2021 AMC 10 A Problem 23

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To help you master 2021 AMC 10 A Problem 23, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 A Problem 23

Solution

Solution 1

State-by-State Probability Method

Group the squares by their structural type (Center, Edge, Corner) to simplify the grid. Calculate the probability of being in each type of square after each hop, keeping in mind that the Corner is a stopping point.

When locations on a grid have identical properties (like the 4 edges on a wrap-around board), you can group them into a single ‘state’ to drastically reduce the number of paths you need to track.

Step 1: Define the states and transition probabilities

Group the 9 squares into 3 types: Center (1 square), Edge (4 squares), and Corner (4 squares). From the Center square, all 4 directions lead to an Edge square, so the probability of moving from Center to Edge is 1. From an Edge square (for example, the top-middle square), the 4 directions lead to: – Down: Center square (1 choice) – Up: Wraps around to the bottom-middle square, which is another Edge square (1 choice) – Left and Right: Corner squares (2 choices) Thus, from an Edge, the probabilities are $\frac{1}{4}$ to Center, $\frac{1}{4}$ to Edge, and $\frac{1}{2}$ to Corner. From a Corner square, Frieda stops, so we just accumulate the probability of being there.

Step 2: Track probabilities for Hops 1 and 2

Frieda starts at the Center (Hop 0). For Hop 1, she must move to an Edge. For Hop 2, she distributes her probability from the Edge according to the rules established above.

$$P(\text{Center}{)}_2=1\times \frac{1}{4}=\frac{1}{4}$$

$$P(\text{Edge}{)}_2=1\times \frac{1}{4}=\frac{1}{4}$$

$$P(\text{Corner}{)}_2=1\times \frac{1}{2}=\frac{1}{2}$$

Step 3: Track probabilities for Hop 3

Calculate the new probabilities by analyzing where she was at Hop 2. The probability of being at an Edge comes from both the previous Center state (all of it moves to Edge) and the previous Edge state (one-quarter of it stays on an Edge).

$$P(\text{Center}{)}_3=P(\text{Edge}{)}_2\times \frac{1}{4}=\frac{1}{16}$$

$$\begin{array}{rl}& P(\text{Edge}{)}_3=\left(P(\text{Center}{)}_2\times 1\right)\\ & +\left(P(\text{Edge}{)}_2\times \frac{1}{4}\right)=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}\end{array}$$

$$\begin{array}{rl}& P(\text{Corner}{)}_3=P(\text{Corner}{)}_2+\left(P(\text{Edge}{)}_2\times \frac{1}{2}\right)\\ & =\frac{1}{2}+\frac{1}{8}=\frac{5}{8}\end{array}$$

Step 4: Track probabilities for Hop 4

We only need to find the probability that she has reached a Corner by the 4th hop. This includes the accumulated probability that she was already at a Corner after Hop 3, plus the new chance she just arrived from an Edge.

$$P(\text{Corner}{)}_4=P(\text{Corner}{)}_3+\left(P(\text{Edge}{)}_3\times \frac{1}{2}\right)$$

$$\begin{array}{rl}& P(\text{Corner}{)}_4=\frac{5}{8}+\left(\frac{5}{16}\times \frac{1}{2}\right)=\frac{20}{32}\\ & +\frac{5}{32}=\frac{25}{32}\end{array}$$

Correct answer: D. $\frac{25}{32}$
The probability is $\frac{25}{32}$, which corresponds to choice D. We can verify the sum of probabilities at Hop 3: $\frac{1}{16}+\frac{5}{16}+\frac{5}{8}=\frac{1+5+10}{16}=1$, confirming no probability was lost.

Solution 2

Complementary Path Counting Method

Instead of tracking fractions, count the number of 4-hop paths that NEVER visit a corner. Subtract this from the total possible paths to find the probability of reaching a corner.

Complementary Probability: $P(\text{Event})=1-P(\text{Not Event})$. This is particularly powerful when the ‘Not Event’ (never hitting a corner) entirely avoids the absorbing state, keeping the number of choices consistent at every step.

Step 1: Reframe the problem using the complement

Instead of stopping at a corner, imagine Frieda always makes exactly 4 hops. Since she has 4 choices for each hop, there are $4^4=256$ total possible sequences. The event ‘reaches a corner and stops’ is the exact complement of ‘makes 4 hops and NEVER touches a corner’. We will count the paths that strictly visit Center and Edge squares.

Step 2: Establish the path transitions

Let $C_n$ and $E_n$ be the number of valid paths (avoiding corners) ending at a Center or Edge square after $n$ hops. From a Center square, all 4 choices go to an Edge, creating 4 paths. From an Edge square, exactly 1 choice goes to a Center, and exactly 1 choice goes to an Edge. (The other 2 choices go to a Corner, which we discard).

$$C_{n+1}=E_n\times 1$$

$$E_{n+1}=(C_n\times 4)+(E_n\times 1)$$

Step 3: Count valid paths for Hops 1 and 2

We start with 1 path at the Center at Hop 0. We update the counts using the transition rules.

$$C_1=0$$

$$E_1=1\times 4=4$$

$$C_2=E_1\times 1=4$$

$$E_2=(C_1\times 4)+(E_1\times 1)=0+4=4$$

Step 4: Count valid paths for Hops 3 and 4

We continue applying the transition rules for the 3rd and 4th hops.

$$C_3=E_2\times 1=4$$

$$E_3=(C_2\times 4)+(E_2\times 1)=16+4=20$$

$$C_4=E_3\times 1=20$$

$$E_4=(C_3\times 4)+(E_3\times 1)=16+20=36$$

Step 5: Calculate the final probability

Sum the valid paths ending at either safe square after 4 hops, divide by the total possible paths, and find the complement probability.

$$\text{Total valid paths}=C_4+E_4=20+36=56$$

$$P(\text{no corner})=\frac{56}{256}=\frac{7}{32}$$

$$P(\text{reaches corner})=1-\frac{7}{32}=\frac{25}{32}$$

Correct answer: D. $\frac{25}{32}$
The probability of reaching a corner is $\frac{25}{32}$, matching choice D. Counting paths with integers avoids fractional arithmetic errors, providing a solid verification of our previous method.

Similar Problems in Real Exams

The following five AMC 10 A problems are closely related to AMC 10 A 2021 Question 23. These real AMC 10 A problems test the same key concept—Probability. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2024 AMC 10 A Problem 24

2023 AMC 10 A Problem 7

2020 AMC 10 A Problem 13

2025 AMC 10 A Problem 14

2025 AMC 10 A Problem 16

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• State Grouping in Symmetric Grids:

Requires structural symmetry, such as the wrap-around topology of a torus where all edges and all corners behave identically. By recognizing that all 4 Edge squares have the same probabilities of reaching a Corner or Center, we reduce a 9-square grid into a simple 3-state system.

• Complementary Probability:

P($\text{Event}$) is the probability of the target outcome occurring. P($\text{Not Event}$) is the probability of it failing to occur. Because the frog stops when she reaches a corner, calculating the probability of reaching a corner in at most 4 steps is exactly the same as 1 minus the probability of surviving 4 steps without ever touching a corner.

• Absorbing States and Path Counting:

Applies when all outward choices from non-absorbing states have equal probability. Counting the whole-number paths that stay strictly in the Center and Edge squares is mathematically equivalent to the fractional probability calculation, but it is often less prone to arithmetic mistakes.

Key Takeaway: Remember This Strategy for Next Time

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