5 cards in sequence, don't understand answer Find the probability that a 5 card poker hand will be 5 cards in sequence regardless of suit, consider ace high, not low.
given answer:
\begin{align}
\frac{{9 \choose 1} {4 \choose 1} ^ 5}{52 \choose 5}
\end{align}
I understand this:
${9 \choose 1}$:  9 is the number of possible 5-element sequences.
the five ${4 \choose 1}$'s are the possible suits for each of the 5 cards.
But, in the given answer i dont see how, once picked the first card of the sequence, is the player restricted to pick the remaining 4 cards so they fit in such sequence. 
As far as i understand the answer, the player is able to pick the remaining 4 cards freely without having to restrict himself, or, less probably, herself, to the sequence initiated in ${9 \choose 1}{4 \choose 1}$.
An explanation of what i am missing here is greatly appreciated.
Thank you very much.
 A: I wouldn't see it as if I was picking really. I'd just say as you did there are 9 different sequences, for those 9, there are 4 different suits for every card as you say. So that would 4x4x4x4x4 for a total of 4^5 possibilities for those 9 sequences. So 9x4^5 is the total number possibilities to fulfill your criteria.
Then (52 5) is just the total number of sequences you can get. What you need to understand is that all these cards sequences are equiprobable meaning that you have as much of a chance of getting that specific royal flush as of getting those 4 twos and a five of hearts let's say. Think about it you'll see what I mean. So with equiprobable probabilities, the probability is just the number of results that satisfy your criteria divided by the total number of possible results. Hence your answer. I'm no great teacher but I tried feel free to ask more questions :)
A: Both numerator and denominator count hands, not $5$-card sequences.
Let us count the number of favourables. The low number in our straight hand can be any of $2,3,4,5,6,7,8,9,10$, for a total of $9$ possibilities.
For every choice of low number, there are $4$ ways to choose the suit of the lowest card. For every way of choosing the suit of the lowest card, there are $4$ ways to choose the suit of the next lowest card, and so on, for a total of $9\times 4^5$.
