combinations/probability question please help!!

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

this is my solution but it is wrong:

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

I hope that your explaination you will elaborate:

1. why my method is wrong

2. the actual solution.

Thank you very much for even considering my post!!!:)

Adhil

Re: combinations/probability question please help!!

P(at least 1) = 1 - P(None). This is probably more useful here.

So there are possible ways to select 4 distinct cards. Then there are different suits for those cards. Finally there are different cards.

So you have

Re: combinations/probability question please help!!