Number of Ways to Draw a Pair in a Poker Deck

Question

I am confused over calculating the number of ways in which I can select a pair out of a deck of 52 cards. This is how I go about solving the problem:

Following the definition of a pair in card games, its like 2 cards having the same ranks for instance two 6's or two Aces etc. Hence I can select a first card out of the 52 cards in $C(52,1) = 52$ ways and now that I have selected this card, I only have 3 other cards to select my 2nd card which will be of a different suit but of the same rank so $C(3,1)$ ways to select this card, hence the total number of ways is $52*3 = 156$, but according to this link - Link looking at solution (4) I am not able to get why is this such a huge number. Can someone explain it succinctly where am I going wrong?

Answer 1

The correct way to proceed is the following: there are $C(52,2)$ possible pairs of cards that can be taken out from the deck. The number of pairs in which the two cards have the same rank is $C(4,2)\times 13$. So the probability of taking two cards from the deck and getting a pair of cards with the same rank is $(C(4,2)\times 13)/C(52,2)$.

Answer 2

First, to clarify the problem, we are drawing two cards and hoping for a pair. The discussion at the link posted in the question is about drawing five cards and hoping for a pair.

There are two counting strategies floating around this page.

One strategy is to first choose any card from the available fifty-two cards. (Perhaps you draw A$\spadesuit$.) Regardless of your draw, there remain in the deck exactly three cards that will complete your first card to a pair. (For example, A$\clubsuit$, A$\heartsuit$, and A$\diamondsuit$.) This strategy counts every pair twice, however. (We wouldn't consider A$\spadesuit$ A$\heartsuit$ as a different hand than A$\heartsuit$ A$\spadesuit$.) We can correct this error by dividing by two, which gives us a final count of $\frac{52 \cdot 3}{2} = 78$.

We can avoid the overcounting and subsequent correction with a different strategy that describes the properties of a pair rather than draws the cards. Every pair is uniquely identified by its rank and two distinct suits of that rank. We first choose one of the thirteen ranks. (Perhaps we choose Ace.) Next, we need two of the four suits of that particular rank (such as $\spadesuit$ and $\heartsuit$) in such a way that order does not matter. This can be done in $\binom{4}{2}$ ways (or $C(4,2)$, if that notation is more familiar). This gives us a total count of $13 \cdot \binom{4}{2} = 78$.