HMMT November, 2015

Solution:

At every second, each ant can travel to any of the three vertices they are not currently on. Given that, at one second, the three ants are on different vertices, the probability of them all going to the same vertex is $\frac{1}{27}$ and the probability of them all going to different vertices is $\frac{11}{27}$, so the probability of the three ants all meeting for the first time on the $n^{th}$ step is $\left(\frac{11}{27}\right)^{n-1} \times \frac{1}{27}$. Then the probability the three ants all meet at the same time is $\sum_{i=0}^{\infty}\left(\frac{11}{27}\right)^{i} \times \frac{1}{27}=\frac{\frac{1}{27}}{1-\frac{11}{27}}=\frac{1}{16}$.