What is the probability of arrive either A or B at starting point K?

Question

There are two points which are $A$ and $B$. The distance between $A$ and $B$ is $50$ meter. One person goes to $A$ with probability $\frac{1}{6}$, he goes to $B$ with probability $\frac{3}{6}$. And he goes nowhere with probability $\frac{2}{6}$. If he arrives either point $A$ or point $B$, then game over. The distance between his location (point $K$) and the point $A$ is $15$ meter. And the distance between his location (point $K$) and the point $B$ is $35$ meter. What is the probability of arrive either $A$ or $B$ at starting point $K$?

Answer 1

Model: Set $A:=0$, $B:=50$. Let $\xi_j \sim \frac{1}{6} \delta_{-1} + \frac{1}{3} \delta_0 + \frac{1}{2} \delta_1$, $j \geq 1$, be independent identically distributed random variables. Then

$$X_n := k + \sum_{j=1}^n \xi_j$$

is the position of the person after $n$ steps if the person starts at $X_0=k$. We would like to calculate the probability that the hitting time

$$\tau := \inf\{n \in \mathbb{N}; X_n =0 \, \text{or} \, X_n =50\}$$

is finite, i.e.

$$\mathbb{P}^k (\tau<\infty);$$

Calculations: Problems of this type are usually solved by deriving a difference equation. To this end, we set

$$v_k := \mathbb{P}^k(\tau<\infty), \qquad k \in \{0,\ldots,50\},$$

i.e. $v_k$ is the probability that the person arrives at $A=0$ or $B=50$ if she starts at $X_0=k$. It follows from the Markov property of $(X_n)_{n \in \mathbb{N}}$ that

$$v_k = \frac{1}{6} v_{k-1} + \frac{1}{3} v_k + \frac{1}{2} v_{k+1}. \tag{1}$$

Moreover, we have

$$v_0 = \mathbb{P}^0(\tau<\infty)=1 \qquad \quad v_{50} = \mathbb{P}^{50}(\tau<\infty)=1. \tag{2}$$

Equation $(1)$ is equivalent to

$$v_{k+1}-v_k = \frac{1}{3} (v_k-v_{k-1}).$$

Iteration yields

$$v_{k+1}-v_k = \frac{1}{3^k} (v_1-v_0). \tag{3}$$

On the other hand, by $(2)$,

$$0 = v_{50}-v_{0} = \sum_{k=0}^{49} (v_{k+1}-v_k) \stackrel{(3)}{=} (v_1-v_0) \sum_{k=0}^{49} \frac{1}{3^k}.$$

Since $\sum_{k=0}^{49} \frac{1}{3^k} >0$, this proves

$$v_1-v_0 =0,$$

i.e. $v_1 = v_0 \stackrel{(2)}{=} 1$. Now it follows from $(3)$ that $v_k = 0$ for all $k \in \{0,\ldots,50\}$. This means that the person arrives at $A$ or $B$ (with probability $1$) in any case.