1. ## Card probability question

I am having trouble with a exam review question. The question is

How many 5 card poker hands are there? what is the probability of a hand containing 3 aces and 2 queens?

For the first part there are $52 \choose{5}$ possible hands

For the second part I am getting something like this

$13 \choose{1}$ $4 \choose{3}$ $12 \choose{1}$ $4 \choose{2}$ / $52 \choose{5}$
Is this correct? Thanks for any help that anyone can provide.

2. Originally Posted by fishguts
I am having trouble with a exam review question. The question is

How many 5 card poker hands are there? what is the probability of a hand containing 3 aces and 2 queens?

For the first part there are $52 \choose{5}$ possible hands

For the second part I am getting something like this

$13 \choose{1}$ $4 \choose{3}$ $12 \choose{1}$ $4 \choose{2}$ / $52 \choose{5}$
Is this correct? Thanks for any help that anyone can provide.
Hi fishguts! (innovative name by the way).

You seem to be going overboard on the numerator.
What exactly are you thinking of by multiplying your 3 aces by 13C1
and your 2 queens by 12C1 ?

Why do you feel that your numerator is not just 4C3 multiplied by 4C2,
which is "any pair of queens can be matched with any triple of aces"
?

3. Originally Posted by Archie Meade
Hi fishguts! (innovative name by the way).

You seem to be going overboard on the numerator.
What exactly are you thinking of by multiplying your 3 aces by 13C1
and your 2 queens by 12C1 ?

Why do you feel that your numerator is not just 4C3 multiplied by 4C2,
which is "any pair of queens can be matched with any triple of aces"
?
Well I am multiplying it by 13C1 and your 2 queens by 12C1 because for aces we have 13 cards of value to choose from, and after we chose the aces we then multiply the queens by 12C1 because thats how many cards of value are left once aces are chosen

4. Originally Posted by fishguts
Well I am multiplying it by 13C1 and your 2 queens by 12C1 because for aces we have 13 cards of value to choose from, and after we chose the aces we then multiply the queens by 12C1 because thats how many cards of value are left once aces are chosen
The numerator is just $\dbinom{4}{3}\dbinom{4}{2}$.
How many ways can one choose three kings?
How many ways can one choose two queens?

5. so If the question was worded what is the probability of choosing 3 cards of the same value and 2 cards of the same value would my first approach apply to that problem?

6. Originally Posted by fishguts
so If the question were worded what is the probability of choosing 3 cards of the same value and 2 cards of a dfferent value (both the same value) would my first approach apply to that problem?
See my corrections. Your method is correct

7. Originally Posted by Plato
See my corrections. Your method is correct
Thanks! Also I should put on my contacts before writing back to your replies sorry for the errors.