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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$\begingroup$ Please explain how you got your factor of $52$. $\endgroup$ – bof May 25 '14 at 1:41
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$\begingroup$ 4 suites of 13 cards each, 4 * 13 = 52. $\endgroup$ – Jaysun May 25 '14 at 1:46
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$\begingroup$ 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.) $\endgroup$ – bof May 25 '14 at 1:47
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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
$${{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}}\over{52\choose 5}}={{(13)(4)(12)(6)}\over2598960}={3744\over2598960}\approx0.00144$$
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$\begingroup$ Thank you for your detailed answer. It really helps. $\endgroup$ – Jaysun May 25 '14 at 2:15
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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.)
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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.
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@Arthur Skirvin's answer is great, but I just want to provide another thinking process for the numerator:
$$(\textrm{choose two kinds})\cdot(\textrm{choose the kind of three cards}\cdot\textrm{choose cards})\cdot(\textrm{choose the kind of cards}\cdot\textrm{choose cards})\\ = {13\choose2}\cdot{2\choose1}{4\choose3}\cdot{1\choose1}{4\choose2}.$$
answered Apr 21 '18 at 3:05
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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 $\frac{C(5,3)\times52\times3\times2\times48\times3}{52\times51\times50\times49\times48}\approx0.00144$.
