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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?

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

2 Answers 2

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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.

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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.

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