The chance of getting a pocket pair in hold'em, or more generally, the chance to get a pair when dealt any two cards, is 5.88%. There are a couple of common ways to arrive at this number.

Permutation Approach

The rank of the first card doesn't matter, it is not relevant which pocket pair is obtained. It is the second card that is of interest. The second card must match the first one. This approach essentially treats card on its own and considers the two cards as a series of dependent events.

Thus the chance to draw the first card correctly is 100%, it can't be wrong. In a standard 52 card deck there are 4 cards of each rank. Since one of those has already been dealt there are only 3 remaining. Additionally, the deck itself now only has 51 cards. To get that second card has a chance of 3 / 51 = 5.8824%.

To calculate the final odds of two dependent events they can simply be multiplied together. 100% x 5.8824% = 5.8824%. This is rather trivial since one event has a 100% chance of occurring.

Combination Approach

Rather than treat the cards on their own, they can instead be considered as a single event of two cards. It is simply a matter of counting how many total sets of two cards are possible, and which ones are the sought pair.

The total number of combinations can be calculated with the binomial coefficient (also know as the choose operator). This is written, in brief form as (NcK). In a deck of cards N = 52 and for the pocket size K = 2. (52c2) = 1326 is the number of ways 2 cards can be selected from a set of 52 such that the order of the cards does not matter.

How many of these are a pair? There are 4 cards of the same rank in a deck. A pair can be formed from these 4 cards by taking any 2 of them. This is again the choose operator, (4c2) = 6. That is only for a single rank however. There are 13 ranks thus 6 x 13 = 78 possible combinations for a pair.

The chance to get a pocket pair is thus the chance to get any of the pair combinations out of the total number of combinations. That is, 78 / 1326 = 5.8824%.

Common Mistakes

Calculating 12 pairs of a rank

There are 4 cards of any given rank. So if one of them has been dealt there are 3 left of that rank. That gives us 4 * 3 = 12 possible pairs. There are of course only 6 pair combinations.

The error here is using a permutation instead of a combination. There are indeed 12 distinct orders in which two cards of the same rank can be dealt. The order however doesn't matter; several of those orders result in the same two cards. For example 5♣ 5♥ has a different order than, but is the exact same pair as 5♥ 5♣.