# Thread: Poker Probability (3 of a kind)

1. ## Poker Probability (3 of a kind)

I'm having trouble trying to figure some of this problem out. I'll provide what I know under the question below.

What is the probability that a five card poker hand will contain three of a kind? (Note: This should not include a full house. That is three of one kind and two of another)

|S| = C(52,5) = 2,598,960
|T| = C(13,1) x C(4,3) x C(48,1) x C(44,1) = 109,824

I know that i'm having trouble with |T|. I've heard that all I have to do is multiply it by 1/2 but I do not know why. Someone please explain throughly. Thanks!

Final Answer will be |S|/|T| = 88/4165

2. Hello dhardin07

Welcome to Math Help Forum!
Originally Posted by dhardin07
I'm having trouble trying to figure some of this problem out. I'll provide what I know under the question below.

What is the probability that a five card poker hand will contain three of a kind? (Note: This should not include a full house. That is three of one kind and two of another)

|S| = C(52,5) = 2,598,960
|T| = C(13,1) x C(4,3) x C(48,1) x C(44,1) = 109,824

I know that i'm having trouble with |T|. I've heard that all I have to do is multiply it by 1/2 but I do not know why. Someone please explain throughly. Thanks!

Final Answer will be |S|/|T| = 88/4165
What you have heard is correct: you do need to divide by $2!$. The reason is that in your working so far, you have selected the final two cards in order. Therefore every possible pair will occur $2!$ times.

If you divide your expression for $|T|$ by $2$, then:
$\frac{|T|}{2}= 54912$

$\Rightarrow \frac{|T|}{|S|}=\frac{88}{4165}$
An alternative method for selecting the final two cards is initially to ignore the restriction: select $2$ from $48$; and then subtract from this the number of ways of selecting two of a kind (so that's selecting $1$ from $12$ and then $2$ from $4$). This gives:
$\binom{48}{2}-\left(\binom{12}{1}\times\binom{4}{2}\right)=1056= \frac{\binom{48}{1}\times\binom{44}{1}}{2!}$