as card #4. 47312=32 cards to select from for card #5. 5) the total # of ways to be dealt three of a kind where others are different is 13*4*48*32
6) the probability that we are dealt three of a kind is 13*4*48*32/52C5
I can't understand why this way of solving is not working...



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