In a $3\times3$ grid, what's the probability that they will meet? (Solution Verification) We recently had this question in our math exam:

In a $3\times3$ grid, Alice walks from the top left of the grid to the bottom right using the edges of the squares, using the shortest path. (She only moves right or down.) Similarly, with the same speed, Bob walks from the bottom right to the top left also using the edges and taking the shortest path. What is the probability that they will meet during their travels?

My Solution: To find the probability, we need to find the ratio
$$ \frac{\text{(The cases in which they meet at a point)}}{\text{(All cases)}}. $$
The total number of cases should be $20 \cdot 20 = 400$ since Alice has $20$ paths that she can take, and so does Bob. since their walks are independent, we multiply them.
To find the number of cases in which they meet, we first find the points at which they can actually meet. If we denote the bottom right corner as $(0, 0)$, then these points are $(0, 3)$, $(2, 1)$, $(1, 2)$, $(3, 0)$. The number of ways Alice and Bob can visit $(0, 3)$ then complete the path is $1 \cdot 1$ So the total number of cases is $1\cdot 1 = 1$. This also applies for $(3, 0)$.
For the other two points, the ways in which Alice and Bob can visit them and complete their paths are $3 \cdot 3$, and so the total number of cases is $9 \cdot 9 = 81$.
If we add all these up, we get $81 + 81 + 1 + 1 = 164$. And so the probability should be $\frac{164}{400}$.

However, this approach seems to be false. Where did I go wrong?
Edit: I'm aware of a similar question on the site, but I still don't get why this approach is wrong, hence I opened a new question.
 A: Disclaimer: no guarantee that this is correct... It's a suggested approach which is too long for a comment.
In an entire travel, Alice must walk 6 steps, of which 3 must be Right and 3 must be Down. If Alice and Bob meet, it is always after 3 steps. So Alice is allowed to take any Right or Down in those first 3 steps, so she has $2^3=8$ possible paths to a meet with Bob. The same applies for Bob, so there are 64 possible paths before a possible meet. Now we must find out how many of those 64 result in a meet.
Looking at the grid, that is also easy. A meet occurs if $N_D+N_U=3$ (number of Alice going Down and number of Bob going Up) and $N_R+N_L=3$ (number of Alice going Right and number of Bob going Left). But if one is fulfilled, the other is automatically also fulfilled because Alice and Bob make 6 steps together which is the sum of both requirements. So you can make a table with $N_D$ (for Alice) and $N_U$ (for Bob) leading to the required result (sum=3):
N_D    N_U   combinations
-------------------------
0        3        1*1=1 
1        2        3*3=9 
2        1        3*3=9
3        0        1*1=1

For example: $N_D=1$: in how many 'locations' can you put a single $N_D$ in the three steps? That's 3. Also: $N_U=2$ is the same as $N_L=1$, so also 3.
So I get 20 possible paths that lead to a meet, out of 64. Probability is then 20/64=5/16.
