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?
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$\begingroup$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
$${{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}}\over{52\choose 5}}={{(13)(4)(12)(6)}\over2598960}={3744\over2598960}\approx0.00144$$
$\endgroup$ 1 $\begingroup$@Arthur Skirvin's answer is great, but I just want to provide another thinking process for the numerator:
$$(\textrm{choose two kinds})\cdot(\textrm{choose the kind of three cards}\cdot\textrm{choose cards})\cdot(\textrm{choose the kind of cards}\cdot\textrm{choose cards})\\ = {13\choose2}\cdot{2\choose1}{4\choose3}\cdot{1\choose1}{4\choose2}.$$
$\endgroup$ $\begingroup$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$\frac{C(5,3)\times52\times3\times2\times48\times3}{52\times51\times50\times49\times48}\approx0.00144$.
$\endgroup$ $\begingroup$Not quite. I would break the counting up into the following steps.
- How many ways are there to choose the face value for the triple?
- How many ways are there to choose the suits for the triple? (You already have this.)
- How many ways are there to choose the face value for the pair?
- 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.
$\endgroup$ $\begingroup$The number of ways of picking 5 cards in a pack of 52 cards is $ {52 \choose 5} = {^{52}C_{5}}$
To achieve full house, one needs to pick one suit among the four suits in $ {^{4}C_{1}} $ ways, then choose a different suit in $ {^{3}C_{1}} $ ways.
There are thirteen cards each of suit, and one needs to pick 3 cards in that suit in $ {^{13}C_{3}} $ and then pick 2 cards in the second suit in $ { ^{13} C_{2}} $ ways.
Since the job can be done in $ {^{4} C_{1}} {^{13} C_{3}} $ ways in the first suit and then $ {^{3} C_{1}} {^{13} C_{2}} $ ways in the second suit, then by product rule there are $ {^{4} C_{1}} {^{13} C_{3}} {^{3} C_{1}} {^{13} C_{2}}$ ways to get a full house.
The probability of getting a full house is $$\frac{{^{4} C_{1}} {^{13} C_{3}} {^{3} C_{1}} {^{13} C_{2}}}{{^{52}C_{5}}}=0.1030012$$
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