Consider a deck of cards, where the backs are marked in such a way that you can distinguish a pip card (Ace, 2,…,10) from a face card (King, Queen, or Jack). You deal a poker hand to a friend. Given that the friend has three pip cards and two face cards, what is the probability that your friend holds a full house?
Poker hand conditional probability
combinatoricsconditional probabilitypokerprobability
Related Solutions
There are indeed $\binom{52}{13}$ different 13-card hands and this will indeed be the size of our sample space and thus our denominator when we finish our calculations.
For the numerator, we need to pause for a moment and understand what the problem is actually asking, since this appears to be where you got stuck.
We are asked to find the probability that in our hand of thirteen cards, there is at least one suit for which we have all three face cards. For example $(A\spadesuit,2\spadesuit,3\spadesuit,\dots,10\spadesuit,J\spadesuit,Q\spadesuit,K\spadesuit)$ has all three of the face cards for spades. Similarly if all those cards happened to be hearts instead it would also count since we would have all of the face cards for hearts. Similarly still, a hand like $(J\spadesuit,Q\spadesuit,K\spadesuit,J\heartsuit,Q\heartsuit,K\heartsuit,J\diamondsuit,Q\diamondsuit,K\diamondsuit,\dots)$ would count since we have all of the face cards from spades (we also happen to have all of the face cards from hearts and diamonds too).
Let $\spadesuit$ represent the event that we have have all of the face cards from spades. Similarly, let $\diamondsuit, \heartsuit, \clubsuit$ represent the event that we have all of the face cards from diamonds, hearts, and clubs respectively.
You are asked to find $Pr(\spadesuit\cup\diamondsuit\cup \clubsuit\cup \heartsuit)$
To do this, let us apply inclusion exclusion. We expand the above as:
$Pr(\spadesuit\cup \diamondsuit\cup\clubsuit\cup\heartsuit) = Pr(\spadesuit)+Pr(\diamondsuit)+\dots-Pr(\spadesuit\cap \diamondsuit)-Pr(\spadesuit\cap \clubsuit)-\dots+Pr(\spadesuit\cap \diamondsuit\cap \clubsuit)+\dots-Pr(\spadesuit\cap\diamondsuit\cap \clubsuit\cap \heartsuit)$
Now, let us calculate each individual term in the expansion.
The calculation you did before is relevant. Indeed, we calculate $Pr(\spadesuit)=\dfrac{\binom{3}{3}\binom{49}{10}}{\binom{52}{13}}$. This is again merely the probability that we have all of the face cards from the spades and is not the final probability that we were tasked with calculating.
We continue and calculate more terms:
For example $Pr(\spadesuit\cap \diamondsuit)=\dfrac{\binom{6}{6}\binom{46}{7}}{\binom{52}{13}}$
We then notice what symmetry there is in the terms and can simplify some. Finally, we write the final expression for our final answer (and get an exact number only if actually requested or required, usually opting to leave the answer in terms of binomial coefficients without additional simplification).
In $7$-card poker, the best five cards are selected. Therefore, you can obtain a full house in three ways:
- A three of a kind, a pair, and two singletons
- A three of a kind and two pairs
- Two three of a kinds and a singleton
Also, you failed to distinguish between the rank of the three of a kind and the rank of the pair. By selecting two additional cards at random, you created the possibility of a four of a kind, which is not permitted. The first error made your outcome too small. The second error would have led to an over count had you not made the first error, particularly since the four of a kind could have been paired with a pair or a three of a kind.
A three of a kind, a pair, and two singletons: There are $13$ ways to select the rank of the three of a kind, $\binom{4}{3}$ ways to select three of the four cards of that rank, $12$ ways to select the rank of the pair, $\binom{4}{2}$ ways to select two cards of that rank, $\binom{11}{2}$ ways to select the ranks of the two singletons, and $4$ ways to select a card from each of those ranks.
There are $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}\binom{11}{2}\binom{4}{1}^2$$ such hands.
A three of a kind and two pairs: There are $13$ ways to select the rank of the three of a kind, $\binom{4}{3}$ ways to select three of the four cards of that rank, $\binom{12}{2}$ ways to select the ranks of the two pairs, and $\binom{4}{2}$ ways to select two cards from each of those ranks.
There are $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{2}^2$$ such hands.
Two three of a kinds and a singleton: There are $\binom{13}{2}$ ways to select the ranks of the three of a kinds, $\binom{4}{3}$ ways to select three of the four cards of each of those ranks, $11$ ways to select the rank of the singleton, and $4$ ways to select a card of that rank.
There are $$\binom{13}{2}\binom{4}{3}^2\binom{11}{1}\binom{4}{1}$$
Since these three cases are mutually exclusive and exhaustive, the number of favorable cases is found by adding the three results. Dividing that sum by $\binom{52}{7}$ yields the desired probability.
Addendum: As aschepler pointed out in the comments, it is not possible to combine a full house with a straight or a flush with only seven cards, so the cases above are indeed exhaustive.
Best Answer
The three pips have to match, and the two face cards have to match. The first condition is the same as saying that three cards, chosen from $40$, are either all As, all 2s, . . ., or all 10s. That probability comes out to: $$10\cdot\frac{\binom{4}{3}}{\binom{40}{3}}=\frac{1}{247}$$ Similarly, the probability that the face cards match is:$$3\cdot\frac{\binom{4}{2}}{\binom{12}{2}}=\frac{3}{11}$$ The probability of both of these things happening is the product of those two fractions, $\frac{3}{2717}\approx0.11\%$, or about $1$ in $906$.