SOLUTION: Adam the ant starts at (0,0). Each minute, he flips a fair coin. If he flips heads, he moves 1 unit up; if he flips tails, he moves 1 unit right. Betty the beetle starts at (1

Question 1210227: Adam the ant starts at (0,0). Each minute, he flips a fair coin. If he flips heads, he moves 1 unit up; if he flips tails, he moves 1 unit right.

Betty the beetle starts at (1,1). Each minute, she flips a fair coin. If she flips heads, she moves 1 unit down; if she flips tails, she moves 1 unit left.

If the two start at the same time, what is the probability that they meet while walking on the grid?

Enter your answer as a fraction in simplest form.

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Solution:
Let Adam's position at time $t$ be $(A_x(t), A_y(t))$ and Betty's position at time $t$ be $(B_x(t), B_y(t))$.
Adam starts at $(0,0)$ and moves up or right with equal probability each minute.
Betty starts at $(1,1)$ and moves down or left with equal probability each minute.
For them to meet at time $t$, their positions must be the same: $A_x(t) = B_x(t)$ and $A_y(t) = B_y(t)$.
Let $U_A(t)$ be the number of heads (up moves) Adam makes in $t$ minutes, and $R_A(t)$ be the number of tails (right moves). Then $U_A(t) + R_A(t) = t$.
Adam's position at time $t$ is $(R_A(t), U_A(t))$.
Let $D_B(t)$ be the number of heads (down moves) Betty makes in $t$ minutes, and $L_B(t)$ be the number of tails (left moves). Then $D_B(t) + L_B(t) = t$.
Betty's position at time $t$ is $(1 - L_B(t), 1 - D_B(t))$.
For them to meet at time $t$:
$R_A(t) = 1 - L_B(t) \implies R_A(t) + L_B(t) = 1$
$U_A(t) = 1 - D_B(t) \implies U_A(t) + D_B(t) = 1$
Since $R_A(t), L_B(t), U_A(t), D_B(t)$ are non-negative integers, the only possible solutions are:
Case 1: $R_A(t) = 1, L_B(t) = 0$ and $U_A(t) = 0, D_B(t) = 1$. This implies $t=1$.
Adam's position at $t=1$: $(1, 0)$ (moved right). Probability = 1/2.
Betty's position at $t=1$: $(1-0, 1-1) = (1, 0)$ (moved down). Probability = 1/2.
Probability of meeting at $(1,0)$ at $t=1$ is $(1/2) \times (1/2) = 1/4$.
Case 2: $R_A(t) = 0, L_B(t) = 1$ and $U_A(t) = 1, D_B(t) = 0$. This implies $t=1$.
Adam's position at $t=1$: $(0, 1)$ (moved up). Probability = 1/2.
Betty's position at $t=1$: $(1-1, 1-0) = (0, 1)$ (moved left). Probability = 1/2.
Probability of meeting at $(0,1)$ at $t=1$ is $(1/2) \times (1/2) = 1/4$.
The probability that they meet at time $t=1$ is $1/4 + 1/4 = 1/2$.
For them to meet at a later time $t > 1$, the total number of steps taken by Adam to the right and Betty to the left must be 1, and the total number of steps taken by Adam up and Betty down must be 1. This is only possible at $t=1$.
The probability that they meet while walking on the grid is the probability that they meet at time $t=1$.
Final Answer: The final answer is $\boxed{\frac{1}{2}}$
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