What is the chance two bugs will meet within X timesteps given specific movement patterns?
Two bugs are walking on a line in random fashion and are 10 cm apart from each other. At each time step, each of them has a 1/2 chance to move a centimetre to the left, and the same chance to move a centimetre to the right.
The question is what is the chance that the bugs have either passed through the same point or have met after 7 time steps? They do not have to be at the same place at the same time, but their paths must intersect at some point. They also do not have to meet at exactly the 7th time step, as long as their paths intersect at any point in time before or at the 7th time step.
My initial instict was (1/2)^7 * (1/2)^7 = 1/16384 ≈ 0.00006 but that doesn't factor in the distances. And frankly the number seems a bit lower than I might expect.
Note: this answer requires that the bugs are at the same place at the same time. OP has clarified that passing through the same point at different times also counts as success, so the chance is higher.
You can fix one bug, say the left one, and have the other stay put with probability $1/2$ and move left or right $2$ cm with probability $1/4$ each. They meet if either
- the bug takes five steps left in a row
- the bug takes six steps, five left and one stationary, but we cannot count all left first as we got that in 1
- the bug takes seven steps, six left and one right, but not the first five left
- the bug takes seven steps, five left and two stationary, with the stationary ones among the first six
This gives $$\frac 1{4^5}+5\frac 1{2\cdot 4^5}+5\frac 1{4^7}+{6 \choose 2}\frac 1{2^2\cdot 4^5}=\frac {121}{16384}\approx 0.00739$$