I understand this question as:

Given 6 cards, what is probability there is at least one pair.

He have the following cases,

Where "M" stands for match and "X" stands for any. There are 14 cases if you write out the list. In each of these cases the probability is:Code:MMXXXX MXMXXX ...

(52/52) ---> The first card being any card.

(3/51) ---> The next card got to match.

(50/50) ---> The next can be anything.

(49/49) ---> The next can be anything.

(48/48) ---> The next can be anything.

(47/47) ---> The next can be anything.

Multiply then out to get,

3/51 = 1/17

Now multiply this by 12 because that is the number of disjoint cases:

(12/17)