In a pack of cards there are 4 sets and 13 cards in each set $\rightarrow$ 1(ace),2,3,4,,5,6,7,8,9,10,J(11),Q(12),K(13).
The possible cases for straight hand can be (13,12,11,10,9) , (12,11,10,9,8) , (11,10,9,8,7) .....(5,4,3,2,1)
= ways of choosing $k$ consecutive integers from $n$ integers
=$(n-k+1)$
=$(13-5+1)$
=$9$ cases.
and Probability of selecting cards for each case = $\frac{4}{52} * \frac{4}{51} * \frac{4}{50} * \frac{4}{49} * \frac{4}{48}$
But this above cases also include the cases where all the $5$ cards can be from same slot and there are $4$ such cases possible
Also the order of selection of cards doesn't matter so we will multiply by $5!$
so now
Probability of selecting cards for each case = $\frac{4^{5}-4}{52*51*50*49*48}*5!$
$\therefore$ Required Probability
= $9*5!* \frac{4^{5}-4}{52*51*50*49*48}$
= $9*5!* \frac{1020}{52*51*50*49*48}$
= $\frac{1101600}{311875200}$
= $0.00354$
