Gracie starts at her house and walks one block east. After each block she walks, she flips a coin. She walks one block east if the coin lands heads; otherwise one block west. She stops her walk the first time she lands back home. What is the probability she stops her walk after walking exactly eight blocks? Express your answer as a common fraction.

1 answer

Let \( p_n \) be the probability that Gracie returns home after exactly \( n \) steps. We want to compute \( p_8 \).

First, the possible scenarios for Gracie's movement need to be established. If Gracie walks \( n \) blocks, and she is back home, she must take an equal number of steps east and steps west. Therefore, out of \( n \) steps, let \( k \) be the number of steps east, and thus \( n - k \) must be the same (steps west). This gives us the equation: \[ k = \frac{n}{2} \] which indicates that \( n \) must be even. For \( n = 8 \), we have \( k = 4 \).

To find the total number of valid sequences of 8 steps that lead back to the origin, we need to count how many ways we can arrange 4 heads (east) and 4 tails (west) in 8 flips of the coin. The number of such arrangements is given by the binomial coefficient: \[ \binom{8}{4} \]

Calculating this: \[ \binom{8}{4} = \frac{8!}{4!4!} = \frac{40320}{24 \times 24} = 70 \]

Next, we also need to consider the total number of possible paths she could take in 8 steps, where each step can either be east or west. The total number of sequences of 8 steps is: \[ 2^8 = 256 \]

Thus, the probability that she returns home after exactly 8 steps is the number of successful paths divided by the total number of paths: \[ p_8 = \frac{\binom{8}{4}}{2^8} = \frac{70}{256} \]

Now we can simplify \( \frac{70}{256} \): The greatest common divisor of 70 and 256 is 2: \[ \frac{70 \div 2}{256 \div 2} = \frac{35}{128} \]

Therefore, the probability that Gracie stops her walk after exactly 8 blocks is: \[ \boxed{\frac{35}{128}} \]