A man stands one step away from a cliff. With $$2 \over 3$$ probability, he steps to the right (away from the cliff), and with $$1 \over 3$$ probability, he steps towards the cliff. What is the probability that he escapes the cliff?

The solution for this problem is as follows:

Say $$P_1$$ is the probability of falling off the cliff from 1 step away. $$P_2$$ is the probability of falling off the cliff from 2 steps away.

$$P_1 = {1 \over 3} + {2 \over 3}P_2$$

The solution states that from $$2$$, the paths leading to the cliff (point $$0$$) are going from $$x=2$$ to $$x=1$$, and $$x=1$$ to $$x=0$$. The probability of both of these paths is effectively $$P_1$$ each, as we are simply translating a step to the right. Therefore, $$P_2 = P_1 \cdot P_1$$, because $$P_2 = P( \text{particle goes from} \;x=2\; \text{to} \;x=1) \cdot P(\text{particle goes from} \;x=1\; \text{to} \;x=0)$$. Thus,

$$P_1 = {1\over 3} + {2\over 3} P_1\cdot P_1$$.

Here's my question: Why do we not consider that $$P_2$$ can also go from $$x=2$$ to $$x=3$$? (Then from $$x=3$$, we would also have another chance to go from $$x=2$$, and/or $$x=1$$). Shouldn't $$P_2 = P(\text{particle goes from} \;x=2\; \text{to} \;x=1)\cdot P(\text{particle goes from}\; x=1\; \text{to}\; x=2) + {2\over 3}P(3)$$, or something like that?

• If he steps to the left, does that mean he falls over? Or is he "0 steps" away? Commented Jul 11, 2020 at 21:12
• Going left means he falls over Commented Jul 11, 2020 at 21:29
• James, the obvious words to use are forward and backward. Left and right are just confusing. Commented Jul 11, 2020 at 23:15

We already account for that, when we say P(particle goes from x=2 to x=1) = $$P_1$$. This because $$P_1$$ meant probability to go from x=1 to x=0 after any number of times i.e. considering even the case like 1->2->3->2->1->0. Hence we can directly use this and say probability to go from x=2 to x=1 after any number of times and all cases included is $$P_1$$. Note that $$P_1$$ is not simply $$\frac{1}{3}$$. It is more than that as we consider all the cases. If you are still not satisfied try computing it case by case i.e. probability of reaching the cliff after n steps. I think you will get a binomial expansion like form.
The proposed solution is poorly worded, which I think is what is confusing you. You can think of $$P_1$$ as the probability of ever hitting $$x=0$$ starting from $$x=1$$ and $$P_2$$ as the probability of ever hitting $$x=0$$ starting from $$x=2$$. To go from $$x=2$$ to $$x=0$$, you must first hit $$x=1$$ starting from $$x=2$$, and then you must hit $$x=0$$ starting from $$x=1$$. The probability of hitting $$x=1$$ starting from $$x=2$$ is the same as the probability of hitting $$x=0$$ starting from $$x=1$$. So, if instead of in the proposed solution, you write $$P_2 = P(\text{particle ever hits x=1 starting from x=2}) \cdot P(\text{particle ever hits x=0 starting from x=1}),$$ perhaps the thinking becomes clearer. Both the probabilities on the right-hand side are equal to $$P_1$$.
You can also get the same answer by writing down more equations, as you suggest, but it is a lot messier. In addition to the first equation you wrote, you have for any $$i \geq 2$$, $$P_{i} = \frac13 P_{i-1} + \frac23 P_{i+1},$$ and you can use this and your first equation to get $$P_i$$ for any $$i \geq 1$$. In general, if the probability of stepping away from the cliff is $$p$$, and the probability of stepping towards it is $$q = 1-p$$, then $$P_i = (q/p)^i$$ for $$i \geq 1$$, which in your problem gives $$P_1 = \frac13/\frac23 = \frac12$$. See the gambler's ruin problem for details, which is the general version of this problem.
With that said, the first solution is much simpler if you only need $$P_1$$ and not $$P_n$$ for general $$n \geq 1$$. That solution is hopefully clear if you reinterpret the probabilities as mentioned above.