That's a lotta decks!

First, we calculate the probability (P) that none of the decks will be sorted as required.

Probability that a deck is sorted in a specific arrangement = $\frac {1}{52!}$

Hence, probability that all decks are not sorted according to that arrangement = $(1-\frac {1}{52!})^{52!}=P$

Now, since $52!$ is an astronomically massive number, we can consider it to tend to $\infty$ . Hence,

$P =\displaystyle \lim _{x\rightarrow ∞} (1-\frac {1}{x})^x$ = $\frac {1}{e}≈ 0.367$

The required probability is $1-P=0.632$

For $N$ cards, the probability of a given distribution is $\dfrac{1}{N!}$

For N combinations the probability of having none of them correct is $(\dfrac{N-1}{N})^N$

Therefore, the probability that at least one will be sorted is $1 - (\dfrac{N-1}{N})^N$

For large $N$ like, for example $N = 52!$ this goes $1 - \dfrac{1}{e} = 0.63$