This is a classic Markov chain problem. Recall that in the Markov chain section of the mathematics course we found the expected time to reach the absorbing state (the opposing vertex) from some other state $i$ is $$ t_i = 1 + \sum_{k} t_k p_{ik} $$ where t_k is the expected time to reach the absorbing state from some other connected state $k$, and $p_ik$ is the probability of transition from state $i$ to state $k$. $$ $$ Let us then define the states. We begin on the starting vertex, which is 3 edges away from the opposing vertex. From this vertex we will always travel to a vertex which is 2 edges away from the opposing vertex. $$ t_3 = 1 + t_2 $$ From the state $t_2$, there are two neighbouring vertices which are 1 edge away and there is the edge which returns us to our starting vertex. As a result, $$ t_2 = 1 + \frac{2}{3} t_1 + \frac{1}{3} t_3. $$ Finally, from the vertex which is 1 edge away from the opposing vertex, there are two neighbouring vertices which are 2 edges away, and one edge which will lead us to the opposing vertex. Therefore, $$ t_1 = 1 + \frac{2}{3} t_2 + \frac{1}{3} t_0. $$ Since $t_0$ is the time to reach the absorbing state from the absorbing state, it is equal to zero. We solve these simultaneous equations for $t_3$ and obtain the result $$ t_3 = 10. $$ It takes the ant on average 10 seconds to reach the opposing vertex.