Problem Solving

Tracy 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 a pair of cards that have the same value?

a)1/33
b)16/33
c)17/33
d)32/33
e)1

OA = B

Solution :

Let's say that there are 6 CARDS A B C D E F; TWO SUITS - HEARTS AND SPADES

HEARTS: A B C D E F
SPADES: A B C D E F

Method 1 :

The first card could be chosen in 12C1/12C1 ways.
Second card - 1C1/11C1 (Because, there are only two same valued cards)
Third Card - 10C1/10C1
Fourth card - 8C1/9C1

Now, combinations of AABC = 4!/ (2!) = 12

Therefore, probability =

12x1x10x8
=---------- X 12 = 32/33 which is incorrect.
12x11x10x9

Method 2:

I tried another method:

We can choose one number from 1 to 6 in 6C1 ways. If we have to choose another number for pair, combinations = 6C1X 1C1 (Because if we choose, say, "2", then there is only one more number left to choose. Therefore, 1C1)

we can then choose 1 number out of the remaining 5 numbers in 5C1 X 2 (there are two sets of 5 such numbers)
Similarly, the last number can be chosen in 4C1 X 2 ways.

Therefore, probability =
6c1 x 1c1 x (5c1 x 2 ) x (4c1 x 2)
= ------------------------------------
12c4

= 32/33 which is again incorrect.

Can someone please help me? I got the same answer using two different methods, both of which are wrong


Thanks