Was working on this problem wherein two random walkers starting at $0$ and $4$ respectively take ten steps (either $+1$ steps or $-1$ steps with equal likelihood). What is the probability that they will end up in each of their original spots and have never met along the way?
The probability that they will get back to their original spot is : $\frac{{10 \choose 5} {10 \choose 5}}{4^{10}}$ but I can't figure out in what fraction of the paths intersect. (Also if they meet but don't cross it is still considered an intersection)
I tried with the case where the steps are $4$ and $6$.
For $4$ steps there is only one scenario where they cross, so the total paths is: $\frac{{4 \choose 2} {4 \choose 2} - 1}{4^{4}}$
When I get to $6$ steps is when I get lost. Of the total ${6 \choose 3} {6 \choose 3}$ possible paths which will get them back to their respective origins, the number of of those potential paths wherein they meet after $2$ steps is ${4 \choose 1} {4 \choose 1}$ but then I need an expression for the paths where they meet after 3 steps and 4 steps (notice they can't meet at the 5th step if they are to make it back to their original points) and what I am really struggling with is double counting. As in how many paths do they meet at the third step where they didn't meet at the 2nd. And likewise how many paths do they met at the 4th step where they didn't meet at the 3rd or 2nd.
Since $6$ is not too large a number I can count these by hand, but I am having difficulty coming up with a general expression.
Any help would be appreciated