Revisit : Probability of the Ball being on the $M$-th Stair $\underline{\textbf{Overview}}$
This is an awkward self-answer question of
this original question.
I strongly suspect that the original question will soon be closed and then deleted.

A child is throwing a ball on a staircase with $N$ stairs.
Each step of the staircase is slightly slanted, so when the ball lands on it,
it rolls down to the next step. From the bottom step, the child throws the ball
back to a randomly chosen step.


If the game is played for a long time, what is the probability that at a
random moment, the ball will be on the $M$th step?


The child chooses the step to throw the ball back to with an arbitrary
(not necessarily uniform) distribution.

I regard the situation as awkward because the original problem has ambiguities that have to be resolved
before the problem can be attacked.  As part of the posting of this question itself,
I specify certain assumptions and try to justify them.
I regard the underlying question as non-trivial.  In addition to considering the
analysis in my self-answer, which is based on my stated assumptions, I invite
alternative assumptions that you might regard as just as reasonable, along with
your associated answer, based on these assumptions.
$\underline{\textbf{Initial Ambiguity Resolution}}$

*

*Assume that each of the elements in the set $\{1, 2, \cdots, N\}$ is associated
with one of the steps.  Further, assume that the lowest step is $(1)$ and the highest
step is $(N)$.


*Assume that the throw of the ball to a step is instantaneous, so that
(in effect) the ball is always on the collection of steps.


*Assume that the ball always stays on a specific step for exactly $1$ second,
and then transistions (i.e. falls) to the step below.


*Assume that the transistion from one step to a step below is instantaneous, so
that at any given moment, the ball is associated with one of the elements in
$\{1,2,\cdots, N\}.$


*Assume (similarly) that when the ball is on (1), it stays there for $1$ second,
then instantaneously transistions to the child, and then instaneously transitions
(i.e. is thrown) back to one of the steps.
$\underline{\textbf{Controversial Ambiguity Resolution}}$
This is the portion of my assumptions that I regard as iffy.  I don't see how
the original problem can be attacked without making assumptions (at least) similar
to those in the previous section.
Suppose, for example, that the last targeted step is $(T) \in \{1,2,\cdots,N\}$.
Then, it is impossible for the ball to be at any step above $(T)$, and each of
the steps in the subset $\{1,2,\cdots,T\}$ are equally likely.
Therefore, I don't see how the problem can be attacked without first considering
the probability that a specific step $T$ is the target.  In that regard, I have no
idea what to do with:

The child chooses the step to throw the ball back to with an arbitrary
(not necessarily uniform) distribution.

Initially, I assumed;

*

*Each time that the child throws the ball, the child chooses some element
at random from the set $\{1,2,\cdots,N\}$ and then throws the ball to the corresponding
step.


*Consequently, at any given moment in time, the probability that the last
targeted step before that moment is $(T)$ is $(1/N)$, regardless of the value of
$(T)$.
While I continue to accept the first assumption above as inevitable, I now think that
the second assumption above is untenable.
Consider the targetings (i.e. ball throwings) as a stream of instantaneous
actions, each of which is interspersed by k seconds, with $k$ dependent on the
specific step targeted.
That is, if the top step is targeted, then it will be N seconds of time before
the next targeting (i.e. ball throwing), while if the bottom step is targeted,
then it will be 1 second  of time before the next targeting.
Therefore, if you choose a random moment in time, at that moment, it is
(for example) N times more likely that the previous target was the top step
rather than the bottom step.
Consequently, 
setting $\displaystyle W = (1 + 2 + \cdots + N) = \frac{N(N+1)}{2}$, 
I am assuming that the probability of the last targeted step prior to a given moment
being step $(T)$ is: 
$\displaystyle p(T) = \frac{T}{W}.$
$\underline{\textbf{My Background}}$
About $50$ years ago I took a Probability course in college and did ok.  I have
forgotten much of the theory, and usually rely exclusively on intuition to attack
Probability problems.
If relevant, some decades ago I survived but have forgotten much of:

*

*"Real Analysis : Volume 1 : 2nd Ed." (Apostol, 1966).


*The first $(2/3)$ of "Elementary Number Theory" (Uspensky and Heaslett, 1938)
[through quadratic reciprocity].
$\underline{\textbf{My Work}}$
See my self - answer.
 A: You can approach this problem nicely using states.
Let $S_j$ the event that the ball is on the $j^{\text{th}}$ step. Then $P(S_{j-1}|S_j)=1$ for $j\in \{2,...,N\}$ and $P(S_j|S_1)=\frac{1}{N}$ for $j\in \{1,...,N\}$. These probabilities unveil the following $N\times N$ state transition matrix...
$$P=\begin{pmatrix}\frac{1}{N}&\frac{1}{N}&...&\frac{1}{N}&\frac{1}{N}\\ 1&0&...&0&0\\ 0&1&...&0&0\\ 0&0&...&0&0\\ 0&0&0&1&0\end{pmatrix}$$ You can verify that the stationary distribution $\pi=(\pi_1,\dots ,\pi_N)$ for this matrix is $$\pi=\frac{2}{N(N+1)}(N,N-1,...,1)$$ This stationary vector is unique since there is only one communicating class.
The components $\pi$ reveal exactly what you're looking for, namely the probability that the ball belongs to a particular state in the long run. For example, the probability the ball resides on the second step is approximately $\pi_2 =\frac{2(N-1)}{N(N+1)}$ if we played this game for a long time. So, after a long time, we see that $$P(S_M)\approx \pi_M=\frac{2(N-j+1)}{N(N+1)}$$
A: Repeating some of my pertinent assumptions:

*

*Each of the elements in the set $\{1, 2, \cdots, N\}$ is associated
with one of the steps.  The lowest step is $(1)$ and the highest
step is $(N)$.


*$\displaystyle W = (1 + 2 + \cdots + N) = \frac{N(N+1)}{2}$


*The probability of the last targeted step prior to a given moment
being step $(T)$ is: 
$\displaystyle p(T) = \frac{T}{W}.$


*Prior to a given moment, if the last targeted step was step $(T)$, then
the probability that the ball is now on step $M$, is 
$(0) ~: ~$ if $M \in \{T+1, T+2, \cdots, N\}$. 
$\displaystyle \left(\frac{1}{T}\right) ~: ~$ if $M \in \{1,2,\cdots, T\}.$
Therefore, the probability that at any given moment, the ball is on step $M$ is
$$\sum_{T=M}^N \left[p(T) \times \frac{1}{T}\right]. \tag1 $$
The expression in (1) above represents:
$$ \left[\frac{M}{W} \times \frac{1}{M}\right] + 
\left[\frac{M+1}{W} \times \frac{1}{M+1}\right] + 
\cdots + 
\left[\frac{N}{W} \times \frac{1}{N}\right].\tag2 $$
The expression in (2) above may be simplified to
$$\frac{1}{W} + \frac{1}{W} + \cdots + \frac{1}{W} ~: ~~(N + 1 - M) ~~\text{terms}. \tag3 $$
Therefore, the probability that the ball is on step $(M)$ is
$\displaystyle \frac{N + 1 - M}{W}.$
Edit
The same conclusion may be reached somewhat informally, by assuming that during a period
of $W$ seconds, the ball is thrown $N$ times, with each of the steps being targeted once.
Then, during these $W$ seconds, the ball will be associated with step $(M)$
for a total of $(N + 1 - M)$ seconds, since $(N + 1 - M)$ equals the number of
steps at or higher than step $(M)$.
