Thread: 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

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 . The reason is that in your working so far, you have selected the final two cards in order. Therefore every possible pair will occur times.

If you divide your expression for by , then:

An alternative method for selecting the final two cards is initially to ignore the restriction: select from ; and then subtract from this the number of ways of selecting two of a kind (so that's selecting from and then from ). This gives:

Grandad