A frog sitting on the second of three stones jumps to left with probability $0.4$, what is the probability that it will reach the zeroth stone first?

Question

A blind frog sits on the second of three stones in a line. At each minute, it jumps either to left (with probability $0.4$) or right. To the right of third stone there is a lake in which an alligator is waiting for the frog. To the left of the first stone there is a stork waiting for the frog. Find the probability that the stork eats the frog.

I tried to solve this by recurrence relations, but couldn't establish a proper definition of $a_n$. I mean $n$ can denote the stone number that the frogs is sitting, but then I have find $a_0$, and I don't know whether that kind of configurations are allowed.

In another case, I can choose the second stone as my $n=0$, and $n$ can denote, again, the stone numbers, but then I need to include the negative $n$s, which, again, I don't know whether it is eligible, so I'm stuck.

Answer 1

Hint: Consider the sequence $1, p_1, p_2, p_3, 0$ such that each $p_k$ is a weighted average of left and right, with weights $0.4, 0.6$. Solve for $p_2$.

Answer 2

Hint. After two jumps frog can either be eaten by either alligator (with which probability?) or stork (with which probability?), or return to the second stone and make at least two more jumps with the same set of choices with the same probabilities.

Answer 3

The probability is $p^2\frac{1}{1-2p(1-p)}$ for $p=0.4$, then the probability is $\frac{4}{13}$.

I call $P_N({n})$ the probability of being at stage n in step N. With 0 the initial stage and 2 the stork stage. It's easy yo check that $P_{N-2}({0})=2p(1-p)P_{N-4}({0})$ and the solution to that recurrence relation gives that $P_N({2})=p^2(p(1-p))^{\frac{N-2}{2}}$ and summing up all of those gives my result.