it is well-known that:
For a balanced random walk on a circle with n sites, except for the start point A, all the other points share the same probability: 1/(n-1), to be the last visited site.
Now I'm wondering if I change the scenario from balanced to a biased case?
Say for all sites, the probability of going clockwise is p, counter-clockwise is (1-p).
Can we build a martingale to solve this problem?
Edit: Consider the following 2 different scenarios.
- 100 people sitting around a big table. Everyone wants to try an amazing sauce. Starting from you, everyone will pass the sauce bottle to the left or right next seat with probability $\frac{1}{2}$.
What is the probability that the person next to you, on your left becomes the last one to receive the sauce?
How about the person on you right (next to you)?
How about the person at any chosen seat?
- The same setup and the same questions, but now everyone will pass the bottle to the left with probability p and to the right with probability (1-p).
Many thanks to @user51547! The answer is here:
The idea is the same to a biased Gambler's Ruin.
The only difference is that, in classical Gambler's Ruin, one only needs to consider one direction. Here we need to consider from 2 directions. See the link for detailed explanation.
