1D random walk-probability to go back to origin

Question

Suppose there is a random walk starting at the origin such that the probability to move right is $\frac13$ and the probability to move left is $\frac23$.

What is the probability to return to the origin?

Answer 1

Let $P_{i\ge 0}$ be the probability of ever reaching position $x+i$ when starting from position $x$ (this is independent of $x$, since the transition probabilities are). Clearly $P_0=1$, and $P_i\rightarrow 0$ as $i\rightarrow\infty$ provided that the right-hop probability $q < 1/2$ (in this case $q = 1/3$). Otherwise, $$ P_i = qP_{i-1} + (1-q)P_{i+1}, $$ You can guess the solution to be of the form $P_i = \alpha^i$ for some $0 < \alpha < 1$. This turns out to satisfy the conditions if $$ \alpha = q + (1-q) \alpha^2, $$ which has the solution $\alpha = q/(1-q)$ in this case.

For the problem specified, you want to know the total probability of returning to the origin after the first step. If the first step is to the right (which happens with probability $q$), then you must return to the origin; if it is to the left (with probability $1-q$), then you will return to the origin with probability $P_1 = \alpha = q/(1-q)$. So the solution is $$ P_{\text{return}} = q + (1-q)\alpha = q + (1-q)\frac{q}{1-q} = 2q $$ for general $q<1/2$, and $P_{\text{return}} = 2/3$ in this case.

Answer 2

You can find a detailed answer for your question on WolframMathWorld.

To give an exact answer, you should define the number of steps you are interested in returning to the origin. Of course the number of steps should be even.

I have made a little "simulation" (rather: test run) in R:

results <- NULL
for (i in 1:50) {
    N=i*2
    n1=N/2
    p=1/3
    q=1-p
    results <- c(results, ((factorial(N)/(factorial(n1)*factorial(N-n1)))*(p^n1)*(q^(N-n1))))
}

And got the following results (sorry for ugly table):

  N| probability
---+--------------
  1 0.4444444444
  2 0.2962962963
  3 0.2194787380
  4 0.1707056851
  5 0.1365645481
  6 0.1112748170
  7 0.0918458807
  8 0.0765382339
  9 0.0642543198
 10 0.0542592034
 11 0.0460381120
 12 0.0392176509
 13 0.0335193598
 14 0.0287308798
 15 0.0246872745
 16 0.0212584864
 17 0.0183406549
 18 0.0158499487
 19 0.0137180842
 20 0.0118890063
 21 0.0103163865
 22 0.0089617094
 23 0.0077927908
 24 0.0067826142
 25 0.0059084106
 26 0.0051509221
 27 0.0044938086
 28 0.0039231662
 29 0.0034271337
 30 0.0029955687
 31 0.0026197805
 32 0.0022923080
 33 0.0020067342
 34 0.0017575319
 35 0.0015399328
 36 0.0013498176
 37 0.0011836238
 38 0.0010382665
 39 0.0009110715
 40 0.0007997183
 41 0.0007021917
 42 0.0006167398
 43 0.0005418386
 44 0.0004761612
 45 0.0004185515
 46 0.0003680018
 47 0.0003236328
 48 0.0002846770
 49 0.0002504641
 50 0.0002204084

As you can see here, I computed the probabilities of getting back to the original point for all possible steps (N), where N<=100. That is done by computing the probability for all even N, where the number of steps to the right is equal to the number of steps to the left with the following formula: $$P(n_{1}|N)=\frac{N!}{n_{1}!^2}p^{n_{1}}(1-p)^{n_{1}}$$

where N is the number of steps, $n_{1}=\frac{N}{2}$ and $p=\frac{1}{3}$.

Note: the sum of all above is 1.998366. That is higher than 1, as the probability of arriving back to the original point could also include a lower N's probability.

That's all I could do, I like to play with simulations, but not good at all with Markov chains extrapolated to infinite :)

Answer 3

The problem you refer to is the study of return probability for a random walk on a lattice. In your case, the lattice is the 1-D lattice of integers. This problem is very well studied on general lattices. A famous theorem by Polya says that while the probability of return is $1$ for symmetric random walks (i.e., all moves are equally likely unlike in this question) in $1$ and $2$ dimensional lattices, it is strictly less than $1$ in all higher dimensions. See here for more details.

The solution posted by mjqxxxx is very clever and is perfectly valid. If you are interested in other solutions, a very systematic way of studying this problem is through the use of generating functions. See this lecture notes for more information (In fact, these notes have the solution to the problem you posed).