# Poker hand question (Combination)

• April 7th 2008, 12:09 PM
Frostking
Poker hand question (Combination)
Using a regular deck of 52 cards playing five card hands,how many hands will contain four of a kind?
I think maybe it is 13 times 48 since there are 13 different cards to make a four of a kind from and then 48 cards left any one of which could be the fifth card.

How many hands will contain cards of all four suits?

How many hands will have consecutive numbers Assume that ace is low that is equal to one for this question.

How many hands will have two of one number and two of another number and one of a third number? I have done several of this type of question but these three are stumping me!!! Thanks for taking a look.
• April 7th 2008, 07:21 PM
awkward
Quote:

Originally Posted by Frostking
Using a regular deck of 52 cards playing five card hands,how many hands will contain four of a kind?
I think maybe it is 13 times 48 since there are 13 different cards to make a four of a kind from and then 48 cards left any one of which could be the fifth card.

$13 \binom{4}{4} \binom{48}{1}$

Quote:

How many hands will contain cards of all four suits?
$\binom{4}{3} \binom{13}{1}^3 \binom{13}{2}$

Quote:

How many hands will have consecutive numbers Assume that ace is low that is equal to one for this question.
$9 \cdot 4^5$

Quote:

How many hands will have two of one number and two of another number and one of a third number? I have done several of this type of question but these three are stumping me!!! Thanks for taking a look.
$\binom{13}{2} \binom{4}{2}^2 11 \binom{4}{1}$
• April 7th 2008, 08:26 PM
Soroban
Hello, awkward!

I agree completely . . . Good work!

• April 8th 2008, 11:45 AM
Frostking
Poker question thanks
Akward and Soroban thanks very much for helping!!! It is great to know that people are willing to help when others are in need!!!!