Bill has a small deck of 12 playing cards

[helloriteshranjan]

plz don't forget to mention the source.
I saw this question somewhere and solved it correctly. but i dont remember where did i saw it.. this time I went wrong way....ahhh...

plz tell the source.
thanks

[Ben Ku]

This problem is titled "A Beautiful Pair" and is in our CAT exam.

[sunny.jain]

Another method of solving this is:-

Total combinations of picking 4 cards:- 12C4
Total combinations of picking 4 cards with 2 cards with same numbers is:-
6*10C2 - 6C4

You subtract 6C4 to remove the double counting.

So, (6*10C2 - 6C4 )/ 12C4 = 17/33

This is an interesting and clever method. I had to puzzle over it for a while to figure out where you got the 6C4, and others may do the same.

6C4 is the same as 6C2, both equal 6!/4!2! = 6*5/2 = 15.

So I see 6*10C2 as (# of options for which pair is made)*(ways to select the other 2 cards from the remaining 10 cards).

But you want to avoid double counting when the "other two" are a pair themselves. For example, so far, we have counted [1,1,2,2] and [2,2,1,1] as two different selections. So how many sets of 4 cards could be made of 2 matched pairs? 6 pairs are possible, and we would include 2 of them: 6C2 = 6!/2!4!

I still dont understand, Why you are subtracting this double counting, I think we don't need.

Overall we have 6 pairs or 12 cards.
Atleast one pair = { one pair + 2 different card } and { two pairs }

so when we write this:
6C1 * 10C2 ==> this include both the condition.
Then why we are subtracting 6C2 or 6C4 from it.

Please help.

[esledge]

You may want to read this other thread about the same question.

a-beautiful-pair-t6176.html

Another explanation of this "total combos - certain combos" is discussed there. Check it out and let us know.

[mehdi80]

Just encountered the same problem in a preparation course:
I think the key to the problem is that the cards are shuffled so there's no categorization applied, therefore:
Total # of taking 4 cards out of 12 cards (and not two 6 cards) is
4C12.
Total # of 4 chosen cards with at least a pair:
the cards are "
1 2 3 4 5 6
1 2 3 4 5 6 "
so there are 6 ways of choosing the first pair, for the second pair (since there are two removed already) there is 2C10 for the next two: so 1C6*2C10
to Eliminate the double we have 2C6 ways of choosing exactly Two pairs. so
P = (6*2C10-2C6)/(4C12)= 17/33

cheers
Mehdi

[shellysgomez]

The question says "atleast one"

here 6*10C2 - 6C2 = "exactly one pair"

Should we not compute

"exactly one pair" + " two pairs" ?

[mschwrtz]

Overall we have 6 pairs or 12 cards.
Atleast one pair = { one pair + 2 different card } and { two pairs }

so when we write this:
6C1 * 10C2 ==> this include both the condition.
Then why we are subtracting 6C2 or 6C4 from it.

Because 6C1*10C2 counts TWICE every set of two pairs. See Emily's account higher in this thread.

But this whole approach is WAY too complicated. It's so much easier to calculate the odds that you will NOT get a pair, and to subtract that probability from 1:

probability that second card does not match first=10/11
probability that third does not match either first or second (assumes that those are different)=8/10
probability that fourth does not match first, second, or third (assumes that those are different)=6/9

product of those probabilities=16/33
1-16/33=17/33

[socrates]

The method shown in the official explanation is the easiest way (probability 1-X). You can use prob / comb here but you really don't want to given the other method. Did you read that explanation? Do you have any questions about it?

Hi Stacey
I am bringing this up after really long but I have a question regarding the official explanation. Can you confirm if the method i followed is alright?

Ans 1 - Probability that no pair
1 - p

For p,
Using fundamental principle of counting for denominator
1st card - 12
II card - 11
III card - 10
IV card 9
Possibilities = (12*11*10*9)/(4*3*2*1)

Numerator -
1st card : any card, hence can be drawn in 12 ways
2nd card : out of the remaining 11 cards, we remove the chances of 1 card having the same value, hence 10 ways
Similarly 3rd card: 8 ways
4th card: 6 ways

=> numerator = 12*10*8*6

p= (12*10*8*6)/[(12*11*10*9)/(4*3*2*1)]
= 16/33

Ans: 1-[16/33] = 17/33

[jnelson0612]

sukriteez, well done! I actually like your approach more than some of the others given on this thread.

[contactsrikar]

Hi,

sukriteez's method seems very intuitive, but it doesn't look like it will work.

His calculation of 'p' does not result in 16/33

Can somebody please correct this? I really like his approach but am unable to find the error

[contactsrikar]

I think I got it now.

the numerator should be (12*10*8*6) / 4! because order does not matter

[jnelson0612]

I think I got it now.

the numerator should be (12*10*8*6) / 4! because order does not matter

Great. :-)

[ridzee]

Another method of solving this is:-

Total combinations of picking 4 cards:- 12C4
Total combinations of picking 4 cards with 2 cards with same numbers is:-
6*10C2 - 6C4

You subtract 6C4 to remove the double counting.

So, (6*10C2 - 6C4 )/ 12C4 = 17/33

This is an interesting and clever method. I had to puzzle over it for a while to figure out where you got the 6C4, and others may do the same.

6C4 is the same as 6C2, both equal 6!/4!2! = 6*5/2 = 15.

So I see 6*10C2 as (# of options for which pair is made)*(ways to select the other 2 cards from the remaining 10 cards).

But you want to avoid double counting when the "other two" are a pair themselves. For example, so far, we have counted [1,1,2,2] and [2,2,1,1] as two different selections. So how many sets of 4 cards could be made of 2 matched pairs? 6 pairs are possible, and we would include 2 of them: 6C2 = 6!/2!4!

Hi Emily

I am trying to solve this problem using slot method to find number of combinations for exactly 1 pair

According to me that should be 12*1*10*8/4!

Basically 1st card can be chosen in 12 ways , corresponding pair in 1 way, 3rd card out of the 10 remaining cards, 4th card can be chosen in 8 ways. Then divide it by 4! as suggested by slot method. Two pairs can be 6C2
Total combinations are 12C4 .
I cant understand where I m going wrong in this method. On back solving I know that probability of 1 pair should be 240.But slot method doesnt give that answer :(

[StaceyKoprince]

The slot method that you set up assumes that you get the cards exactly in that order:
1st card then pair of that first card then second card then non-pair of that second card

but what if the pair of the first card actually comes on your 3rd pick? Or your 4th pick?

Let's look at a concrete example. Let's say the first card is the 6 of clubs, next is the 6 of spades, next is the 5 of clubs and finally anything other than the 5 of spades - let's say we pick the 3 of clubs.

We'll call that: 6C, 6S, 5C, 3C

After you pick that first card (the 6 of clubs), there are 3! ways that the remaining cards could occur:
6C, 6S, 5C, 3C
6C, 6S, 3C, 5C
6C, 5C, 6S, 3C
6C, 5C, 3C, 6S
6C, 3C, 6S, 5C
6C, 3C, 5C, 6S

So you actually have to multiply your slot-method result by 3!, or 6. 40 * 6 = 240. :)

[kam.santiago]

I tried to solve it as below:

1-(combinations of four cards with distinctive values/combinations of four random cards)

combinations of 4 cards with different values:
1st card: 1 out of 12, hence: 1C12;
2nd card: 1 out of the remaining 10 (since the 11th card has the same value as the 1st card), hence 1C10;
3rd card: 1 out of the remaining 8, hence 1C8;
4th card: 1 out of the remaining 6, hence 1C6.

Thus, combinations of 4 cards with different values: 1C12*1C10*1C8*1C6.

combinations of four random cards
1C12, 1C11, 1C10 and 1C9 (without regards to the identical value pairs - that's what underlies the "four random cards", right?)

1-(1C12*1C10*1C8*1C6/1C12*1C11*1C10*1C9)=17/33