problem with cards

Cards from an ordinary deck of 52 playing cards are turned face up one at a time. If the 1st card is an ace, or the 2nd a deuce, or the 3rd a three, or . . ., or the 13th a king, or the 14 an ace, and so on, we say that a match occurs. Note that we do not require that the (13n + 1)th card be any particular ace for a match to occur but only that it be an ace. Compute the expected number of matches that occur.

My attempt is: Let $X_i$ denote the indicator variable for the event that the i-th card is matched, 0 otherwise. $N=\sum_{i=1}^{52} X_i$=number of matches that occur

$$\begin{align}E[N]=E[\sum_{i=1}^{52} X_i]&=\sum_{i=1}^{52} E[X_i]\\&=E[X_1]+E[X_2]+...E[X_{52}]\\&=\frac{4}{52}+...\frac{4}{52}+\frac{3}{52}+....\frac{3}{52}+\frac{2}{52}+...+\frac{2}{52}+\frac{1}{52}+...\frac{1}{52}\\&=\frac{4}{52}*13+\frac{3}{52}*13+\frac{2}{52}*13$+\frac{1}{52}*13\\&=\frac{10}{4}\\&=2,5\end{align}$$ because the probability of having a match is $\frac{4}{52}$ for the first 13 cards, $\frac{3}{52}$ for the second 13 cards, $\frac{2}{52}$ for the third 13 cards, $\frac{1}{52}$ for the last 13 cards. But in my book the final result is 4 and the probability is $\frac{1}{52}$ for all the 52 cards. Why?

Imagine all the 52 cards arranged in a row.

As you have said,
let $X_i$ denote the indicator variable for the event that the i-th card is matched, $0$ otherwise.

By linearity of expectation, we have $E\Sigma (X_i) = \Sigma E( X_i)$

The expectation of an indicator random variable is just the probability of the event it indicates, so $E(X_i) = P(i_{th}\; card\; matches) = \frac4{52}$,

and $\Sigma E(X_i) = \frac4{52}\cdot 52 = 4$