A Beautiful Pair

[viralmehta28]

This question is from MGMAT CAT Exam #1, Question 13

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, there are 2 cards in the deck that have the same 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?

A 8/33

B 62/165

C 17/33

D 103/165

E 25/33

Please help me identify my error.

My approach: Probability (Bill finds atleast one pair of cards) = 1 - Probability(Bill finds no pair)

Prob (Bill finds no pair) = n(Bill finds no pair) / total no of ways 4 cards can be drawn from 12

Denominator = 12C4 = 12! / (4! *8!) = 12*11*10*9 / (4*3*2*1)

Numerator Calculation: 1st card can be any card, hence can be drawn in 12 ways
2nd card can be any card, except the one that has the same value as the first card, hence 10 ways
Similarly 3rd is 8 ways and 4th is 6 ways
Therefore numerator = 12*10*8*6

Probability(Bill finds no pair) = 12*10*8*6 / 12*11*10*9 / (4*3*2*1) which is more than 1
(Answer is off by 4! in the denomiator)

Could you please tell me if the logic is incorrect. May be we need to multiply it by 4! because the 4 cards can be inter-changed, but I think that might not be the case.
Looking forward to your comments, thank you.

[sonia.chawla]

Hi

The mistake is that you are applying combinations in numerator while you are using fundamental principle of counting in denominator. While it may sound strange, this often gives the error that you got. (I case you needed to divide the numerator by 4!)

following your logic
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!
Numerator - as you said 12*10*8*6/4!

p= (12*10*8*6)/12*11*10*9 (the four factorials cancel out)
= 16/33

Ans 1- 16/33 = 17/33

[JonathanSchneider]

A separate approach:

Your first card could be anything. Since any card works, the probability of your first card working = 1.

Your second card has a 1/11 chance of matching the first card. If it matches, you're good. That would be 1/11.

There is, however, a 10/11 chance that your second card does not work. You then look at the third card. There is a 2/10 chance that this card will match one of the two preceding cards. Thus, the third card will match (10/11)(2/10), or 2/11 of the time.

Your third card, of course, has an 8/10 chance of not matching either of the first two, in which case you would have a 3/9 chance of the last card matching one of the first three. The total probability in this case is: (10/11)(8/10)(3/9) = 8/33.

Now, simply add 1/11 + 2/11 + 8/33 = 17/33.

Not necessarily the fastest method, but something this can be simpler to follow than the combinatorics route.

[haoyang_qu]

A separate approach:

Your first card could be anything. Since any card works, the probability of your first card working = 1.

Hi Can you explain why the probability of your first card working is 1 and not 1/12? Thanks

[JonathanSchneider]

When Bill chooses his first card, there is nothing yet to match to. As a result, all cards are valid. There is no card that will result in a "loss" of the game.

[brij.jhu]

there is another approach to do this.
First, we compute the denominator: number of ways to 'select' 4 cards out of 12
= 12C4 = 495

Now, for the numerator, we will use the 1-x approach.
So, the number of ways to get zero pairs = ?
I simplified this problem as below:
We group the 12 cards into 6 groups, each group having 2 cards with same value(1 to 6).
Now, if we select at most one card from each group there can never be a pair.
So, the number of ways to select 4 out 6 grps = 6C4.
Now each group has 2 cards, so either card will work.
So the total number of ways of selection is = 6C4 * 2 * 2 * 2 * 2
= 6C4 * 2^4 = 240.

so 1-x = 1 - 240/495 = 17/33.

(Note - i am interpreting 'turning over 4 cards' as a simple 'selection' (instead of permutation where order matters) I still get the same answer)

[esledge]

neat method, brij.jhu. Clever use of 1-x and selection of matched pairs almost as if they are glued together.