Each of the two ants has 4 edges to choose. This gives a total of 4

^{2}= 16 possible choices. Of those choices, we have to discard the four cases in which the ants choose the same edge, because then they would meet between their starting vertices after walking 1/2 edge each.

We are left with 12 possible choices. This result can also be obtained by noting that if we let one of the ants choose the edge it likes, then the second ant is only left with three possible choices: 4*3 = 12.

After walking their first edge, in 4 of the 12 possible cases, the ants find themselves still on opposite vertices, while in the other 8 cases, they are at two ends of the same edge.

Now, if they are at the two ends of an edge, it is for them impossible to meet after walking exactly a further edge without one of them having to walk back to its vertex of origin. As the question states that the ants never double back, these possibilities have to be discarded.

As a result of the considerations above, the only cases that satisfy all the conditions are those in which the ants move from a pair of opposite edges to a different pair of opposite edges. The probability is 4/16 = 1/4.

In its new position, each ant can choose one of 3 possible edges (not 4, because ants never double back). Therefore, there are 3

^{2}= 9 possibilities. Of these, only 2 lead to a meeting satisfying the required conditions: those in which both ants walk to one of the two vertices that were so far unoccupied by either one of them. The probability is therefore 2/9.The total probability is given by the product of the probabilities calculated for the two edges: P = 1/4 * 2/9 = 1/18 = 0.0555 .