The long way around does not count

Question

Sixteen tiles form a ring, numbered 0 to 15. Only tiles 0 and 4 are exits. A walker starts on tile 1 and each step moves to an adjacent tile, either direction with equal chance.

What is the probability the walker exits at tile 0 rather than tile 4?

Accepted Answer

The two exits sit close together, splitting the ring into a short arc (0, 1, 2, 3, 4) and a long arc (tiles 5 through 15). The walker starts on the short arc at tile 1. To reach the long arc it would have to step onto exit 0 or exit 4 first, ending the walk, so the long way around is completely irrelevant. The problem is fair gambler's ruin on the short line from 0 to 4, starting at 1. The chance of reaching 4 first is 1/4, so the chance of exiting at 0 is 1 - \frac14 = \frac34. Takeaway. Only the arc containing the start counts; here that is the short 0-to-4 stretch, giving exit-at-0 probabi…

The long way around does not count

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