# poker hands

• Nov 30th 2011, 09:24 AM
ehpoc
poker hands
I am doing the following question.

"5. How many poker hands are there that contain at least one of 10@, QB, AC, or 6D ?

For my solution I did

A=52*(51C4)
B=52*51*(50C3)
C=52*51*50*(49C2)
D=52*51*50*49*(48C1)

then

A+B+C+D=E

Does this seem to make sense?
• Nov 30th 2011, 06:59 PM
jar
Re: poker hands
think of what the question is really asking. "at least one of or "

and what does A + B + C + D mean.

good luck,
• Nov 30th 2011, 07:56 PM
ehpoc
Re: poker hands
The question asks at least 1.

So that means how many ways are can you get one of them or two of them or three or all 4 of them.

A=the amount of ways to get 1
B=the amount of ways to get 2
C=the amount of ways to get 3
D=the amount of ways to get 4

edit: I did mess up though. It should be

A=52*(51C4)
B=(13C2)(4)(4)(50C3)
C=(13C3)(4)(4)(4)(49C2)
D=(13C4)(4)(4)(4)(4)(48C1)
• Dec 1st 2011, 03:22 PM
ehpoc
Re: poker hands
could anyone confirm if they agree with this solution?
• Dec 1st 2011, 03:37 PM
Plato
Re: poker hands
Quote:

Originally Posted by ehpoc
could anyone confirm if they agree with this solution?

Well, as usual with most of your postings you steadfastly refused to tell anyone what the totally non-standard notations means.
For example, we have no idea what "10@, QB, AC, or 6D" means.
If you want help then explain what that weird string means.
• Dec 1st 2011, 03:44 PM
ehpoc
Re: poker hands
10 of some specific suit, queen of some specific suit, Ace of some specific suit, 6 of some specific suit
• Dec 1st 2011, 05:57 PM
jar
Re: poker hands
Quote:

Originally Posted by ehpoc
A=the amount of ways to get 1
B=the amount of ways to get 2
C=the amount of ways to get 3
D=the amount of ways to get 4

Agree with the above thinking.

Quote:

Originally Posted by ehpoc

A=52*(51C4)
B=(13C2)(4)(4)(50C3)
C=(13C3)(4)(4)(4)(49C2)
D=(13C4)(4)(4)(4)(4)(48C1)

Please look into what does it really give you from each of the above statements.

As Dr. Kocay said sometime you can pull it out from the hat.(Headbang) Personally, I think mathematics should not be pulling out from a hat.
• Dec 1st 2011, 06:09 PM
Soroban
Re: poker hands
Hello, ehpoc!

I hope I understand the problem . . .

Quote:

$\displaystyle \text{How many poker hands are there that contain at least one of }10\heartsuit,\:Q\spadesuit,\:A\clubsuit\text{ or }6\diamondsuit\,?$

There are:.$\displaystyle {52\choose5} \:=\<2,\!598,\!960$ possible poker hands.

How many hands contain none of the four cards?

There are:.$\displaystyle {48\choose5} \:=\:1,\!712,\!304$ such hands.

Answer: .$\displaystyle 2,\!598,\!960 - 1,\!712,\!304 \:=\:886,\!656$