What is the Probability of Different Poker Hands?

[gimpy]

I think my brain is freezing. I can't see how to get the answers to these questions.

A poker hand is defined as drawing 5 cards at random without replacement from a deck of 52 playing cards. Find the probability of each of the following poker hands.

(a) Full House (one pair and one triple of cards with the same face value).

(b) Three of a kind (three equal face values plus two cards of different values).

(c) One pair (one pair of equal face value plus three cards of a different kind).

So for (a) i got [tex]\frac{C(26,2)C(11,1)}{C(52,2)}[/tex]. But I am so lost. To much studying this week :S.

The answers in the back f the book are
(a) 0.00144
(b) 0.02113
(c) 0.42257

Can anyone help me out?

 

[NateTG]

Is a full house also a three of a kind? Is a full house also a pair?

Consider that there are [tex]\binom{52}{5}[/tex] possible different poker hands, all of which are equaly likely. So if you can figure out the number of different hands that are , for example a four of a kind, then you're home free.

So, let's look at the possibilities for a four of a kind:
There are 13 different card values that the four of a kind can have, and there are 48 other cards that can be in the four of a kind, so the odds of getting a four of a kind are:
[tex]\frac{13*48}{\binom{52}{5}}[/tex]

 

[gimpy]

Thanks for your reply,

Well i understand that. That was actually one of the questions which i got. Say for:
(a) Full House (one pair and one triple of cards with the same face value).

there are 13 different three of a kind that you can get just like 4 of a kind. Then you are left with 49 cards which equals 24 pairs which one left over and only 12 distinct pairs. So does this mean:
[tex]\frac{13*24}{\binom{52}{5}}[/tex]
or take the pair first, therefore 26 pairs and then 12 ways of selecting 3 of a kind.
[tex]\frac{26*12}{\binom{52}{5}}[/tex]

Both of those answers doesn't correspond to the answer 0.00144 in the book.

Im so confused

 

[NateTG]

Well, there's a little bit more to it:

There are [tex]13*\binom{4}{3}[/tex] three of a kinds, and [tex]12*\binom{4}{2}[/tex] remaining pairs. So for a full house it's:
[tex]\frac{13\binom{4}{3}*12\binom{4}{2}}{\binom{52}{5}}[/tex]