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Bill has a small deck of 12 playing cards

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Re: Bill has a small deck of 12 playing cards [#permalink] New post 17 Sep 2012, 21:33
Just edited my original post
Meant to take 5C2 not 10C2,

so my doubt

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical


for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 17 Sep 2012, 22:25
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stne wrote:
Just edited my original post
Meant to take 5C2 not 10C2,

so my doubt

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical


for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .


Responding to a pm:

There is a difference between this question and the other one you mentioned.

In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits.


Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead.

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different values from the 5 remaining to form the singles *This is correct *
2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits)

Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6.

Probability = 255/495 = 17/33


Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator.

Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct)
Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct)

Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct)

Probability = 255*4!/495*4! = 17/33
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 17 Sep 2012, 22:43
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stne wrote:
Just edited my original post
Meant to take 5C2 not 10C2,

so my doubt

one pair

6C1*5C2(4!/2!)= 720

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical


for two pairs

6C2 * 4!/2!2! = 90

6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .


one pair
6C1*5C2(4!/2!)= 720


NO
Choose one pair out of 6 - 6C1 - and you don't care for the order in which you choose the two cards
Choose two pairs out of the remaining 5 pairs - 5C2 - and then, from each pair, you have 2 possibilities to choose one of them, therefore 5C2*2*2
Here you stop! Don't care about any order. Period.

for two pairs
6C2 * 4!/2!2! = 90


NO
You choose two pairs - 6C2 - and you stop here. From each pair you take both cards, and don't care about any order.
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 17 Sep 2012, 23:17
VeritasPrepKarishma wrote:
Responding to a pm:

There is a difference between this question and the other one you mentioned.

In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits.


Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead.

6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different values from the 5 remaining to form the singles *This is correct *
2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits)

Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6.

Probability = 255/495 = 17/33



Awesome , this was going to be my next question, I was going to ask that the singles can be selected first and then the pair or we could have different arrangements as the question does not say that the pair should be together ( adjacent)so we could have B3A1A5A2
where A1A2 is the pair of a single suit and B3A5 are the single cards.

Below explanation already clarifies that question. Thank you for hitting the bulls eye. This was exactly what was bothering me.

VeritasPrepKarishma wrote:
Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator.

Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct)
Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct)

Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct)

Probability = 255*4!/495*4! = 17/33


This was enlightening .


EvaJager wrote:
one pair
6C1*5C2(4!/2!)= 720

NO
Choose one pair out of 6 - 6C1 - and you don't care for the order in which you choose the two cards
Choose two pairs out of the remaining 5 pairs - 5C2 - and then, from each pair, you have 2 possibilities to choose one of them, therefore 5C2*2*2
Here you stop! Don't care about any order. Period.

for two pairs
6C2 * 4!/2!2! = 90

NO
You choose two pairs - 6C2 - and you stop here. From each pair you take both cards, and don't care about any order.



Thank you Eva ..

But the question I was going to ask next to you was that, " the pair can be selected first or the singles can be selected first or the pair can be in between the 2 singles etc " , the question does not say that the pair has to be together and singles together or that that the pair has to be selected first and then the singles , so this thought was confusing me.

I was basically getting confused with this sum, digit-codes-combination-103081.html#p802805

Karishma has very clearly answered that very doubt.Just wanted to know " why I was doing , what I was doing , ".
Its now clear, thanks to both of you , hope I am in a better position to understand such questions now.
Karishma and Eva thank you so much for this.
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 28 Nov 2012, 05:02
Hi Bunuel

No doubt that your method is right, but this is my thought process

an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A

the number of patterns that A,B,C,A hand appears = 4C2

Therefore, number of ways A is chosen= 12
when the first A is chosen, it locks in the value for the second A
number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
number of ways C is chosen = 9

P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33

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Re: Bill has a small deck of 12 playing cards [#permalink] New post 05 Mar 2013, 11:34
Hi Bunuel,

My doubt here is that:

6C4 # of ways to choose 4 different cards out of 6 different values;
2 ^4 -as each of 4 cards chosen can be of 2 different suits.

Im not clear why we did 6C4 even though we selected out of 12 cards and also the purpose of doing 2^4.

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Re: Probability [#permalink] New post 29 Apr 2013, 02:41
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Bunuel wrote:
maheshsrini wrote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33


Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}.



Responding to a pm:

How do we obtain 6C4*2^4?

Think of what you have:
6 cards numbered 1 to 6 of 2 different suits. Say you have 1 to 6 of clubs and 1 to 6 of diamonds. A total of 12 cards.

You want to select 4 cards such that there is no pair i.e. no two cards have the same number. This means all 4 cards will have different numbers, say, you get a 1, 2, 4 and 6. The 1 could be of clubs or diamonds. The 2 could be of clubs and diamonds and so on for all 4 cards.

This means you must select 4 numbers out of the 6 numbers in 6C4 ways.
Then for each number, you must select a suit out of the given two suits.
That is how you get 6C4 * 2*2*2*2

This is the total number of ways in which you will have no pair i.e. two cards of same number.
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Re: Probability [#permalink] New post 26 May 2013, 06:09
Bunuel wrote:
maheshsrini wrote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33


Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}.

C^4_6 - # of ways to choose 4 different cards out of 6 different values;
2^4 - as each of 4 cards chosen can be of 2 different suits;
C^4_{12} - total # of ways to choose 4 cards out of 12.

So P=1-\frac{16}{33}=\frac{17}{33}.

Or another way:

We can choose any card for the first one - \frac{12}{12};
Next card can be any card but 1 of the value we'v already chosen - \frac{10}{11} (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left);
Next card can be any card but 2 of the values we'v already chosen - \frac{8}{10} (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left);
Last card can be any card but 3 of the value we'v already chosen - \frac{6}{9};

P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}.

So P=1-\frac{16}{33}=\frac{17}{33} - the same answer as above.

Answer: C.

Hope it helps.


Hi Bunnel,

I used combinatorics, is this way correct.

Prob (at least 1 pair) = fav/tot
fav = 12c1*1c1*10c10*9c8 + 12c1*1c1*10c10*1c1
tot = 12c4

Prob = (12*10*9/8+12*10) / (12*11*10*9/4*3*2)
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 06 Aug 2013, 03:53
I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

Could somebody point out the error in this solution and why?
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Re: Bill has a small deck of 12 playing cards [#permalink] New post 19 Oct 2014, 12:26
hey bunuel!
i thought of this problem like this:
I tried to find the probability of this guy not getting a pair, but i am not getting the required final result for this no pair case.
For the first draw, number of options=12,
for the second one, he can draw from 10 of the remaining(so as to not repeat the face value).
for the third card, he has 8 options,
and similarly 6 for the last card.
So i went on to multiply these outcomes, and got 12*10*8*6.
These were the number of outcomes to not get a pair.
Now i divided it by 12C4, to get the probability.(to get the final answer i will surely subtract it from 1)
But for the answer, the denominator is taken only as 12*11*10*9, i don't get it as to why it is not 12C4 , what am i missing?
thanks in advance :)
Re: Bill has a small deck of 12 playing cards   [#permalink] 19 Oct 2014, 12:26
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