What is the probability that a five-card poker hand contains a straight flush, that is, five cards of the same suit of consecutive kinds (i.e. <1-2-3-4-5, 2-3-4-5-6, …, 9-10-J-Q-K, 10-J-Q-K-1>?
My attempt: We need to chose 5 cards in the same suit, each suit contains 13 different cards so $\binom{13}{5}$, we only need it to be in the same suit so we chose one suit out of 4, that is $\binom{4}{1}$, total number of outcomes is $\binom{52}{5}$, Now there are only 13 situation where we have consecutive kinds, and total number of how we can chose 5 out of 13 regardless of order is $\binom{13}{5}$
Hence, the probability is: $$\frac{\binom{13}{5}\binom{4}{1}}{\binom{52}{5}} \cdot \frac{13}{\binom{13}{5}} = 2.00 \times10^{-5} $$
However, i think there's something missing or i have made a mistake, can anyone verify my solution ?