Reading through my Probability book brushing up on some stuff:
What is the probability that a poker hand is a full house? (A full house is defined as a hand with three cards of one denomination and two cards of another denomination; ex. three Queens and two 4's.)
The solution (this is an example) is stated as:
The number of different poker hands is $52\choose5$. To count the number of full houses, let us call a hand of type (Q,4) if it has three queens and two 4's, with similar representations for other types of full houses. Observe that (Q,4) and (4,Q) are different full houses, and types such as (Q,Q) and (K,K) do not exist. Hence there are $13 \times 12$ different types of full houses. Since every particular type, say (4,Q), there are $4\choose3$ ways to select three 4's and $4\choose2$ ways to select two Q's, the desired probability is
$$ \frac{13 \cdot 12 \cdot \binom{4}{3} \cdot \binom{4}{2}}{\binom{52}{5}}$$
I don't understand how the $13 \times 12$ comes about, and that's where I need some clarification.
Thank you in advance.
Best Answer
There are 13 ways of picking the first denomination and for every one of those there are 12 ways of picking the second (since you can't pick one denomination twice).
There's a rule that says if there's $n_1$ ways of picking the first element of an ordered pair and $n_2$ ways of picking the second, the number of possible pairs is $n_1n_2$.
More generally, when you have $n$ objects and you want to pick an ordered combination of size $k$, $P_{k,n}=\frac{n!}{(n-k)!}$. This is called a permutation of size $k$ of the objects.
In this case, $n=13$ and $k=2$, so $P_{2,13}=\frac{13!}{11!}=\frac{13 \cdot 12 \cdot 11!}{11!}=13 \cdot 12$.