Probability that a 5-card poker hand contains exactly four cards of the same suit?

Question

I'm not that confident in my ability, so I've come here.

This is what I have:

$\frac{C(4,1) * C(13,4) * C(3,1) * C(13,1)}{C(52,5)}$

My thinking is that $C(4,1)$ selects one suit from the 4 suits and $C(13,4)$ selects 4 cards from that suit. Then $C(3,1)$ selects another suit from the remaining 3 suits and $C(13,1)$ selects one card from that suit.

Then divide by $C(52,5)$.

Is this correct?

Answer 1

Yes, that is correct. As a sanity check, an alternative approach is to choose the two suits, choose which one contains the four cards, and then choose four cards from that one suit and one from the other suit: $$\binom{4}{2}\binom{2}{1}\binom{13}{4}\binom{13}{1}=111,540$$

Answer 2

Yes. You seek the probability for obtaining one from four suits with four from its thirteen cards, and one from the remaining three suits with one from its thirteen cards, when selecting any five from all fifty-two cards.$$\def\cbinom#1#2{\mathop{^{#1}\mathrm C_{#2}}}\dfrac{\cbinom 41\cbinom {13}4\cbinom 31\cbinom{13}1}{\cbinom{52}5}$$