I have a simple problem involving probability of drawing at least 1 pair of cards in a four card hand. I am not getting the right answer but I dont understand the flaw in my logic. Can anyone explain to me why my approach is wrong?
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?
Probability of 1 or more pair = ((6) * (10 choose 2))/(12 choose 4). i.e. the # of hands with at least 1 pair are 6 (the number of ways to create a pair) * (10 choose 2) (the # of ways to select 2 cards from the remaining 10 cards after the pair). This simplifies to 6/11, however, the correct answer is 17/33.
Any help understanding would be greatly appreciated.