Let the probability of falling into the pool on the \(k^\text{th}\) step be \(f_k\). By definition, the probability of taking a step toward the pool is \(\frac{1}{3}\), and thus when starting 1 step away from the pool, \(f_1 = \frac{1}{3}\). For this section and going foreward, we mostly will focus on a general case where the \(\mathbb{P}(\text{stepping toward the pool}) = p\)
Now, since we no longer need to deal with even numbers of steps per Lemma 1, we will institute a change of notation; let \(n=2k+1\), where \(n\) is the \(n^\text{th}\) instance of falling into the pool after \(2k+1\) steps and let \(F_n = f_{2k+1}\) for more convenient notation. This would imply that \(f_1\), the probability of falling in on the first step, is equivalent to \(F_0\).
Next, falling in requires two pieces of information; the number of paths that lead to falling in, and the number of steps in that path given the probability of those steps taking place. We will first tackle the latter consideration. Consider that at the \(n^\text{th}\) opportunity to fall that falling in could occur would require that \(n\) steps were taken away from the pool, and \(n+1\) steps were taken toward the pool. Given the probability of stepping toward the pool is \(p\), then the probability of falling in on a single path at the \(n^\text{th}\) opportunity to fall is \[ p^{n+1}(1-p)^{n} = p\big(p(1-p)\big)^n \]
If this were all we had to calculate, we could deal with this problem with out too much hastle. However, there might be more than one way to enter the pool. Consider that if the robot entered the pool after taking 7 total steps (cooresponding to \(n=3\)) it might have steped away three times and then toward the pool four times (AAATTTT), or it could have gone away, toward, away, toward, away, toward, and finally toward (ATATATT); so I know that there are at least 2 ways to fall into the pool in 7 steps. The next activity explores this idea further.
This would imply that falling in at the \(n^\text{th}\) opportunity to fall, \(F_n\), will be \(C_n\cdot p \big(p(1-p)\big)^n\) where \(C_n\) is the number of paths that lead to the pool's \(n^\text{th}\) opportunity to fall. Figuring out the values of \(C_n\) will take the most of the Background & History section,, but in the meantime we now have the tools to do define what the exact probability looks like.
If we add up all of the possible ways the robot could fall in, we find the total probability of falling in. That probability is: \begin{align*} \mathbb{P}(\text{falling in} | p) &= \mathbb{P}(\text{falling in on the 1st step} | p)\\ &\qquad + \mathbb{P}(\text{falling in on the 3rd step} | p) + \mathbb{P}(\text{falling in on the 5th step} | p) + \dots \\ &= p\cdot C_0 + p^2(1-p)\cdot C_1 + + p^3(1-p)^2\cdot C_2 + \dots\\ \mathbb{P}(\text{falling in} | p) &= \sum_{n=0}^{\infty} C_n\cdot p \big(p(1-p)\big)^n \end{align*}