Problem: Pick 7 cards from the regular 52 English cards deck, were they must be in order and suits don't matter (aces can loop).
As stated here for this similar problem, I suppose the answer would be something like:
$\frac{\binom{13}{1}\cdot\binom{4}{1}^7}{\binom{52}{7}}$
And I understand the reason for the top part of that fraction, where the $\binom{13}{1}$ corresponds to every one card I pick, times 1 suit out of 4 in total; 7 times (for each card).
The thing is that I don't quite get why I need to divide this number by $\binom{52}{7}$ and what it means for the problem.
Thanks for any help, sorry for my English.
Appears you are not using standard definition of a straight.
If you allow loop on ace the there are 13 unique straights. $\binom{13}{1}$
Of the 7 cards they can be any suit. ${\binom{4}{1}^7}$
${\binom{52}{7}}$ is simply the number of 7 card hands taken from 52 cards