Problem Solving

Question - 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

My approach :

We have four cards that are drawn. Let's call them A B C D

Now we can only have two pairs at max because four cards are drawn.

Option (i) Two cards = 1 pair : => A and B are same. C and D are different
The first card can be drawn in 12 ways.
The second card can be drawn in only 1 way because we need a pair and I am assuming that A = B
The third and the fourth can be drawn in 10, 9 ways respectively.

Therefore, total number of combinations for 1 pair = 4C2 *12*1*10*9 {4C2 is used because A and B can be permuted}.......(U)

Option (ii) 2 pairs : A and B are same. C and D are same too.... But, A != C
Using the same logic as above A B C D = 12*1*10*1 * 4C2 .............(V)

Total number of combinations = 12*11*10*9 ................(W)

Therefore, probability = (U+V)/W = 20/33

Something is wrong here. Any help is greatly appreciated.....


Thanks
Voodoo