One-One Correspondences 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 $(2,4)$. 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?
How do I go about solving this problem, and what's the answer?
 A: Adam and Betty may meet only after three steps from the start, in the points $(2,1),(1,2)$ or $(0,3)$.
After three steps, Adam is in $(2,1)$ with probability $\frac{3}{8}$, in $(1,2)$ with probability $\frac{3}{8}$ and in $(0,3)$ with probability $\frac{1}{8}$.
After three steps, Betty is in $(2,1)$ with probability $\frac{1}{8}$, in $(1,2)$ with probability $\frac{3}{8}$ and in $(0,3)$ with probability $\frac{3}{8}$.
So the probability that Adam and Betty meet is:
$$ \frac{3}{8}\cdot\frac{1}{8}+\frac{3}{8}\cdot\frac{3}{8}+\frac{3}{8}\cdot\frac{1}{8}=\color{red}{\frac{15}{64}}.$$
A: I assume (correctly?) that the question means "what is the probability that, on finishing a round of coin tosses, the two find themselves on the same grid point."  If that's correct, it is not hard to see that they can only meet at three points: (1,2), (2,1), or (0,3). Let's analyze (1,2), as an example.  For Adam to get to (1,2), he needs to throw 2H and 1T in any order, probability 3/8.  For Betty to get there she also needs to throw 2H and 1T in any order, also 3/8 probability.  Getting both has a 9/64 probability.  The other two possibilities can be handled similarly.
