1

$n>1$ points numbered $0,1,2,...,n-1$ are placed on a circle.

A random walker starts his journey at point $0$ and at each step, he steps randomly on the circle to one of the two closest points.

For $0 \le i \le n-1$, we'll say that $p_i$ is the probability that the moment this man reaches the point $i$, he already visited the rest of the points. Meaning, he traveled all rest the points already and now he finaly visited the point $i$ for the first time. And that's where his journey comes to an end.

So, what is $p_i$ ?

The answer is:

$$p_i = \begin{cases} \frac{1}{n-1}, & \text{for $i\neq0$} \\ 0, & \text{for $i=0$} \end{cases} $$

And i don't get it! Does it means that $p_i$ is uniformly distributed or something?

The way i see it, the point $i$ may never even get reach! If some point, let's say point $2$, is placed far from everyone, why would this man reach it at at? he'll go around in the same routine 'till infinity and never decide to go to $2$.

I believe i didn't get the question right. Or maybe i'm wrong with my thinking. I just hope i translated this question correctly so you can help me understand what is going on here or solve this.

RobPratt
  • 50,938
ryden
  • 583
  • The thing is, these routines where he never reaches 2 after a large number of steps have very low probability. – Alan Abraham Jan 10 '22 at 19:25
  • Also I believe the beginning of Fnacool's answer in https://math.stackexchange.com/questions/4273163/simple-random-walk-on-cycle-graph-ending-on-specific-vertex-after-cover-time/4274257#4274257 should help you in reaching the formula for $p_i$. – Alan Abraham Jan 10 '22 at 19:29
  • Yes it means $p_i$ is uniform (and that the starting point is visited before the first step - you would get a different answer otherwise, for example consider $n=2$ or $3$). The probability of not visiting all the points is $0$ – Henry Jan 10 '22 at 19:39
  • @DavidC.Ullrich I think you are missing the part that it is on a circle. – Alan Abraham Jan 10 '22 at 19:55
  • @AlanAbraham Oops. Thanks... – David C. Ullrich Jan 10 '22 at 21:13

0 Answers0