A frog on a ring of stones, which exit?

Question

Twelve stones sit in a circle, numbered 0 to 11. Stones 0 and 8 are exits: if the frog lands on either, it leaves. A frog starts on stone 3 and each minute hops to an adjacent stone, clockwise or counterclockwise with equal chance.

What is the probability the frog leaves through stone 0 rather than stone 8?

Accepted Answer

The two exits cut the ring into two arcs. The frog starts at stone 3, which lies on the arc 0, 1, 2, \dots, 8. It can never reach the other arc (stones 9, 10, 11), because getting there would mean first stepping onto exit 0 or exit 8, and then it is gone. So only the arc from 0 to 8 matters, and on that arc the frog does an ordinary fair random walk with absorbing ends. That is exactly gambler's ruin on a line from 0 to 8, starting at 3. The chance of reaching the far end 8 first is (distance from 0) divided by (length) = 3/8. So the chance of leaving through 0 is 1 - \frac38 = \frac58. Takea…

A frog on a ring of stones, which exit?

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