At each vertex of a tetrahedron, there sits an ant. At a given moment, each of them starts to move along a randomly chosen edge and crosses over to the adjacent vertex. What is the probability that two ants meet either on the way or at the end of their journey?
Solution. Let's number the vertices of the tetrahedron from 1 to 4. A position change can be represented by a sequence of four numbers from 1 to 4. For example, the sequence $(2,3,2,1)$ means that the ant at vertex 1 moves to vertex 2, the ant at vertex 2 moves to vertex 3, the ant at vertex 3 moves to vertex 2, and the ant at vertex 4 moves to vertex 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_285e0b62e7c742f165e1g-1.jpg?height=301&width=369&top_left_y=337&top_left_x=867)

The total number of possible position changes is $3^{4}=81$, since each ant can move in 3 different directions. Let's count how many of these position changes result in the ants not meeting. This happens if each ant moves to a different vertex (so they won't meet at the vertices), which means that the sequence of numbers contains all different numbers (e.g., $(2,3,4,1)$ is such a sequence). To ensure that the ants do not meet on any edge, it is necessary that no two ants move towards each other. For example, if the ant at vertex 1 moves to vertex 2 (i.e., the first number in the sequence is 2), then to avoid meeting, the ant at vertex 2 cannot move to vertex 1 (i.e., the second number in the sequence cannot be 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_285e0b62e7c742f165e1g-1.jpg?height=301&width=371&top_left_y=1026&top_left_x=866)

The following sequences satisfy both conditions:

$$
(2,3,4,1) ; \quad(2,4,1,3) ; \quad(3,1,4,2) ; \quad(3,4,2,1) ; \quad(4,3,1,2) ; \quad(4,1,2,3)
$$

Out of the 81 position changes, 6 result in the ants not meeting. Therefore, the probability that two ants meet is $\frac{75}{81}=\frac{25}{27}$.