# Poker Probability: Getting a Single Pair in Five Cards

I'm having some trouble with combinatorics and would like some help resolving this confusion. The problem is as follows.

You are drawing five cards and get the following 3 in your hand: 4h, 4c, Qs. What is the probability that you will draw exactly one pair?

To solve this problem, I reasoned it out like this: There are ${49 \choose 2}$ total ways of drawing your last two cards. However, you can only pick cards that won't pair up your hand; removing the two remaining '4's and the three remaining 'Q's, you are left with 44 cards. After choosing the next card, you must remove the other three of the rank you just chose from the remaining deck, resulting in 40 cards. Therefore, you get ${44 \choose 1}{40 \choose 1}$ ways to draw into a pair. Probability will be ${44 \choose 1}{40 \choose 1}/{49 \choose 2}$.

The given answer says that there are 11 remaining ranks in the deck to choose from, and out of each rank, there are 4 suits. So the number of ways to draw a pair will be $4^2{11 \choose 2}$.

These two answers are not equivalent, and are in fact off by a factor of 2. That is, ${44 \choose 1}{40 \choose 1} = 2\cdot 16{11\choose 2}$. What mistake did I make originally and what are some tips to avoid making such errors? I don't understand why these two answers are not the same.

Thanks!

• In the first method, you treat the order of the two additional cards as important, which means you should divide by $49\cdot 48$ not $\binom{49}{2}$. Then you'll see the two are equal. – Thomas Andrews Feb 13 '17 at 15:45
• The hand 4h, 4c, Qs, Ad, 2d is the same hand as 4h, 4c, Qs, 2d, Ad. You have double counted every possible hand, having counted it once when taking the higher rank in the remaining two cards first and again when taking the higher rank of the remaining two cards second. – JMoravitz Feb 13 '17 at 15:45
• If you were to approach via direct conditional probability arguments, given that you've drawn the first three cards being 4h, 4c, Qs, the probability that you continue to have only one pair at the end of drawing the fourth and fifth cards would be $\frac{44}{49}\cdot \frac{40}{48}$ which also agrees with the given answer. – JMoravitz Feb 13 '17 at 15:48

Your calculation is off by a factor $2$ because your $44 \cdot 40$ counts the ordered ways to draw two cards without pairing while your ${49 \choose 2}$ counts the unordered ways to draw two cards from the remaining $49$
They are computing something different from what you are computing, and they are computing it wrong. It looks like they are trying to compute the chance that the two cards you draw form a pair other than fours or queens. Having drawn one of $44$ cards that is neither a four nor a queen there are three cards that can make a pair, so there are $44 \cdot 3$ ordered ways to draw a pair other than fours or queens. You could also say there are ${4 \choose 2}\cdot 11$ unordered ways to draw a pair other than fours and queens because there are $11$ ranks and you choose two of the four of that rank.