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.
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 $9\times 4^5$.