
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Bamal, Halvan, and Zuca are playing *The Game*. 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 *The Game* ends Zuca hasn‘t lost?

*Proposed by Edwin Zhao*

<details><summary>Solution</summary>*Solution.*  $\boxed{\dfrac{29}{90}}$ 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  $$ \dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}} $$</details>

Present the answer in LaTex format: \boxed{Your answer}
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