Nine tiles in a ring, escaping through tile 0

Question

Nine tiles form a ring, numbered 0 to 8. Tiles 0 and 5 are exits. A beetle starts on tile 2 and each step moves to an adjacent tile, either direction with equal chance.

What is the probability the beetle exits at tile 0 rather than tile 5?

Accepted Answer

The exits split the ring into the arc 0, 1, 2, 3, 4, 5 (length 5) and the arc through tiles 6, 7, 8 (length 4). The beetle starts at tile 2, on the first arc, and cannot cross an exit to reach the other side. So the problem is fair gambler's ruin on the line from 0 to 5, starting at 2. The chance of reaching 5 first is 2/5, so the chance of exiting at 0 is 1 - \frac25 = \frac35. Takeaway. Identify the arc holding the start (here 0 to 5) and run gambler's ruin on it: exit-at-0 probability is (distance to 5) over (length) = 3/5.

Nine tiles in a ring, escaping through tile 0

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