# What are the odds of me dealing two cards in a row and having them be a pair?

Here is the problem. I deal two cards face up on the table. What are the odds they a pair? Simple? I thought so too.

Solution 1: Correct? When considering the first card, we know that doesn't matter. There is zero chance of it being a pair! The second card is very valuable. We know that there are only 3 cards left in that deck that actually match that first card. There are 51 cards in the deck in total. This leaves 0/52 + 3/51.

Solution 2: Correct? Let's consider every single possible combination of cards we can get. (The deck will all be divided to four to make our lives easier.) We can get AA | 11 | 12 | 13 | 14 | 15 | 16 ... (There were intentionally two pairs in the front.) WE know that there are 169 possibilities (52 / 4 choose 2). NOW, we know, we see that there are 13 pairs to this entire deck. (One pair for each 'row'. ) That leaves 13 pairs / 169 cards.

Solution 1: ~5% Solution 2: ~7

Here is another example (to clear up confusion) with a much simpler deck.

We have a deck of sick cards. (1, 2, 3, 1, 2, 3). Now, let's try solution one.

Solution 1: 0/6 + 1/5. (1/5 because there is only one card left to actually match those other first card. There are also only five cards left in the deck because the first one ask use.)

Solution 2: Every combination

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

We see 12 pairs out of 36 possible combinations. This is 12/36.

Solution 1: 20 Solution 2: 33

Someone please give the rights answer and explain why this is wrong>?

• Your Solution 2 is not very clearly explained. I buy your Solution 1. It's not clear, e.g., what you mean by "divided by 4". Feb 15, 2023 at 23:57
• @AdrianKeister The reason we are dividing by for is so that we don't need to list out every single possibility four times over. If we divide EVERYTHING by four, the proportions should stay the same. Feb 16, 2023 at 0:04
• I have added the self-study tag. Please check its wiki for its scopes and requirements. Feb 16, 2023 at 2:23

+1 Doctor Milt's answer. A convenient way to look at these problems is use the binomial coefficient $${n \choose k}$$, read as "n choose k". We choose out of 52 cards 2 of them, so the total number of possible hands is $${52 \choose 2}$$, from those hands we want a pair (2) that is made out of 4 cards, i.e. $${4 \choose 2}$$, and they are 13 ranks we can use either one, i.e. $${13 \choose 1}$$. This brings us our probability to have a pair in the first two cards as $$p = \frac{{4 \choose 2}{13 \choose 1}}{{52 \choose 2}}$$ $$=$$ $$\frac{6 \times 13}{1326}$$ $$=$$ $$0.0588235294...$$ which equates to $$\frac{3}{51}$$. Finally, just to note, the odds of an event are computed as the ratio of the probability $$p$$ over one minus $$p$$; i.e. in this case the odds will be below $$1$$ (here $$0.0526315789...$$). The odds shown in the main post appear inverted.