Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular
hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost?

[i]Proposed by Edwin Zhao[/i]

[hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$
Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other
two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$.

At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is
$$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$
[/hide]
1. **Initial Setup and Parity Argument**:
   - Consider the vertices of the hexagon $ABCDEF$ colored alternately in black and white.
   - Since the players start at random distinct vertices, there are two possible scenarios:
     - All three players start on vertices of the same color (same parity).
     - One player starts on a vertex of a different color (different parity) than the other two.

2. **Probability of Different Parity**:
   - The probability that Zuca starts on a vertex of different parity from Bamal and Halvan is $\frac{3}{10}$.
   - The probability that all three players start on vertices of the same parity is $\frac{1}{10}$.

3. **Game Dynamics**:
   - If Zuca is on a different parity vertex, he will always remain on a different parity vertex because he can only move to adjacent vertices or the opposite vertex, which are all of the same parity.
   - Bamal and Halvan, moving to adjacent vertices, will also remain on vertices of the same parity.

4. **Collision Analysis**:
   - If Zuca is on a different parity vertex, he will never collide with Bamal or Halvan, so he wins with probability $\frac{3}{10}$.
   - If all three players are on the same parity, we need to analyze the possible moves:
     - There are $2 \cdot 2 \cdot 3 = 12$ possible moves (2 for Bamal, 2 for Halvan, and 3 for Zuca).
     - Out of these 12 moves, 3 will lead to the same arrangement, so we disregard those, leaving 9 possible moves.
     - Among these 9 moves, Zuca wins in exactly 2 cases (when Bamal and Halvan collide and Zuca moves to a neighboring vertex).

5. **Combining Probabilities**:
   - The probability that Zuca wins when all three players are on the same parity is $\frac{2}{9}$.
   - Combining the probabilities:
     \[
     \text{Total probability that Zuca wins} = \frac{3}{10} + \left(\frac{2}{9} \cdot \frac{1}{10}\right) = \frac{3}{10} + \frac{2}{90} = \frac{27}{90} + \frac{2}{90} = \frac{29}{90}
     \]

The final answer is $\boxed{\frac{29}{90}}$