1. Probability combinations help!

What is the probability of having two aces, two kings, and a queen in a five‐card poker hand?

,Thanks

2. Re: Probability combinations help!

Originally Posted by lucak
What is the probability of having two aces, two kings, and a queen in a five‐card poker hand?
There are $\displaystyle \binom{4}{2}\binom{4}{2}\binom{4}{1}$ ways to have two aces, two kings, and a queen in a five‐card poker hand.

3. Re: Probability combinations help!

Originally Posted by Plato
There are $\displaystyle \binom{4}{2}\binom{4}{2}\binom{4}{1}$ ways to have two aces, two kings, and a queen in a five‐card poker hand.
I understand how you got that that part I get but the answer in the back of the book was (4combination2)(4combination2)(4combination1) that was in the numerator. Then in the denominator it was 52combination5 and the answer was 3 over 216580. I don't understand how they got that answer

4. Re: Probability combinations help!

Originally Posted by lucak
I understand how you got that that part I get but the answer in the back of the book was (4combination2)(4combination2)(4combination1) that was in the numerator. Then in the denominator it was 52combination5 and the answer was 3 over 216580. I don't understand how they got that answer
Do you understand that $\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}~?$

5. Re: Probability combinations help!

Originally Posted by Plato
Do you understand that $\displaystyle \binom{N}{k}=\frac{N!}{k!(N-k)!}~?$
Yes I understand that, but how did they get 3 as the top numerator?

6. Re: Probability combinations help!

Originally Posted by lucak
Yes I understand that, but how did they get 3 as the top numerator?
You do not. Your textbook is wrong. $\displaystyle \dfrac{\dbinom{4}{2}\dbinom{4}{2}\dbinom{4}{1}} {\dbinom{52}{5}}=\frac{10673}{192629862}}$.