Probability of obtaining unordered five-card poker hand

Question

What is the probability of obtaining a five-card poker hand consisting of two spades and one card of each of the other three suits?

Answer 1

Throughout we assume cards are dealt without replacement.

There are $\binom{13}{2}=78$ ways of choosing two spades, then $\binom{13}{1}=13$ ways of choosing a card from each of the three other suits respectively. To find the probability you must divide by the total number of possible five card poker hands $\binom{52}{5}=2598960$, if we assume all hands are equally likely.

$$P(2\spadesuit,1\heartsuit,1\clubsuit,1\diamondsuit)=\frac{\binom{13}{2}\binom{13}{1}^3}{\binom{52}{5}}=\frac{78\cdot 13^3}{2598960}=\frac{2197}{33320}$$

Alternatively, assume we pick the cards in the order $\spadesuit_1$, $\spadesuit_2$, $\heartsuit$, $\clubsuit$, $\diamondsuit$. Then the probabilities are
$$P(\spadesuit_1)=\frac{13}{52}, \,P(\spadesuit_2\,|\,\spadesuit_1)=\frac{12}{51}, \,P(\heartsuit)=\frac{13}{50}, \,P(\clubsuit)=\frac{13}{49},\,P(\diamondsuit)=\frac{13}{48}$$ Now there are $5!=120$ permutations of this hand, but since the order in which the spades $\spadesuit_1$ and $\spadesuit_2$ are dealt doesn't matter we get $\frac{5!}{2!}=60$ total hands.

$$P(2\spadesuit,1\heartsuit,1\clubsuit,1\diamondsuit)=\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{13}{50}\cdot\frac{13}{49}\cdot\frac{13}{48}\cdot\frac{5!}{2!}= \frac{12\cdot13^4\cdot 60}{52\cdot51\cdot50\cdot49\cdot48}=\frac{2197}{33320}$$