Validating Probability Theory analysis about average place in a group Overview 
In this posting I present a recent mathSE question.  In the recent question, there
were (in effect) three different approaches taken to answer the question.  All
three approaches reached the same conclusion.  I understood two of the approaches,
but not the third.
In this posting, I am going to discuss the original question as well as the
three approaches.
My question is 
Why is the third approach valid?

The Original Question 
The question was originally posted
here.
The question is 

$43$  equally strong sportsmen take part in a ski race; $18$ of them belong to club A,
$10$ to club B and $15$ to club C. What is the average place for the best participant
from club B?

You will notice that I omitted the 2nd half of the original question as not being
relevant to the issue that I wish to discuss.

My Background
50 years ago, as a Math major, I took a Probability course.  I understood it at
the time.  Now, I rely almost exclusively on intuition to attack Probability
problems.  I have been able to understand the proof of Linearity of Expectation
with respect to dependent events, as proven in
this article.

The First Approach 
Especially Lime and I each posted a separate answer to the original question.  Both of our answers (in effect) used the
same computational approach.  The idea was that the best person from B might
occupy any of positions $(1)$ through $(34)$.  Therefore, the expected position
can be computed as
$$\sum_{k=1}^{34} \left[k \times p(k)\right]$$
where $p(k)$ is the probability that the best person from B occupied place $(k)$.
The computation that I gave was:
$$\left\{1 \times \frac{10}{43}\right\} ~+~
\left\{ ~\sum_{k=2}^{34} \left[k \times 
\frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!} \times \frac{10}{44 - k}\right]
~\right\}.
$$
This Wolfram Alpha Link
indicates that the above summation results in an answer of $(4)$.

The Second Approach 
In a comment following the answer of Especially Lime, in the original question, the original poster (i.e. the OP)
of the original question, RainCatalyst, posted the following analysis:
Consider the $33$ people who are not in B.  For $i \in \{1,2,\cdots,33\}$, let
$x_i$ denote an indicator variable assigned to one of the people that is not in B.
Define the value of $x_i$ as follows:

*

*$x_i = 1 ~: ~$ if the $i$-th player has a better result than everyone from group B

*$x_i = 0 ~: ~$ otherwise.

Then $\displaystyle \left\{ ~X = \sum_{i=1}^{33} x_i ~\right\}~$ represents the total number of players better
than everyone from B. If you consider any random person not in B, along with the
$(10)$ people in B, you have a group of $(11)$ people, anyone of whom is equally likely
to have the highest position.  Therefore, for each variable $x_i$, the probability
that $x_i$ = $(1)$, rather than $(0)$ is $\frac{1}{11}.$
Therefore, the expected value of $X$ is $\sum_{i=1}^{33} \left[\frac{1}{11}\right].$ 
Since the expected value of $X$ is $(3)$, the expected value of the
person in B who finished the highest is therefore $(4)$.  This is because you expect that there are
$(3)$ people from A or C that finished ahead of the best person from B.
I understand this analysis, which depends on Linearity of Expectation, and it does make sense to me.

The Third Approach
This is the one that I have trouble with.  It comes from Mike Earnest's answer to the original question. 
I will try to paraphrase his thinking.  However, it is probably best if you examine
his answer directly, as well as his comments following his answer.

His approach (in effect) is to label the people in B as $b_1, b_2, \cdots, b_{10}$,
where $b_1$ finished the best, and $b_{10}$ finished the worst.  Then, there are
$11$ positional gaps created: before $b_1$, and after each of $b_1$ through $b_{10}$.


Each of the $(33)$ people that are not in B are equally likely to be positioned in any of
the $(11)$ postional gaps.  Therefore, you would expect each positional gap to contain (on average) $(3)$ people
who are not in B.  Under this presumption, you expect that the first positional gap contains $(3)$ people who
are not in B.  Therefore, the best person from B is expected to finish in position $(4)$.

I certainly don't regard it as coincidental that his answer matches the first two approaches.  However, I
have trouble understanding why his analysis leads to the right answer.
Simply because the most likely distribution of the 33 people in the 11 positional gaps is 
$3 - 3 - 3 - \cdots - 3$, does not imply that other distributions can not occur.  Furthermore,
when examining the possible values of the size of the first positional gap, which
can range from $(0)$ through $(33)$, you have the complication that each size $(k)$
must be given a weight of $(k + 1)$.  This is because if the size of the first positional
gap is $(k)$, then $b_1$ finished in position $(k+1)$.  So, it seems to require that
some consideration be given to the probability assigned to each value of $(k)$, as
$(k)$ ranges from $(0)$ to $(33)$.
Stating the question differently, I understand that you expect that the size
of the first positional gap is $(3)$.  I simply don't understand the full significance of
this expectation.
As a better way of illustrating my uncertainty, consider two related problems:

*

*You roll a die once: what is the expected number that will occur on the die?

*You roll a die once; what is the square of the expected number that will occur on the die?

In both of the above problems, you expect a $(3.5)$ to be rolled.  In the first problem, each
of the $(6)$ possible rolls is given a weight equal to the size of the number.  So, with such
a problem, rolling a $(1)$ is given a weight of $(1)$, while rolling a $(6)$ is given a weight
of $(6)$.  These weights increase linearly, thus allowing one to immediately conclude,
without computation, that the answer to the first problem is $(3.5)$.
In contrast, in the 2nd problem, the roll of $(1)$ is given a weight of $(1)$, while the roll
of $(6)$ is given a weight of $(36)$, so the answer to the 2nd problem is not $(3.5)^2$.
The rebuttal argument is that in the original problem, as the size of the positional gap changes from $(k)$ to $(k+1)$,
the effect on the overall computation is linear.
My counter to this rebuttal is that as you change the size of the positional gap from $(k)$ to $(k+1)$,
$p(k)$, as referred to in the first approach, is not constant.
Therefore, I have trouble placing confidence in the validity of the 3rd approach.

Addendum
Responding to the answer of Henry. 
This is weird.  On the one hand, I (still) feel unsure of you analysis, for the same reason that I felt unsure of Mike Earnest's original answer.  That is, I had trouble determining the significance of the expectation that the size of the positional gap following the artifical person is $(3)$.
However, I just thought of a minor variation of your (Henry's) answer that I do feel confident of.
Assume that when you add the (artificial) 11th person from B, his position around the table is unknown.  So, instead of finding and removing the artificial person, you take a different approach.
Each of the $(11)$ people around the table that are from B could be the artificial person from B.  So, you run $(11)$ simulations, where with each simulation, you assume that a specific one of the $(11)$ people from B is the artificial person.  You ask yourself, with each of these $(11)$ simulations, how many people from A or C follow (clockwise) this artificial person.
Since you are running $(11)$ simulations, you will end up counting each person from A or C exactly once.  Therefore, the expected number of people from A or C that follows any given artificial person from B is $(3)$.
Actually, this becomes virtually identical to the 2nd approach by RainCatalyst.  So, it seems that I can't get away from his approach.
 A: If you understand the second approach, you should understand the third approach. They are the same logic, just different ways of phrasing.
While the second approach says "$x_i=1$", the third approach says "person number $i$ is in the leftmost gap," but they mean the same thing. They both argue that the probability of this happening is $1/11$, and there are $33$ other people, so on average, we get $33/11=3$. This is valid because of linearity of expectation, though the third approach does not mention this explicitly.
When you say.

"I understand that you expect the size of the first gap to be $3$,"

then you must understand my approach. The size of first gap is three, on average. The position of $B$'s best player is one plus the size first gap. Therefore, letting $X$ be the number of people in the first gap, since
$$
E[X+1]=E[X]+1=3+1=4,
$$
we get the expected position is $4$.
I cannot tell what your complaint is with the contribution of $(k+1)$ and $p(k)$, etc has to do with my method. My approach does not even refer to $p(k)$ at all. At no point do I take a weighted average of the possible of the gap sizes  between $0$ and $33$; I avoid this entirely by using linearity of expectation. Therefore, I think your complaint is in the wrong ball park.
A: Here is a variety of Mike Earnest's answer.
Suppose we instead sit $18$ As, $11$ Bs (not $10$ because we have added a special extra B) and $15$ Cs around a circular table.

*

*Going clockwise, in the gap between any of the $11$ successive pairs of $B$s there can be $0$ to $33$ As and Cs

*By symmetry, the distributions of the numbers of As and Cs in each gap are identical.  They are not independent, but that does not matter for expectations

*The expected number of As and Cs in each gap is therefore $\frac{33}{11}=3$
Now spot and remove the special extra B and unwind the remaining people so they are in line.  The expected number of As and Cs in each gap is still $3$ as are the expected number before the first B and after the last B
So in the original problem, the expected number before the fastest $B$ is $3$, meaning the expected position of the fastest B is $4$th
