Loop of 20, reaching gate 10 before gate 0

Question

Twenty nodes form a ring, numbered 0 to 19. Nodes 0 and 10 are absorbing gates. A token starts at node 7 and each step moves to an adjacent node, either direction with equal chance.

What is the probability the token reaches gate 10 before gate 0?

Accepted Answer

The two gates cut the ring into two arcs. The token starts at node 7, on the arc 0, 1, 2, \dots, 10, and cannot cross a gate to reach the far arc (nodes 11 through 19). So the problem is fair gambler's ruin on a line from 0 to 10, starting at 7. The chance of reaching the far end 10 before 0 is (distance from 0) over (length): \frac{7}{10} = 0.7. Takeaway. On the reduced arc from 0 to 10, reaching node 10 first has probability 7/10; starting nearer a gate makes that gate the likely exit.

Loop of 20, reaching gate 10 before gate 0

Your answer

Discussion

Learn the concepts

Related questions