2021 AMC 10A Problem 23

1. There are only three kinds of square: the center (call it $C$), the four edge-middle squares ($E$), and the four corners ($K$). All edge squares behave the same way, so we only need to track which kind of square Frieda is on.
2. Work out the hops from each kind:
- From $C$, every hop lands on an edge square.
- From an edge square, say the top middle: left and right reach corners, down goes to the center, and up wraps around to the bottom middle, another edge square. So from $E$: corner with probability $\frac12$, center with $\frac14$, edge with $\frac14$.
3. Now follow the probabilities hop by hop. The columns for $C$ and $E$ are the chances she is still hopping and standing there.

How hop $3$ works: the $\frac14$ at $C$ all moves to $E$. The $\frac14$ at $E$ splits: half of it ($\frac18$) reaches a corner, a quarter of it ($\frac1{16}$) goes to $C$, and a quarter ($\frac1{16}$) stays on an edge.
On hop $4$, the $\frac1{16}$ at $C$ cannot reach a corner, and half of the $\frac5{16}$ at $E$ does.
4. Add the chances of reaching a corner:
$$\frac12 + \frac18 + \frac5{32} = \frac{16}{32} + \frac4{32} + \frac5{32} = \frac{25}{32}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{25}{32}}$.

On a symmetric board, squares of the same kind behave the same, so a random walk shrinks to a few "states". Writing one rule per state and following the chances for four hops is much faster and safer than listing all $4^4 = 256$ hop sequences.

Find the chance she never reaches a corner and subtract from $1$. After hop $4$ she is still hopping only if she was at $C$ after hop $3$ (chance $\frac1{16}$), or at $E$ after hop $3$ and then missed (chance $\frac5{16} \times \frac12 = \frac5{32}$). That is $\frac2{32} + \frac5{32} = \frac7{32}$, so the answer is $1 - \frac7{32} = \frac{25}{32}$.