The simplest way to do this is to ask the opposite questions:

1) What is the probability of drawing 13 cards with NO pair?

The first card can be anything. The second card can be any of the 48 cards that do not pair the first card: probability 48/51. The third card can be any of the 44 cards that do not match either of first two cards: probability 44/50. Do you see the pattern? For each card we subtract 4 from the numerator and 1 from the denominator. Multiplying all the fractions together, the denominator will be 51(50)(49)(48)...(39)= 51!/38! and the numerator is 48(44)(40)...(4)= 4^{12}(12)(11)(10)...(1)=4^{12}12!. The probability of getting 13 cards with no pair is .

2) What is the probability of drawing 13 cards with exactly one pair?

First calculate the probability that the first two cards pair and the other 11 do not. The first card can be anything. The second must match that- probability 3/51. The third can be anything that does not match those: probability 48/50. The fourth can be anything other than the first two or that: 44/49 and so on: (3/51)(48/50)(44/49)... (8/39).

The probability of drawing 13 cards with at most one pair is sum of those two and the probability of drawing 13 cards withat leasttwo pair is 1 minus that probability.