Poker game full house

Question

I'm dealing with an exercise which deals with the poker game. I need to calculate the probability of getting a full house.

Full house is getting 3 cards of the same type and 2 cards of the same type.

I've made a research, but I cannot understand why the combination for getting a full house is

$13 \choose 1$$4 \choose 3$$12 \choose 1$$4 \choose 2$ Can someone explain me in details why we multiply those combinations? I mean, explain why we consider these numbers.

Answer 1

How many ways could you get a full house of kings over queens? You need 3 of the 4 kings and 2 of the 4 queens. That explains the $4 \choose 3$ and the $4 \choose 2$ factors. But of course you could have a lot of other types of full houses. How many options do you have for the three of a kind? $13 \choose 1$. Then of course you only have 12 other denominations left for the pair - that's the $12 \choose 1$.

Answer 2

• First we select which type's 3 cards we want:

  There are 13 types(or ranks) of cards: $A,K,Q,J,10,9,8,7,6,5,4,3,2,1$.We select one of them.

$$\binom{13}1$$

• Select three cards from it:

  There are only four cards of different suites of same type: $K\heartsuit,K\diamondsuit,K\spadesuit,K\clubsuit$, we select any three of them

$$\binom{4}3$$

• Select second type who's 2 cards you want:

  From the remaining $13-1=12$ types we select one type.

$$\binom{12}1$$

• Select two cards out of it:

  Similarly we have $4$ suits, we select two of them of a single type.

$$\binom{4}2$$

• So total ways:

  Product of all ways, multiplication theorem.

$$\binom{13}1*\binom{4}3*\binom{12}1*\binom{4}2$$

Answer 3

There are four suits and 13 cards in each suit. A full house has 3 cards of the same value, and 2 more cards of the same value.

Of the 13 card values choose 1 value for your 3 of a kind: $13 \choose 1$. Now of the 4 suits choose 3 of the cards with this value: $4 \choose 3$.

For the 2 of a kind, there are 12 card values left to choose from (you can't choose the same value from the 3 of a kind, since that would require 5 suits): $12 \choose 1$. Now of the 4 suits choose 2 of the cards with this value: $4 \choose 2$.

Multiply them all together to get the number of combinations.