Pair game - One puzzle a day - Puzzle Buddies

Puzzle:Ganesh 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. Ganesh 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 Ganesh finds at least one pair of cards that have the same value?
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution:The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.

First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).

Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. We have 11 cards left in the deck.

Probability of NO pairs so far = 10/11.

Third card: Since we have gotten no pairs so far, we have two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. We have 10 cards left in the deck.
Probability of turning over a third card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 8/10.

Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11.

Fourth card: Now we have three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. We have 9 cards left in the deck.
Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 6/9.

Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 - 16/33 = 17/33.
Winner: Nonone