See above re: birthday paradox. I overcame some precision issues when trying to calculate the numbers as here’s the final outcome:
If one had a pool of previously shuffled decks to choose from, the chances that any pair of those decks are the same rise above 50% would require a sample space size of approximately 10^34. This follows from the issues in the birthday problem.
However, in the context of shuffling your own deck and then checking if your deck matches a deck that was previously shuffled is a different probability and one that requires a much larger sample space to match against to have a higher than 50% probability of finding a match. Given the approximation n=ln(.5)/ln((52!-1)/52!), the required n is still on the order of 10^67 or more than half of the total available sample space.
52! => 80658175170943878571660636856403766975289505440883277824000000000000
n=ln(.5)/ln((52!-1)/52!) ~ 55907986708849934223332477802005437614613669480434146509789214142941
So the nuanced answer is that, if you wanted a 50-50 chance to find any duplicate shuffles, you’d need to search through 10^34 shuffles
If you wanted a 50-50 chance to figure out if a particular shuffle matched another shuffle, it’s still a 10^67 search space.
This makes sense, as it is related to the birthday problem when you want to find out how many people have to enter a room before there is a 50% chance that someone in that group has the same birthday as you do: ln(.5)/ln((365-1)/365) ~ 253
