Let $P_i$ be the probability that the man falls off the cliff starting from position $i$. We want to find $P_{i_0}$ given that $i_0 \geq 1$. Note that $P_0 = 1$ since the man has already fallen off the cliff when he starts at position $0$. Also note that $P_i = 0$ for all $i \leq 0$ since the man has already fallen off the cliff.

We can now set up a recurrence relation for $P_i$. If the man is currently at position $i$, then he has two options for his next step: he can either move to position $i+1$ with probability $p$, or he can move to position $i-1$ with probability $1-p$. Thus, we have

$$P_i = p P_{i+1} + (1-p) P_{i-1}$$

for all $i \geq 1$. This is a second-order linear recurrence relation with constant coefficients. We can solve it using the characteristic equation:

$$r^2 – (1-p)r – p = 0.$$

The roots of this equation are

$$r_1 = \frac{1-p + \sqrt{(1-p)^2 + 4p}}{2} \quad \text{and} \quad r_2 = \frac{1-p – \sqrt{(1-p)^2 + 4p}}{2}.$$

Note that $r_1 > 1$ and $r_2 < 0$. Therefore, the general solution to the recurrence relation is

$$P_i = A r_1^i + B r_2^i$$

for some constants $A$ and $B$. We can find $A$ and $B$ by using the boundary conditions $P_0 = 1$ and $P_i = 0$ for all $i \leq 0$. This gives us the system of equations

\begin{align*} A + B &= 1, \ Ar_1^i + Br_2^i &= 0 \quad \text{for all } i \leq 0. \end{align*}

Solving for $A$ and $B$, we get

$$A = \frac{r_2}{r_2 – r_1} \quad \text{and} \quad B = -\frac{r_1}{r_2 – r_1}.$$

Therefore, the probability that the man falls off the cliff starting from position $i_0 \geq 1$ is

$$P_{i_0} = \frac{r_2^{i_0}}{r_2^{i_0} – r_1^{i_0}}.$$

Note that this expression makes sense since $r_1 > 1$ and $r_2 < 0$, so $r_1^{i_0}$ grows exponentially while $r_2^{i_0}$ decays exponentially as $i_0$ gets large. This means that $P_{i_0}$ approaches 1 as $i_0$ increases, which makes sense since the man is more likely to fall off the cliff the further he is from it.

Let $S$ be a set of size $n$ and let $k$ be an integer such that $1 \leq k \leq n/2$. We will define a bijection $f$ between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$ that satisfies the given property.

Let $I$ be a $k$-element subset of $S$. We will define $f(I)$ as follows: first, let $J = S \setminus I$ be the complement of $I$ in $S$. Since $|I| = k$, we have $|J| = n-k$. Next, let $T$ be the $k$-element subset of $J$ that contains the $k$ smallest elements of $J$ (with respect to some fixed total ordering of $S$). Note that $T$ is well-defined since $|J| = n-k \geq k$.

Finally, we define $f(I) = T \cup I$. We claim that $f(I)$ is an $(n-k)$-element subset of $S$, and that $f$ is a bijection between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$.

To see that $f(I)$ is an $(n-k)$-element subset of $S$, note that $|f(I)| = |T \cup I| = |T| + |I| – |T \cap I|$, and $|T| = k$, $|I| = k$, and $|T \cap I| = 0$ (since $T$ and $I$ are disjoint). Therefore, $|f(I)| = 2k \leq 2(n/2) = n$, so $f(I)$ is a subset of $S$ with at most $n$ elements. Moreover, $|f(I)| = n-k$ since $|T| = k$ and $|J| = n-k$, so $f(I)$ is indeed an $(n-k)$-element subset of $S$.

To see that $f$ is a bijection, we will define its inverse. Let $A$ be an $(n-k)$-element subset of $S$, and let $J = S \setminus A$. Since $|A| = n-k$, we have $|J| = k$. Let $T$ be the $k$-element subset of $J$ that contains the $k$ smallest elements of $J$ (with respect to the same fixed total ordering of $S$ as before). Note that $T$ is well-defined since $|J| = k \geq k$. Finally, we define $f^{-1}(A) = T \cup A$. We claim that $f

To prove the given statement, let us first denote by $a_n$ the number of set-partitions of $[n]$ such that no consecutive numbers belong to the same block, and by $b_n$ the number of set-partitions of $[n-1]$.

Now, let us consider a set-partition $\pi$ of $[n]$ such that no consecutive numbers belong to the same block. We will show that we can obtain a set-partition $\pi’$ of $[n-1]$ in a one-to-one fashion by removing the element $n$ from $\pi$ and merging the block containing $n-1$ with the block containing $n$.

Formally, let $B_n$ be the block of $\pi$ containing $n$, and let $B_{n-1}$ be the block of $\pi$ containing $n-1$. Since $n$ and $n-1$ do not belong to the same block, we have $B_n \neq B_{n-1}$. We can define a new set-partition $\pi’$ of $[n-1]$ by setting $\pi'(i) = \pi(i)$ for $i=1,\ldots,n-2$, and by merging $B_n$ and $B_{n-1}$ into a single block in $\pi’$.

Conversely, given a set-partition $\pi’$ of $[n-1]$, we can obtain a set-partition $\pi$ of $[n]$ such that no consecutive numbers belong to the same block by adding the element $n$ to $\pi’$ as a new block. Formally, we can define a new set-partition $\pi$ of $[n]$ by setting $\pi(i) = \pi'(i)$ for $i=1,\ldots,n-1$, and by setting $\pi(n) = {n}$.

It is easy to see that these constructions are one-to-one, so the number of set-partitions of $[n]$ such that no consecutive numbers belong to the same block is equal to the number of set-partitions of $[n-1]$. Thus, we have $a_n = b_{n-1}$, as required.