Probability of getting a full house

If five cards are selected at random from a standard 52 card deck, what is the probability of getting a full house.

This is what I am thinking. $(52*\binom{4}{3}*\binom{4}{2})/_{52}C_5$

Is that right?

• Please explain how you got your factor of $52$. – bof May 25 '14 at 1:41
• 4 suites of 13 cards each, 4 * 13 = 52. – Jaysun May 25 '14 at 1:46
• You're off by a factor of $3$. Your $52$ in the numerator should be $13\cdot12=156$ because you have $13$ choices for the rank of the trips (aces full, kings full, queens full, etc.) and then $12$ choices for the rank of the pair (over aces, over kings, over queens, etc.) – bof May 25 '14 at 1:47

A full house has three cards of one kind and two of another, so think about it like this: first you choose a type of card (13 choices), then you choose three out of four of those cards, then you choose a second type of card, and finally you choose two of those four cards. Thus you have ${13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}$ possible full house hands. So the probability is then

$${{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}}\over{52\choose 5}}={{(13)(4)(12)(6)}\over2598960}={3744\over2598960}\approx0.00144$$

• Thank you for your detailed answer. It really helps. – Jaysun May 25 '14 at 2:15

Not quite. I would break the counting up into the following steps.

1. How many ways are there to choose the face value for the triple?
2. How many ways are there to choose the suits for the triple? (You already have this.)
3. How many ways are there to choose the face value for the pair?
4. How many ways are there to choose the suits for the pair? (You already have this.)

Unfortunately, it's a little difficult to parse your question without LaTeX formatting, which compounds with the fact that I never use that particular notation for binomial coefficients. For those who don't know, a full house is a hand of $5$ cards such that $3$ of them share the same rank and the remaining $2$ also share the same rank.

You have $13$ choices (2-10, J, K, Q, A) for the rank of the triple, and once that has been chosen, you have $12$ remaining choices for the rank of the pair. Remember you also have to choose $3$ suits of the possible $4$ for the triple, and likewise $2$ of the possible $4$ for the pair. Express this information in terms of binomial coefficients to get the total number of possible full house hands.

Then simply divide by the total number of unrestricted $5$-card hands for the probability.

@Arthur Skirvin's answer is great, but I just want to provide another thinking process for the numerator:

$$(\textrm{choose two kinds})\cdot(\textrm{choose the kind of three cards}\cdot\textrm{choose cards})\cdot(\textrm{choose the kind of cards}\cdot\textrm{choose cards})\\ = {13\choose2}\cdot{2\choose1}{4\choose3}\cdot{1\choose1}{4\choose2}.$$

You could also think about it this way, where I assume the card choices to be order dependent in both the numerator and the denominator.

The total number of possible choices is $$52\times51\times50\times49\times48$$.

To find the number of full house choices, first pick three out of the 5 cards. For the 3 cards you have $$52\times3\times2$$ cases (once you pick the first card, the rest have to have the same number), and for the two cards you have $$48\times3$$ cases (you cannot pick from the previously selected class).

Thus the probability would be $$\frac{C(5,3)\times52\times3\times2\times48\times3}{52\times51\times50\times49\times48}\approx0.00144$$.