First, will the ants stop at the next vertex, or will they continue to go in the original direction, or will they randomly choose a direction at each vertex and then go forward; Suppose it is the first case.

The second is whether the forward speed of ants is the same. Assuming the same.

Now answer:

Because every ant has two choices at the beginning, there are eight possibilities. If B is on the left of A, there are two possibilities for A and B to meet, namely, A is on the left, B is on the right, and C is on the left or right, that is, the probability of A and B meeting is 1/4, and the probability of ac and bc meeting is 1/4 respectively. The remaining 1/4 means not meeting, which is also two possibilities: all to the left or all to the right.

Assuming that it is the second case of condition one, that is, the ants keep walking, it can be considered that as long as two ants are in different directions, they can meet, so the probability of any two ants meeting is 1/2, and the probability of not meeting is still 1/4. Then why do all possible probabilities add up to more than one? Because in addition to not meeting, one ant will meet the other two ants one after another.

In the third case of condition one, as long as the time is long enough, theoretically each ant can meet at least once.