# How many ways are there to make a “one pair” in poker

I'm confused at the general method of solving this type of problem. The wikipedia page says that there are:

${13 \choose 1} {4 \choose 2} {12 \choose 3} {4 \choose 1}^{3}$ ways to select a pair when 5 cards are dealt. Can someone outline what each calculation means? For example, is $13 \choose 1$ the process of selecting 1 card in the beginning?

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Each card has a rank (e.g.: king, jack, $2$, $6$) and a suit (e.g.: heart, club).

The $\binom{13}1$ is the number of ways to pick the rank of the single pair, and the $\binom42$ is the number of ways to choose the suits of the cards in the pair. To make sure that it's the only pair, then we have to choose other ranks for each of the three other cards. There are only $12$ ranks left to choose from, so $\binom{12}3$ is the number of ways to choose our other three ranks. Each of the other three ranks will have an associated suit, and $\binom41$ is the number of ways to choose the suit of one of the other three ranks. Thus, the number of ways to choose the suits of the other three ranks is $\binom41^3$.

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$13\choose 1$ is the number of ways of choosing the denomination of the pair, whether it is a pair of kings, or a pair of threes, or whatever. Then there are $4\choose 2$ ways to choose suits for the two cards of the pair.

Then there are $12\choose 3$ ways to choose the three different denominations of the remaining three cards from the twelve that are different from the denomination of the pair. Each of those other three cards can be of ${4\choose 1}=4$ different suits, for ${4\choose 1}^3$ choices of three suits in all.

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Why does $13 \choose 1$ represent both cards, though? Why wouldn't we do ${13 \choose 1}^{2}$, say? –  user1038665 Oct 29 '12 at 6:39
Becausr it's a pair. After you choose the denomination for the first card, you don't get to choose the denomination for the other half of the pair, because it has to be the same, or you won't have a pair. –  MJD Oct 29 '12 at 12:37

# Now, the equation...

### First,

out of 13 different values, let's choose one, Ace. (Remember, 1 in 13 = 13)

### Second,

out of 4 different Aces, pick 2. (2 in 4 is 6.)

Now, we know that having any PAIR is 13*6 = 78 different Pairs. Thing is, there could be any other three cards with this pair...

### Third,

you want your next three card to be: one of 4 suits and they CAN be suited, so 1 in 4 suited POWER 3. This renders the possible suit outcomes for the three remaining cards. one in four is 4... 4 power 3 is 64. there are 64 outcomes of suits possible write them down { HHH, HHC, HHS, HHD, ... DDD}.

### Fourth,

but not least, is to decide what can be the value of the three other cards. the last thing you need to consider is that you DO NOT want it to be an Ace NOR ALIKE TOGETHER... you would end up with two pairs, a full or three or four of a kind.

So to solve this it's simple, you need 3 different values out of the 12 remaining (Ace is not possible).3 in 12
12!/(12-3)!3!=
=12*11*10*9*8*7*6*5*4*3*2*1 / 9*8*7*6*5*4*3*2*1 3*2*1
=12*11*10................... / ................. 3*2*1
=1320 / 6
=220

## FINALLY

We solved 13 * 6 * 64 * 220 and remember, you can place these numbers in any order... so 13*6*220*64 is just as good. And what's 64's cubic root ? right. It's 4...

### answer: 1098240 hands of 2598960 are "single pair" or approx. 42.2569 percent

Hope that helps. Math, not even once! ;)

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Its better to use latex. –  CODE Feb 13 '14 at 9:42