# probability of a pair in cards

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?

The problem:

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?

My solution:

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.

• You make some interesting implicit assumptions in your calculation. First, it focuses only on the possibility that the pair occurs among the first two within a sequence of four cards: there are other ways a pair can occur, so this underestimates the probability. Second, it ignores the sequence among the second two cards, thereby overestimating the probability. Fortunately, the two mistakes do not cancel, revealing the fact there is a problem! – whuber Feb 18 '13 at 22:17
• Realized I was double counting. Cant simply multiply 6 * (10 choose 2). 2 pair instances will be double counted. e.g. pair of 1s in the 6 and then pair of 2s in the 10 choose 2 versus pair of 2s in the 6 and then pair of 1s in the 10 choose 2. The better way to do it is to separate the cases of exactly 1 pair, exactly 2 pair. ways to create exactly 1 pair = 6 (ways to create pair) * ( (10*8)/2 ) + ways to create exactly 2 pair = (6*5)/2 Total # = 255 Denominator = (12 choose 4) = (12*11*10*9)/(4*3*2) = 11*5*9 = 455. Probability = 255/455 = 51/99 = 17/33. – Evan V Feb 18 '13 at 22:24
• Another way to do this is simply to remove the double counting directly. There are 6*(10 choose 2) = 270 ways to create 1 or 2 pairs where the instances of 2 pairs are double counted. If you subtract the # of ways to make exactly 2 pairs (6*5)/2 then you get 270 - 15 = 255. – Evan V Feb 18 '13 at 22:28

To see 0 pairs there are 12 possibilities for the 1st card, but only 10 for the second card (since the 1st chosen card is no longer possible and the card that matches the 1st would form a pair) and 8 cards for the 3rd and 6 for the 4th, this multiplies to 5760. The total number of possible hands (with order mattering) is $12 \times 11 \times 10 \times 9 = 11880$, so the complimentary number which is the number of (ordered) hands that contain at least 1 pair is $11880 - 5760 = 6120$, divide that by 11880 and it reduces to $17/33$. We could also do this with order not mattering, but that would be more complicated and the extra pieces would all end up cancelling each other.