The following question is taken from the book titled "Probability models for Computer Science" written by Sheldon M. Ross.


A particle moves along n + 1 vertices that are situated on a circle in the following manner. At each stage the particle moves one step that is either in the clockwise direction with probability p or in the counter-clockwise direction with probabiity 1-p.

Starting at a specified state let T denote the time of the first return to that state. Now how to find the probability that the particle has visited all the vertices by time T

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    – D.W.
    Jan 7, 2022 at 21:58

2 Answers 2


First suppose that $p \ne 1/2$, and let $q=1-p$. If the first step is clockwise, then the problem is a gambler's ruin problem, and the chance of success (visiting all nodes) is $$\frac{1-q/p}{1-(q/p)^{n+1}} \,,$$ See [1] or [2], section 17.3 or [3]. If the first step is counterclockwise, then the chance of success is $$\frac{1-p/q}{1-(p/q)^{n+1}} \,.$$ Thus overall, the probability that the particle has visited all the vertices by time $T$ is $$p\frac{1-q/p}{1-(q/p)^{n+1}}+q \frac{1-p/q}{1-(p/q)^{n+1}} \,. \tag{*}$$

If $p=1/2$ then again there is a reduction to a gambler's ruin problem, so by [1] or [2] Section 2.1, or [3], the answer is 1/(n+1). This answer can also be obtained by taking the limit as $p \to 1/2$ in $(*)$.

[1] Feller, W. 1968. An introduction to probability theory and its applications, third edition, Vol. 1, Wiley, New York. Chapter XIV

[2] https://www.yuval-peres-books.com/markov-chains-and-mixing-times/

[3] Grinstead, C. and L. Snell. 1997. Introduction to Probability, 2nd revised edition, American Mathematical Society, Providence. Chapter 12.


Here is the idea. Let the vertices be $v_0,\ldots,v_n$, where $v_0$ is the origin and the vertices are arranged in clockwise order. The first step is either clockwise or counterclockwise — let's say clockwise. Then the probability you're after is the probability that you reach $v_n$ before you reach $v_0$, starting at $v_1$. This is random walk on a line, so you should be able to analyze it.


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