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2021 AMC 10A Problem 23

2021 AMC 10A Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Conditional Probability & States · difficulty 4 of 5 · about 7 min · position P21-25

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

A)
$\frac{9}{16}$
B)
$\frac{5}{8}$
C)
$\frac{3}{4}$
D)
$\frac{25}{32}$
E)
$\frac{13}{16}$

Answer

Key insight

Only three states matter: center, edge, corner. From an edge, half the hops reach a corner, a quarter return to the center, a quarter wrap to another edge.

Solution

By symmetry only the type of square matters: center ( $C$ ), edge-middle ( $E$ ), or corner ( $K$ ).

From $C$ , all four hops land on an $E$ . From an $E$ (say the top middle): the two sideways hops reach corners, hopping inward returns to $C$ , and hopping outward wraps to the bottom middle, another $E$ . So from $E$ : $K$ with probability $\frac12$ , $C$ with $\frac14$ , $E$ with $\frac14$ .

Track the probability of being unstopped after each hop.
- After hop 1: $E$ with probability $1$ .
- After hop 2: $C$ with $\frac14$ , $E$ with $\frac14$ (corner reached with $\frac12$ ).
- After hop 3: from $C$ ( $\frac14$ ) we go to $E$ ; from $E$ ( $\frac14$ ) we go to $C$ with $\frac1{16}$ or $E$ with $\frac1{16}$ . So $E$ : $\frac14 + \frac1{16} = \frac5{16}$ , $C$ : $\frac1{16}$ .
- After hop 4: no corner from $C$ ; from $E$ ( $\frac5{16}$ ) she fails to reach a corner with probability $\frac12$ .

Probability of never reaching a corner:
$\frac{1}{16} + \frac{5}{16}\cdot\frac12 = \frac{2}{32} + \frac{5}{32} = \frac{7}{32}.$
So the probability of success is $1 - \frac{7}{32} = \frac{25}{32}$ .

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

Why this works

Random walks on symmetric boards compress to a handful of states; the absorbing state (corner) is easiest to handle by tracking the surviving mass in the other states. Writing one transition rule per state and iterating four times is faster and safer than enumerating $4^4$ hop sequences.

Alternative approach

Add the success probabilities directly: hop 2 succeeds with $\frac12$ ; hop 3 succeeds from the $E$ mass $\frac14$ with $\frac18$ ; hop 4 succeeds from the $E$ mass $\frac{5}{16}$ with $\frac{5}{32}$ . Total $\frac{16 + 4 + 5}{32} = \frac{25}{32}$ .

The trap

Ignoring the wraparound (an edge square's outward hop lands on the opposite edge, not off the board) or forgetting that Frieda stops once she reaches a corner.

Common mistakes

Ignoring the wraparound (an edge square's outward hop lands on the opposite edge, not off the board) or forgetting that Frieda stops once she reaches a corner.
Letting the process continue past a corner and counting later corner visits again, which overcounts the success probability.

Techniques

Count the complement and subtract from the total · Define states/recurrence and iterate