Four bugs on a tetrahedron

The number of ways the bugs can move is $3 \cdot 3 \cdot 3 \cdot 3 = 81$ , since each can choose from $3$ available edges.

In order to not meet on either an edge or a vertex, implies that they move in a cyclic manner. e.g. $A$ goes toward $B$ , $B$ goes toward $C$ , $C$ goes toward $D$ , and $D$ goes toward $A$ .

There are only six ways to do this:

• $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$
• $A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$
• $A \rightarrow C \rightarrow B \rightarrow D \rightarrow A$
• $A \rightarrow C \rightarrow D \rightarrow B \rightarrow A$
• $A \rightarrow D \rightarrow B \rightarrow C \rightarrow A$
• $A \rightarrow D \rightarrow C \rightarrow B \rightarrow A$

Therefore, the probability is $\frac{6}{81} = \frac{2}{27}$

$2+27 = \boxed{29}$

Each bug can choose to move along any three of the available edges, thus the total number of ways is $3 \times 3\times 3\times 3=81$

For no 2 bugs to meet they must move along the contour of a closed polygon of 4 sides.

Such a polygon will be created when we combine any two adjacent faces of the tetrahedron.

this can be done in ${4 \choose 2}$ ways

thus the required probability is $\frac{6}{81}$

which reduces to $\frac{2}{27}$

thus the required solution is $27+2=29$