How many cards are expected to retain the original, i.e., the place before the shuffle, position in a well-shuffled deck of cards?
Let X1 be the event where the first card retains position one after the shuffle. We give a value of 1 if it gets the right spot and 0 otherwise. A shuffled card can occupy any of the 52 positions; therefore, the probability of getting any place (including the first) is 1/52. The expected value for X1 becomes:
E(X1) = 1 x (1/52) + 0 x (51/52) = 1/52
It’s easy to notice that E(X2) also follow the same logic and becomes 1/52, etc.
The expected number for each card to stay in the same spot is:
E(X1 + X2 + X3 … X52) = E(X1) + E(X2) + E(X3) + … E(52) = 52 x (1/52) = 1
You can see that this is true for any number of cards from 1.
