That is correct. So if is the probability of not having any card in its correct place then is the probability of at least one card in its correct place.
Do you know about derangements ?
Actually I'm not a math specialist, I tried to solve it in a easy way and I got p=1.
A card in position 1 in deck 1 has probability 1/52 to be equal to a card in position 1 in deck 2.
And since we have 52 cards, then the probability to get two equal cards at same position is 52 * 1/52 = 1
Could it be p=1?
Two decks are shuffled. The first is the reference deck and I interpret that the question is what is the P that the second deck will have the exact same order as the first. I would answer 52 factorial is the number of possible events.P= 1/8*10^67