$\newcommand{\N}{\mathbb N}$The question can be apparently formalized as follows.
We have a random walk $(X_1,X_2,\dots)$ on the state space $\N:=\{1,2,\dots\}$ starting at point $1$ at time moment $1$ -- that is, $X_1=1$. If the walker is at point $j\in\N$ at a time moment $t\in\N$, then it moves to the right by $1$ with a probability $b_{t,j}\in[0,1]$ or stays at the point $j$ with probability $1-b_{t,j}$:
\begin{equation*}
\begin{aligned}
&P(X_{t+1}=j+1|X_t=j,X_{t-1},\dots,X_1) \\
&=b_{t,j} =1-P(X_{t+1}=j|X_t=j,X_{t-1},\dots,X_1).
\end{aligned}
\end{equation*}
The question is then this:
Given any $n\in\N$ and any probability distribution $\pi$ on the set $[n]:=\{1,\dots,n\}$, do there necessarily exist numbers $b_{t,j}\in[0,1]$ for all natural $t$ and $j$ such that the distribution of $X_n$ is $\pi$?
The answer to this question is yes, which can be proved by induction on $n$. Indeed, for $n=1$ this claim is trivial. To make the induction step, we only have to prove the following:
Take any natural $n\ge2$. Take any nonnegative real numbers $p_1,\dots,p_n$ (corresponding to the desired distribution of $X_n$) such that $p_1+\dots+p_n=1$. Then there exist (i) nonnegative real numbers $q_1,\dots,q_{n-1}$ (corresponding to an appropriate distribution of $X_{n-1}$) such that $q_1+\dots+q_{n-1}=1$ and (ii) numbers $b_1,\dots,b_{n-1}$ in $[0,1]$ such that
\begin{equation*}
p_j=q_{j-1}b_{j-1}+q_j(1-b_j)\text{ for all $j\in[n]$,} \tag{1}\label{1}
\end{equation*}
where $q_0:=0$, $q_n:=0$, $b_0:=0$, and $b_n:=0$.
But such numbers $q_1,\dots,q_{n-1},b_1,\dots,b_{n-1}$ can be explicitly and simply constructed e.g. as follows:
\begin{equation*}
q_1:=p_1+p_2,\ q_2:=p_3,\ \dots,\ q_{n-1}:=p_n,
\end{equation*}
\begin{equation*}
b_1:=\frac{p_2}{p_1+p_2},\ b_2:=1,\ \dots,\ b_{n-1}:=1
\end{equation*}
(if $p_1+p_2=0$, let $b_1:=1$). Indeed, then all the equalities \eqref{1} will hold. $\quad\Box$
So, to get the desired distribution, we are just successively splitting the probability of the state $1$ while carrying the other probabilities to the next time step.
That is, once the walker has left the state $1$, it will march deterministically to the right, one unit step at a time. If at a time moment $t\in\{1,\dots,n-1\}$ the walker still remains in the state $1$, then at the next time moment, $t+1$, it will still be in the state $1$ with the (conditional) probability $\dfrac{p_1+\dots+p_{n-t}}{p_1+\dots+p_{n-t}+p_{n-t+1}}$ (or will move to the state $2$ with the complementary probability $\dfrac{p_{n-t+1}}{p_1+\dots+p_{n-t}+p_{n-t+1}}$).
As an illustration, here is the tree diagram of getting the probabilities $0.2,0.3,0.4,0.1$ for the respective states $1,2,3,4$ at time $t=4$:
The root of this tree, at the top, corresponds to time moment $t=1$, followed by the layers corresponding to time moments $t=2,3,4$. The nodes of the tree corresponding to the time moments $t=1,2,3,4$ are labeled by the probabilities $P(X_t=j)$ for $j=1,\dots,t$.
