2022 AMC 10 B Problem 23

Geometric Probability Method

We can separate the problem into two independent parts: finding the probability of taking exactly 2 or 3 steps, and finding the probability that the distance exceeds 1 in that many steps. We will use the areas and volumes of simple geometric shapes to find these probabilities.

Step 1: Understand the condition for the number of steps

Amelia’s decision to stop depends only on the time durations $t_n$. Because $t_1$ is chosen from $(0,1)$, the elapsed time after the first step is $t_1<1$, so she never stops after step 1. She moves on to step 2. After step 2, the total elapsed time is $t_1+t_2$. If $t_1+t_2>1$, she stops at the end of step 2. If $t_1+t_2\le 1$, she continues to step 3. Since she takes at most 3 steps in all, she will always stop at the end of step 3 if she reaches it. Therefore, she either takes exactly 2 steps or exactly 3 steps.

Step 2: Calculate the probability of taking exactly two or three steps

The probability she stops after 2 steps is the probability that $t_1+t_2>1$. Since $t_1$ and $t_2$ are chosen uniformly and independently from $(0,1)$, we can model their outcomes as a random point $(t_1,t_2)$ in a $1\times 1$ square. The region where $t_1+t_2>1$ is a right triangle with vertices $(1,0),(1,1)$, and $(0,1)$. The area of this triangle is half the area of the square.

Step 3: Find the probability her final position is close to the origin given two steps

It is slightly easier to calculate the probability that her distance is $\le 1$, and then subtract from $1$ at the end. If Amelia takes 2 steps, her final position is $x_1+x_2$. The distances $x_n$ are chosen independently of the times $t_n$. Because $x_1$ and $x_2$ are uniform on $(0,1)$, the probability that $x_1+x_2\le 1$ corresponds to the area of the triangle $x_1+x_2\le 1$ inside the unit square.

Step 4: Find the probability her final position is close to the origin given three steps

If Amelia takes 3 steps, her final position is $x_1+x_2+x_3$. We want the probability that $x_1+x_2+x_3\le 1$. In a 3D coordinate system, this region forms a triangular pyramid (tetrahedron) inside the $1\times 1\times 1$ unit cube. The vertices of this tetrahedron are the origin $(0,0,0)$ and the points $(1,0,0),(0,1,0),(0,0,1)$. The base in the $x_1{x}_2$-plane is a right triangle with area $1/2$, and the perpendicular height along the $x_3$-axis is $1$. Because the entire sample space (the unit cube) has a volume of $1$, the volume of this tetrahedron directly gives the probability.

Step 5: Combine the probabilities to find the final answer

We use the Law of Total Probability to find the overall probability that her final position is $\le 1$. We multiply the probability of each step count by the probability her distance is short enough in that many steps, and add them together. Finally, we take the complement to find the probability her position is greater than $1$.

Order Statistics and Symmetry Method

Instead of visualizing 3D volumes, we can transform the sum of independent variables into cumulative sums. The condition that the sum is $\le 1$ becomes the condition that these cumulative sums are strictly ordered and bounded by 1. We can find this probability using simple permutations and symmetry.

Step 1: Transform the sum into cumulative points

Instead of imagining a 3D shape, let’s track Amelia’s position after each step. Let $y_1=x_1$, $y_2=x_1+x_2$, and $y_3=x_1+x_2+x_3$. Because each distance $x_i$ is strictly positive, these new variables must be strictly increasing: $0<y_1<y_2<y_3$. The condition that her total distance after 3 steps is at most $1$ simply becomes $y_3\le 1$.

Step 2: Use symmetry to find the probability for three steps

We need the probability that $0<y_1<y_2<y_3\le 1$. Because the transformation between the independent distances $x_i$ and the cumulative points $y_i$ is a direct one-to-one mapping that only uses addition, it preserves geometric volume. Thus, this probability is equivalent to picking three independent random numbers from $(0,1)$ and having them land in perfectly increasing order. There are $3!=6$ possible ways to order three distinct numbers. By symmetry, all orderings are equally likely. The probability of this specific strictly increasing ordering is $\frac{1}{6}$.

Step 3: Apply the same logic to the two-step case

By the exact same reasoning, if Amelia takes 2 steps, we want $x_1+x_2\le 1$. Letting $y_1=x_1$ and $y_2=x_1+x_2$, we need $0<y_1<y_2\le 1$. This is the probability that two independent random numbers chosen from $(0,1)$ are in increasing order. There are $2!=2$ possible orderings, so the probability is $\frac{1}{2}$.

Step 4: Calculate the final probability

As in the first method, we know the probability of taking 2 steps is $\frac{1}{2}$ and the probability of taking 3 steps is $\frac{1}{2}$. We combine these with our newly found probabilities.