If one card is removed from the pack there will be 51! permutations of the remaining cards. The removed card is now returned to its original position. Then we would have 52! permutations, 52 of which will contain the card in its original position. So the probability of finding the card in its original position is 1/52 (=51!/52!). This includes finding at least some of the other cards in their same positions, too. The question doesn't ask for the probability of only one card being in the same position.