Bill has a standard deck of 52 playing cards made up of 4 : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 08:40

### 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

# Bill has a standard deck of 52 playing cards made up of 4

Author Message
Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 18 [0], given: 0

Bill has a standard deck of 52 playing cards made up of 4 [#permalink]

### Show Tags

14 Nov 2006, 11:54
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Bill has a standard deck of 52 playing cards made up of 4 suits of 13 cards each. He likes to play a game in which he shuffles the deck and counts the number of cards he must turn over until 2 cards of the same value appear. How many cards must he turn over such that the probability that at least 2 cards have the same value is greater than 1/2?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 10

How do you slove this?
Senior Manager
Joined: 08 Jun 2006
Posts: 340
Location: Washington DC
Followers: 1

Kudos [?]: 48 [0], given: 0

### Show Tags

14 Nov 2006, 14:16
P of 1st card = 1
2nd Card - P of drawing same number is 3/51
3rd Card - P of drawing same number is 3*2/50 [bcos 2 cards of different numbers have already been drawn]
nth card - P of drawing same number is n*3/(52-(n-1))
n*3/(52-(n-1)) > 1/2
Manager
Joined: 04 Dec 2005
Posts: 80
Followers: 1

Kudos [?]: 20 [0], given: 0

### Show Tags

06 Jan 2007, 07:47
Can some one please more indepth explanation for this question. I am not getting the approach!
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

06 Jan 2007, 15:27
withme wrote:
Can some one please more indepth explanation for this question. I am not getting the approach!

Well... The approach is to determine card after card the probability to get a similar card.

At start, the probability is 0. It's the first picked card, no card before.

Then, as anindyat did, we calculate the probability to have a similar card to the first one. Such card appears in the 3 other suits available in the deck. P =3/51

Then, one more time, we have 2 cards taken now and we pick a new one. We calculate the probability to have one of those 2 cards in the other 3 suits. P = 3*2/50

With 3 cards, we have the probaility to get a similar card to the 3 taken card is 3*3/49

Finally, we can observe a pattern : 3*n/(52-(n-1)) with n cards taken.

The original question is thus respresented by the rule :
3*n/(52-(n-1)) > 1/2
<=> 6*n > 53 - n
<=> n > 53 /7 > 49/7 = 7

Senior Manager
Joined: 23 Jun 2006
Posts: 387
Followers: 1

Kudos [?]: 322 [0], given: 0

### Show Tags

06 Jan 2007, 18:02
first let's admit this is no gmat question.... it is way too difficult... (sounds more like a poker question....)
second.. till now i think everybody got it wrong...
do you really think it is E? do you think that drawing 10 cards the probability of getting 2 of the same number is about 0.5 or slightly more?
drawing 14 cards you have probability of 1. the probability of drawing 13 cards in a row each with different value is extremely low.... without any calculations i'd rule E as being too large, and probably D as well (but with hesitations)

i always like to give a similar question to my students....
how many people in a party you need so that the probability that there are at least two people with the same birthday is 0.5? think about it for a minute.... give your guess.... here you choose out of 365 possible values..... and you need only....... 29 .... yes this is enough to have the probability of 0.5 that at least two of them share the same birthday....

so out of 13 possible value you need 10? or 7? something must have gone wrong....

anyway - here is my solution. and yes it is a bit complicated.... i didn't find any simpler way of solving. i think there is none, but i might be wrong.

taking n cards, what is the probability of getting 2 with the same value?
this is what the question really about.. finding the formula as a function of n, then finding the smallest n for which the formula gives something more than 0.5

it is easier to solve it answering the opposite question. taking n cards, what is the probability of having n different values? clearly the answer to this question is the complement of the original question. so if we find n such that the probability for n different values is less than 0.5 then the probability for at least 2 same values will be greater than 0.5

here it is....

there are 52Cn possibilities to choose n cards from a deck of 52.
let's count those that have n different values.
first we count the possible set of values.... which is 13Cn.
then for each value we should choose one card out of 4 different options... hence the total number of n different values cards is:
(4^n)*13Cn

and the probability is: p(n) = (4^n)*13Cn/52Cn

plugging n=1 (p=1) and n=2 (p=48/51) should make you feel more confident that we really do have the correct formula...

bit of arithmetic we can simplify the formula:

p(n) = 52/52 * 48/51 * 44/50 * 40/49 ..... taking n fractions.

now just plugging in different n
n=5 is almost there ( p(5) = 0.50708 ... remember we want p<0.5)
n=6 is already there p(6) = 0.34524

Senior Manager
Joined: 23 Jun 2006
Posts: 387
Followers: 1

Kudos [?]: 322 [0], given: 0

### Show Tags

06 Jan 2007, 18:10
well... after all this work.. there is a simpler way to get to the same result.

the crux is however look at the complement !!!! (this trick helps in many probability questions).
in our case... try to find the probability of not repeating the same value.
so card no.1 - no restrictions. p1 = 1
card no.2 - we can't repeat the value of the already chosent card so:
p2=p1*48/51
card no.3 - we can't choose any of the 2 values...
p3 = p2*44/50 = 48/51*44/50
and so on...
each card "restrict" the choices by 4 cards out of a deck with one less card.

once you arrived to this formula - continue as before to find that n=6 (answer is C)
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

07 Jan 2007, 02:01
hobbit, notice that we say 53/7 aprroxi 7.1.... so it's 8 cards on 52... plausible don't u think?

Now, Pokhran II, perhaps would u give the OA please
Senior Manager
Joined: 23 Jun 2006
Posts: 387
Followers: 1

Kudos [?]: 322 [0], given: 0

### Show Tags

07 Jan 2007, 03:45
Fig wrote:
hobbit, notice that we say 53/7 aprroxi 7.1.... so it's 8 cards on 52... plausible don't u think?

Now, Pokhran II, perhaps would u give the OA please

i've already shown that answer is 6.... and we missed 5 by a fraction...

i can also point to were your reasoning went wrong....
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

07 Jan 2007, 07:23
hobbit wrote:
Fig wrote:
hobbit, notice that we say 53/7 aprroxi 7.1.... so it's 8 cards on 52... plausible don't u think?

Now, Pokhran II, perhaps would u give the OA please

i've already shown that answer is 6.... and we missed 5 by a fraction...

i can also point to were your reasoning went wrong....

Hobbit... I believe that u should simply let the people be with their needs and their wills ... Don't u think?

Perhaps, I need to have the OA, or perhaps I need other people that come to speak and to understand.... and so on...

Let the thing be
Senior Manager
Joined: 23 Jun 2006
Posts: 387
Followers: 1

Kudos [?]: 322 [0], given: 0

### Show Tags

07 Jan 2007, 10:12
apologies for the misunderstanding.....
sure .. i'd be happy to se the OA and the OE even more... i'm curious about that as well and about the source of the question... i myself not 100% confident in my way.... probability questions always seem to be counter intuitive to me....

i only commented because from your response it wasn't clear if you saw that my answer was 6... you just commented that 7 or 8 may be plausible...

sorry again for the misunderstanding. i always welcome different opinions as they reflect actual way people think and grasp those questions....
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

07 Jan 2007, 14:07
hobbit wrote:
apologies for the misunderstanding.....
sure .. i'd be happy to se the OA and the OE even more... i'm curious about that as well and about the source of the question... i myself not 100% confident in my way.... probability questions always seem to be counter intuitive to me....

i only commented because from your response it wasn't clear if you saw that my answer was 6... you just commented that 7 or 8 may be plausible...

sorry again for the misunderstanding. i always welcome different opinions as they reflect actual way people think and grasp those questions....

Thanks for your understanding indeed ... I must say I appreciate your reaction here

I also share the view that it's the interactions of needs from people and expressed from their posts that engender the higher level of understanding
Intern
Joined: 21 Nov 2006
Posts: 44
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

07 Jan 2007, 18:07
Off top of my head, without trying problem. If you pick 14 cards prob is 1 or 100%. So 7 cards would be 50%. Doesnt work that way though? Would be too easy.

I think Hobbit has it although on the gmat I would guess. This would be a time killer.

E
Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 18 [0], given: 0

### Show Tags

08 Jan 2007, 14:13
Thanks all for trying!

The question is from Manhattan CAT.

The OE and OA:
The problem asks us to find the number of cards that must be turned over before the probability is greater than 1/2 that a pair shows up.

This is a difficult problem involving events in succession.
The probability of having a pair on the first card is 0, regardless of which card is selected.
For the second card, there are 51 cards left in the deck and 3 of them will form a pair with the card already on the table, so the probability that there is a pair after two cards is:
(3/51).

For the third card, there are 50 cards left in the deck and 6 of them would pair with one of the two cards already on the table (if a pair has not already been formed). As this involves an OR event, we add the probabilities. The probablity that there is pair after three cards is:
(3/51) + (6/50). This sum is less than 1/2.

For the fourth card, there are 49 cards left in the deck, and 9 of them will pair with one of the three cards already on the table (if a pair has not already been formed). The probability that there is a pair after four cards is:
(3/51) + (6/50) + (9/49). This sum is less than 1/2.

For the fifth card, there are 48 cards left in the deck, and 12 of them will pair with one of the four cards already on the table (if a pair has not already been formed), so the probability that there is a pair after five cards is:
(3/51) + (6/50) + (9/49) + (12/48). This sum is greater than 1/2, as can be quickly estimated given that the numerators add up to 30, which is clearly more than half of the average denominator of approximately 50.
When you are dealt five cards at random, the probability is greater than 1/2 that you will have a pair after five cards have been dealt.

Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 18 [0], given: 0

### Show Tags

08 Jan 2007, 14:15
hobbit wrote:

how many people in a party you need so that the probability that there are at least two people with the same birthday is 0.5? think about it for a minute.... give your guess.... here you choose out of 365 possible values..... and you need only....... 29 .... yes this is enough to have the probability of 0.5 that at least two of them share the same birthday....

hobbit, could you explain how 29 be the answer?
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 149 [0], given: 0

### Show Tags

08 Jan 2007, 14:30
Ok .... Thanks for the full OE
Senior Manager
Joined: 23 Jun 2006
Posts: 387
Followers: 1

Kudos [?]: 322 [0], given: 0

### Show Tags

09 Jan 2007, 01:08
Pokhran II wrote:
Thanks all for trying!

The question is from Manhattan CAT.

The OE and OA:
The problem asks us to find the number of cards that must be turned over before the probability is greater than 1/2 that a pair shows up.

This is a difficult problem involving events in succession.
The probability of having a pair on the first card is 0, regardless of which card is selected.
For the second card, there are 51 cards left in the deck and 3 of them will form a pair with the card already on the table, so the probability that there is a pair after two cards is:
(3/51).

For the third card, there are 50 cards left in the deck and 6 of them would pair with one of the two cards already on the table (if a pair has not already been formed). As this involves an OR event, we add the probabilities. The probablity that there is pair after three cards is:
(3/51) + (6/50). This sum is less than 1/2.

For the fourth card, there are 49 cards left in the deck, and 9 of them will pair with one of the three cards already on the table (if a pair has not already been formed). The probability that there is a pair after four cards is:
(3/51) + (6/50) + (9/49). This sum is less than 1/2.

For the fifth card, there are 48 cards left in the deck, and 12 of them will pair with one of the four cards already on the table (if a pair has not already been formed), so the probability that there is a pair after five cards is:
(3/51) + (6/50) + (9/49) + (12/48). This sum is greater than 1/2, as can be quickly estimated given that the numerators add up to 30, which is clearly more than half of the average denominator of approximately 50.
When you are dealt five cards at random, the probability is greater than 1/2 that you will have a pair after five cards have been dealt.

i'd argue that this OE is not correct.... i'll explain on the first wrong statement in my view:
the probability of having a pair of same value in 3 cards is not 3/51+6/50.
true... for 2 cards probability of a pair is 3/51
but for three cards you have:
3/51 +(48/51)*(6/50)

the reason for the 6/50 to be added it must be, at the same time that the first two didn't yield a pair (which is 48/51).

to help visualize this correction suppose there are only 3 possible values A,B,C each with two cards... so we have A1,A2,B1,B2,C1,C2 cards...
what is the probability to have 2 of the same letter in a three card drawing...

according to the reasoning of the OE it would be:
1/5 (for the second card)+2/4 (for the third card) = 7/10

there are 120 possibilities for 3 card drawings (6*5*4)
lets count all triplets that contain a pair
A1,A2 (4 out of 120)
A2,A1 (4 out of 120)
similarly (B1 B2), (B2 B1), (C1 C2) and (C2 C1) add 4 each.
total 24 triplets whose first 2 are pair (no surprise... we already calculated it would be 1/5 of 120)
lets count those triplets whose first and third form a pair
A1,*,A2 (4)
A2,*,A1 (4)
similarly for (B1 * B2) (B2 * B1) (C1 * C2) (C2 * C1)
total of 24.
and last, we count those triplets whose second and third form a pair - and in a similar manner we have 24...

total - 72 out of 120. that represent a probability of 0.6 not 0.7 !!!!!

hence you can see that the reasoning of the OE is wrong.

i still think that the answer to the original question is 6 (C).

for 2 cards it would be p(2)=3/51
for 3 cards p(3) = p(2)+(1-p(2))*6/50 = 3/51+48/51*6/50
for 4 cards it is p(4)=p(3) + (1-p(3))*9/49
and so forth.... a bit more complicated formula...
my claim was (and still is) that my approach provide a simpler view of the "opposite" of the required formula. hence help to solve the question correctly.
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1264
Followers: 29

Kudos [?]: 298 [0], given: 0

### Show Tags

12 Jan 2007, 18:48
Pokhran II wrote:
Bill has a standard deck of 52 playing cards made up of 4 suits of 13 cards each. He likes to play a game in which he shuffles the deck and counts the number of cards he must turn over until 2 cards of the same value appear. How many cards must he turn over such that the probability that at least 2 cards have the same value is greater than 1/2?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 10

How do you slove this?

Probability that second card is different from first is 48/51
Probability that first three cards are different is 48/51*44/50
"...first four cards are different is 48*44*40/51*50*49=4^3*(132)/((50^2-1)*50) approx=64*66*2/50*50*50=1.28*1.32*0.04 approx=
1.7*0.04=0.68

" first five....different approx 0.68*36/48 approx=0.51

Clearly the probability of drawing 6 cards of different numbers is less than 0.5, and thus the probability that at least 2 will be the same is > 0.5.

I agree with hobbit's comment- mgmat overcounted!
Re: PS: Probability   [#permalink] 12 Jan 2007, 18:48
Display posts from previous: Sort by