TITLE: Expected number of cards in original position in a shuffled deck of $52$ cards?
QUESTION [11 upvotes]: Assume is shuffle is quite good that it randomizes the card order.
We know that E =  $ \sum_{X=1}^n X*P(X) $
We are already know that n=52 and that there are 52! ways to arrange the cards.
So probability that exactly 1 card is in correct position is
$\frac{1}{52!} {52 \choose 1}*$(derangements of remaining cards)
This will be summed over all the 52 cases. This seems a bit complicated. Is there a simpler way?

REPLY [11 votes]: Examine the $6$ permutations of $123$:
$$\color{red}1\color{red}2\color{red}3$$
$$\color{red}1\color{black}3\color{black}2$$
$$\color{black}2\color{black}1\color{red}3$$
$$\color{black}2\color{black}3\color{black}1$$
$$\color{black}3\color{black}1\color{black}2$$
$$\color{black}3\color{red}2\color{black}1$$
with red indicating a correct position.
Then each number is in its correct position exactly twice, and there are three numbers, hence six red numbers, which gives the expected number of correct positions as $\frac66=1$.
In general there are $n$ numbers, and each is correct $(n-1)!$ times, hence $n!$ correct positions, and so the expected number of correct positions is $\frac{n!}{n!}=1$.