Expected number of matches in a deck of cards The problem statement is as such:

Suppose cards from an ordinary deck of $52$ playing cards are turned face up one at a time. If the first card is an ace, or the second a deuce, or the third a three, or ... , or the thirteenth a king, or the fourteenth an ace, and so on, we say that a match occurs. Note that we do note 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.

I have looked through the discussions and answers here: The expected number of matches from a deck of cards?
My initial solution comprised applying the property of Linearity of Expectations as well. Defining a random variable $X_i$ denoting the number of matches on the $i$-th flip such that
$$
X_i = \begin{cases} 1, & \text{if match} \\ 0, & \text{otherwise} \end{cases}
$$
Why is it that from the solutions it seems that we can immediately conclude $P(X_i = 1) = \frac 4 {52} \Rightarrow E[X_i] = \frac 4 {52} = \frac 1 {13}$? I have some difficulty understanding this, as I was under the impression that we would need to account for conditional probabilities, e.g. the probability that we have a match on the 5th card is dependent on how many $5$s we have drawn in the first 4 flips.
Any clarification on the intuitive end would be greatly appreciated.
 A: Even though the probability distributions of matches in different positions are dependent (for example, if the first $51$ cards are matches, the $52$nd must be a match), the overall probability of a match in any position is $1\over13$. Take your example about the $5$th card. While the probability of the $5$th card matching after a particular set of previous draws depends on how many fives were drawn beforehand, the overall probability of a match on the $5$th card is nevertheless $1\over13$.
Perhaps this thought experiment will help you see that it could not be otherwise. Play the following game. I will deal a shuffled deck of cards face-down in a row, and then I will turn over the $5th$ card I dealt. You win if I turn over a five. Would you wager differently because it’s the $5$th card? How can there be special positions in a dealt deck of cards where certain cards are more likely to appear?
Maybe an easier example is useful. I flip a fair coin twice. What do you think the probability is that the second flip is heads? Surely you don’t think heads comes up more (or less) often on the second flip than it does on the first. Similarly, a five comes up in a deal no more (or less often) in the $5$th position than it does in the $7$th, or $11$th, or anywhere else.
