Probability of drawing certain hand, incorrect answer, but why? So I am drawing $5$ cards from a standard deck of $52%$
I want to find the probability that I draw $5$ consecutive cards of same suit with no card looping, and the ace is card $1$.
So the consecutive cards can start on any of $A,2,3,4,5,6,7,8,9$ but can't start on $10$ since that would not complete, e.g $10,J,Q,K$, and then it can't loop so doesn't work.
So we should have $\frac{9*4}{52}$ starting cards, being the four suits and the $9$ possible numbers. Then we should have four appropriate options, being any of the four cards that needs to be grabbed, out of $51$ remained, and then three cards out of $50$
Giving me:
$$\frac{9*4}{52}\cdot\frac{4}{51}\cdot\frac{3}{50}\cdot\frac{2}{49}\cdot\frac{1}{48}=2.77*10^{-6}$$
 A: You ask the probability of first drawing the lowest card of your straight flush, then drawing any of the remaining four card of the tsraight flush, then drawing any of the remaing three cards, ...
It is, however, not given that the lowest card will be the first you draw.
A: Given that order doesn't matter, your universe is all unordered subsets of $\{1,...,52\}$ of size $5$ (so card = $\binom{52}{5}$) and your event is of size $4 \times 9$ (each satisfactory draw is uniquely determined by the choice of a card below $9$ in each of 4 colors).
I get $\frac{4\times 9 }{\binom{52}{5}} = 1.38 \times 10^{-5}$.
Btw, this question is a duplicate: link 
A: 
I want to find the probability that I draw 5 consecutive cards of same suit with no card looping, and the ace is card 1.
@AlexH. No, all five are given, and I pick up the five together. –  Algebra

So you use the face and suit of the lowest card in the 5 as the identity of the hand.  There are $9\times 4$ ways to pick 5 consecutive cards of the same suit when order does not matter.  There are ${52\choose 5}$ ways to select any 5 cards when order doesn't matter.  So the probability is:
$$\frac{9\cdot 4\cdot 5!\cdot 47!}{52!}=\frac{3}{216580}$$
