Let's say that I have a pyramid with 10 coins at the top. I do not have have a photo, but Pascal's Triangle is exactly the kind of pyramid I am talking about, except instead of numbers, there are just empty 'positions' where a coin may sit.
For each coin, I will choose randomly either $1$ or $2$. If I choose $1$, I move the coin down one row and to the left. If I choose $2$, I will move the coin down one row and to the right. The first row is row $0$. The game is over once all coins have moved to their final position in the last row.
In every even row, there is a central position which has an equal number of positions on either side of it (it's the middle position).
Is there an equal number of combinations of flips that return a coin to this central position, or are there collectively more than send a coin anywhere else?
I believe this can be worked out mathematically; for any odd number of rows $n$ you will have a unique combination of moves $(M_1, M_2, M_3...M_{n-1})$ where $M$ is either $L$ or $R$ and signifies a move left or right. Each coin can be assigned one of these combinations, where $M_1$ signifies their first move left or right into the row below, $M_2$ signifies their second, etc. In order for a coin to end up in the center, it must have a set of moves with equal $L$ and $R$.
I found the outcomes for 1, 3, and 5 rows. The percentage of outcomes where a coin ends up in the middle is $2/2$, $2/4$, and $6/16$, respectively. The denominators are powers of two; the number could be quadratic, (cubic?) or exponential. I am not sure.
Given this, I believe that the ratio of coins that end up in the middle vs not in the middle must change depending on the number of rows. Am I correct in this? How could I make a function to model this?
Your reference to Pascal's triangle is more apposite than you appear to realise. The numbers in Pascal's triangle give the number of paths from the top to that cell taking the left-right decisions you describe, so the probability of a coin settling in the centre cell of row $2n$ is $\frac{(2n)Cn}{2^{2n}} = 4^{-n} \binom{2n}{n} = \frac{(2n)!}{4^nn!n!}$. For large $n$ this is approximately $\frac{1}{\sqrt{\pi n}}$.