Expected number of card draws from a standard deck until an ace is drawn The problem: What is the expected number of draws from a standard deck of
52 cards until an ace is drawn?
This question has been asked many times on this forum so I'm not looking for a solution, but I would like to better understand a part of this specific solution:

Consider any card as just it’s suit A, C, D, and S for ace, club,
diamond, and spade, respectively. Since we don’t care about any
particular ace, we can encode the order of the deck as a string of
repeated letters. In fact, we only care about aces, so our deck can be
considered a binary string of 4 A’s and 48 O’s, where O stands for
“other.” The number of such strings is $\binom{52}{4}$
We now consider cases:
The number of strings with A in the first
position is $\binom{51}{3}$
The number of strings with A in the second position is
$\binom{50}{3}$
This continues until the number of strings with A in the 49th
position is $\binom{3}{3}$

I'm looking to understand why the total number of such strings is $\binom{52}{4}$. Specifically, why would combinations be appropriate here and why those parameters specifically? Is there an intuitive reason for this?
I am new to combinations in general but haven't really seen combinations used like this before. Thank you.
 A: Write a sequence of $52$ Os. To make this a valid sequence (with $48$ Os and $4$ As), we just need to change any $4$ Os into As i.e. to choose $4$ positions from $\{1,2,3,\dots,52\}$ to change into As.
That is, we need to choose $4$ distinct items from a set of $52$.
This is precisely the different of "combinations" and its notation.

We can also see more explicitly how the number of possibilities is $$\frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}\left (=\frac{52!}{48!4!}=\binom{52}{4}\right )$$
There are $52$ ways we can choose the first A. To choose the next, there are $51$ Os left and we can choose any of them. Similarly, for the third and fourth choices there are $50$ and $49$ Os to pick.
So, if the order of our choices mattered, there would be $52\times 51\times 50\times 49$ ways to make the choice.
However, choosing $10,26,41,45$ is the same as choosing $41,26,10,45$ or $45,26,10,41$ etc. i.e. the order of the choices does not matter. How many different orders to choose $4$ items in are there? Well, this is the definition of the factorial, or permutation: the answer is $4!$.
So we have counted each possible set of $4$ aces $4!$ times, and dividing by $4\times 3 \times 2 \times 1$ gives us our combination value.
A: There is absolutely an intuitive reason for this.
Imagine you have a deck of 52 blank cards, and you write the letter A on 4 of them. There are $\binom{52}{4}$ ways to do this.
Now imagine you have a string of 52 0's, and you can flip 4 of them into A's. This is the same problem as before, so there are $\binom{52}{4}$ such strings.
