What is the probability that a hand of 13 cards contains no pairs?

Answer 1

Well the short answer is that your chances are pretty slim. The long answer:

The total number of 13-card hands is 52-choose-13, which equals 52! / (13! * 39!). Any hand without any pairs must have exactly one of each value from 2 to ace. All there is to choose is the suit of each card, for 413 possible pairless 13-card hands. The probability of choosing one of these hands is the quotient of these two numbers, or 413 * 13! * 39! / 52!, which equals 4194304/39688347475, approximately 0.00010568 or 1 in 9462. Another explanation:

Say the first card in your hand is an ace. For the second card, you have to avoid the other three aces in the deck. There are 51 cards, and 48 of them are OK, so the probability that the second card doesn't pair the first card is 48/51. Suppose your second card was a king. For the third card, you now have to avoid all remaining aces and kings. There are 50 cards left in the deck and you have to Dodge six of them. So the probability that the third card doesn't pair either of the first two cards is 44/50. Each time you pick a card, there are three more cards you have to avoid, and one fewer card in the deck. So for the 4th card, you have to avoid 9 cards out of 49, giving a probability of 40/49. This continues until the 13th card, when the probability is 4/40. All these things have to happen one after the other, so you have to multiply the probabilities together:

48/51 x 44/50 x 40/49 x 36/48 x 32/47 x 28/46 x 24/45 x 20/44 x 16/43 x 12/42 x 8/41 x 4/40.

This gives the same probability as above (about 1 in 9462).