Assuming that all the shuffles are truly random shuffles, and that approximately 1 trillion shuffles have been shuffled in the history of the world. What are the odds that a shuffle has come up with the same order twice?
I think the answer is: $$ (52!-1)(52!-2)(52!-3)\cdots(52!-n) / (52!)^n , $$ where $n = 10^{12}$
How do I compute this?
Note that your expression is actually about no repeats in $10^{12} + 1$ trials, but it doesn't matter very much. Write $n = 10^{12}$. Note that your probability is certainly less than 1, and we can find a rather coarse lower bound for it by $$ \frac{\prod_{i=1}^{n} 52! - i}{(52!)^{n}} \geq \frac{(52! - n)^{n}}{(52!)^n} \approx 1 - 1.2398 \times 10^{-44}. $$ Here we used WolframAlpha in the last step. This means that under your assumptions, there almost certainly has not been a shuffle resulting in the same order yet.