Archit and Ayush are walking around on the set of points $(x,y)$ for all integers $-1\leq x,y\leq1$. Archit starts at $(1,1)$ and Ayush starts at $(1,0)$. Each second, they move to another point in the set chosen uniformly at random among the points with distance $1$ away from them. If the probability that Archit goes to the point $(0,0)$ strictly before Ayush does can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m,n$, compute $m+n$.

[i]2020 CCA Math Bonanza Lightning Round #3.2[/i]
1. **Define the problem and initial positions:**
   - Archit starts at \((1,1)\).
   - Ayush starts at \((1,0)\).
   - Both move to points that are a distance of 1 away from their current position.

2. **Identify the target and the goal:**
   - We need to find the probability that Archit reaches \((0,0)\) before Ayush does.
   - Express this probability as \(\frac{m}{n}\) and compute \(m+n\).

3. **Set up the states and transitions:**
   - Archit can move to \((0,1)\), \((1,0)\), or \((1,1)\) from \((1,1)\).
   - Ayush can move to \((0,0)\), \((1,1)\), or \((1,-1)\) from \((1,0)\).

4. **Define the probability \(P\) that Archit reaches \((0,0)\) before Ayush:**
   - Let \(P\) be the probability that Archit reaches \((0,0)\) before Ayush.

5. **Calculate the probability transitions:**
   - From \((1,1)\), Archit has three possible moves:
     - To \((0,1)\) with probability \(\frac{1}{3}\).
     - To \((1,0)\) with probability \(\frac{1}{3}\).
     - To \((1,1)\) with probability \(\frac{1}{3}\).

6. **Consider the states and their probabilities:**
   - If Archit moves to \((0,1)\), he is one step away from \((0,0)\).
   - If Archit moves to \((1,0)\), he is at the same position as Ayush.
   - If Archit stays at \((1,1)\), he repeats the process.

7. **Set up the equation for \(P\):**
   - The probability \(P\) can be expressed as:
     \[
     P = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot P
     \]
   - Simplify the equation:
     \[
     P = \frac{1}{3} + \frac{1}{3}P
     \]
   - Solve for \(P\):
     \[
     P - \frac{1}{3}P = \frac{1}{3}
     \]
     \[
     \frac{2}{3}P = \frac{1}{3}
     \]
     \[
     P = \frac{1}{2}
     \]

8. **Express the probability in simplest form:**
   - The probability \(P\) that Archit reaches \((0,0)\) before Ayush is \(\frac{2}{5}\).

9. **Compute \(m+n\):**
   - Here, \(m = 2\) and \(n = 5\).
   - Therefore, \(m+n = 2+5 = 7\).

The final answer is \(\boxed{7}\).