Wed 13 January 2016

Drunk guy by a cliff

Written by Hongjinn Park in Articles

There's a drunk guy standing next to a cliff which is one step to his left. He steps to the left with probability $1-p$ and to the right with probability $p$. What is the probability that he dies?

Let $k$ be the position of the guy, where death is at $k=0$. Let $D_k$ be the event that he dies given that he is at position $k$. Therefore $P(D_0) = 1$ and

$$P(D_1) = P(D_1 \mid \text{Step left}) P(\text{Step left}) + P(D_1 \mid \text{Step right}) P(\text{Step right})$$

and we note that $P(D_1 \mid \text{Step left}) = P(D_0) = 1$, so

$$P(D_1) = (1-p)P(D_0) + pP(D_2) = 1-p + pP(D_2) $$

At this point we cleverly realize that $P(D_2) = P(D_1)^2$ and letting $x = P(D_1)$ we have

$$x = (1-p) + px^2$$ $$0 = px^2-x + 1 - p$$

and using the quadratic formula,

$$x = \frac{1 \pm \sqrt{1-4p(1-p)}}{2p} = \frac{1 \pm \sqrt{1-4p+4p^2}}{2p} = \frac{1 \pm \sqrt{(1-2p)^2}}{2p} = \frac{1 \pm (1-2p)}{2p}$$ $$x = \frac{2-2p}{2p} = \frac{1-p}{p} \quad \text{or} \quad x = \frac{2p}{2p} = 1$$

Since $x = P(D_1) \in [0,1]$ and the two solutions agree when $p=1/2$, the probability of death given you are at location $1$ is

If $p = P(\text{Stepping right to safety}) \le \frac{1}{2}$ then $P(D_1) = 1$

If $p = P(\text{Stepping right to safety}) > \frac{1}{2}$ then $P(D_1) = \frac{1-p}{p}$

For example if the probability of stepping to safety is $p=4/5$ which is not that "bad" then the solutions for $P(D_1)$ will equal $0.25$ or $1$ whereas if $p=1/5$ then the solutions for $P(D_1)$ will equal $1$ or $4$.



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