I have a deck of cards labeled $1, 2, \ldots, 9$. I give them a good shuffle and split them into $3$ piles, each with $3$ cards. Let's call these piles $A, B, C$.
I draw the top card from pile $A$ and see that it's a $1$. I discard this card. Now there are $8$ cards remaining, with pile $A$ having two cards and piles $B$ and $C$ having three cards.
What is the probability that I will draw a $9$ if I take the top card from pile $B$?
To me this question sounds really similar to the Monty Hall problem. At first, each pile had a $\frac 13$ chance of having the $9$ card, and each pile had a $\frac 13$ chance of having the $9$ as the top card, so there was a $\frac 19$ chance you would get the $9$ in a specific pile.
But after we draw the $1$ from $A$, we know that there are only $8$ cards left and $A$ has one less card than the other piles. I think the Monty Hall problem implies that the probabilities for piles $B$ and $C$ don't change (they're still $\frac 19$), but what happens to pile $A$?
You are overthinking this. It doesn't matter how many piles the cards are divided into, or how many cards are in each pile. There is no "Monty" in this situation with hidden knowledge. There are simply eight cards on the table. Drawing one gives you a 1/8 chance of getting the 9.