迷路探险家

The Lost Explorer’s Dilemma

专题: Probability / 概率
难度: L6
来源: QuantQuestion

题目详情

探险家在一条直线上，两个出口分别位于位置 $0$ （陷阱）和 $N$ （自由）。

探险家从位置 $k$ 出发（ $0<k<N$ ）。每一步：

以概率 $p$ 向北（+1），
以概率 $1-p$ 向南（-1）。

到达 $N$ 则逃脱，到达 $0$ 则落入陷阱。

问：从 $k$ 出发逃脱的概率 $P_k$ 是多少？

An explorer is trapped in a mysterious labyrinth with two exits. One exit, located $N$ steps to the north, leads to freedom. The other exit, $N$ steps to the south, leads to a pit. The explorer starts $k$ steps north of a reference point (i.e., between the two exits).

At each decision point, the explorer moves:

North with probability $p$
South with probability $1 - p$

If the explorer reaches the freedom point at position $N$ , they escape. If they reach the pit at position $0$ , they fall in.

What is the probability $P_k$ that the explorer escapes, given that they start $k$ steps north of the reference point?

解析

满足递推

P_k=pP_{k+1}+(1-p)P_{k-1},\quad P_0=0,\ P_N=1.

解得：

若 $p=\frac12$ ，则 $P_k=\frac{k}{N}$ 。
若 $p\ne\frac12$ ，则

P_k=\frac{1-\left(\frac{1-p}{p}\right)^k}{1-\left(\frac{1-p}{p}\right)^N}.

Original Explanation

Let $P_k$ be the probability that the explorer reaches freedom starting from position $k$ .

The recurrence relation is:

P_k = p \cdot P_{k+1} + (1 - p) \cdot P_{k-1}

Boundary conditions:

$P_0 = 0$ (falls into the pit)
$P_N = 1$ (reaches freedom)

🔁 Solution to the recurrence

When $p \ne \frac{1}{2}$ , the solution is:

P_k = \frac{1 - \left( \frac{1 - p}{p} \right)^k}{1 - \left( \frac{1 - p}{p} \right)^N}

When $p = \frac{1}{2}$ (i.e., a fair chance), it simplifies to:

P_k = \frac{k}{N}

✅ Final Answer:

If $p = \frac{1}{2}$ , then $P_k = \frac{k}{N}$
If $p \ne \frac{1}{2}$ , then:

P_k = \frac{1 - \left( \frac{1 - p}{p} \right)^k}{1 - \left( \frac{1 - p}{p} \right)^N}

So, $P_k$ is the probability that the explorer escapes, starting $k$ steps north of the reference point.