Problems(1)
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 21), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability 31). What is the probability that when The Game ends Zuca hasn‘t lost?Proposed by Edwin ZhaoSolution. 9029
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 103. They will all be on the same parity with probability 101.At this point there are 2⋅2⋅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
103+92⋅101=9029
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