Average place within a group in a competition with equal skill 
${43}$ equally strong sportsmen take part in a ski race; 18 of
them belong to club ${A}$, 10 to club  and 15 to club ${C}$. What is the
average place for (a) the best participant from club ${B}$; (b) the
worst participant from club ${B}$?


I've found the possible range of places the participant could get for both cases. In the case (a), the best participant from club ${B}$ can be at any place between $1$ and $34$. As for the case (b), the worst participant from club $B$ can get any place between $10$ and $43$. To find the average place I need to compute the expected (mean) value of the this variable. But I'm not sure how to find the chances for getting each place. I suppose they should be equal, but neither $\frac{1}{33}$ nor $\frac{1}{43}$ seem to give the right answer.
 A: Instead of finding the probability of each place and doing $\sum k\,p(k)$, you can use this trick.
When all the people are lined up in order of place, the ten people in the $B$ group will divide the line into $11$ contiguous sections. These sections are composed of the $43-10=33$ people from groups $A$ and $C$. Furthermore, each person in $A\cup C$ is equally likely to be in any of these $11$ spots. It follows that the expected number of $A\cup C$ members in each section is $33/11=3$. Using this, you should be able to deduce the average positions of the best from the $B$ club and the worst from the $B$ club.
Here is an illustrative picture. $B_1$ is the best, $B_{10}$ is the worst.
$$
\newcommand{\s}{\,\boxed{\,\;3\;\,}\,}
\s B_1
\s B_2
\s B_3
\s B_4
\s B_5
\s B_6
\s B_7
\s B_8
\s B_9
\s B_{10} 
\s
$$

Why is each of the gaps equally likely? Let $X$ be a particular person in clubs $A$ or $C$. We will group all of the orderings into sets of $11$, where $X$ is in a different gap in each set. Therefore, the fraction of orderings where $X$ is in any particular gap is $1/11$.
In each set, all of the people except for the $B$ club and $X$ will retain their positions, while $X$ will swap with some of the $B$ members as follows. Each $\cdots$ obscures a sequence of $A$ and $C$ members is the same for all $11$ orderings.
$$
\#1) \cdots X\cdots B_1\cdots B_2\cdots B_3\cdots B_4 \cdots B_5\cdots B_6\cdots B_7\cdots B_8\cdots B_9\cdots B_{10}\cdots\\\,\\
\#2)\cdots B_1\cdots X\cdots B_2\cdots B_3\cdots B_4 \cdots B_5\cdots B_6\cdots B_7\cdots B_8\cdots B_9\cdots B_{10}\cdots\\\,\\
\#3)\cdots B_1\cdots B_2\cdots X\cdots B_3\cdots B_4 \cdots B_5\cdots B_6\cdots B_7\cdots B_8\cdots B_9\cdots B_{10}\cdots\\\,\\
\vdots\\\,\\
\#11)\cdots B_1\cdots B_2\cdots B_3\cdots B_4 \cdots B_5\cdots B_6\cdots B_7\cdots B_8\cdots B_9\cdots B_{10}\cdots X\cdots
$$
A: First see the answer of Mike Earnest, and the comments following his answer.  Apparently, my computation numerically agrees with his answer.

Alternative approach:
$\underline{\text{Problem 1:}}$
For $k \in \{1,2,\cdots,34\},~$ let
$p(k)$ denote the probability that the highest member of B is in position $k$.
Then, the expected position for the highest member of $B$ will be $$\sum_{k=1}^{34} \left[k \times p(k)\right].$$
Therefore, the problem reduces to computing $p(k)$.
For $k$ to be the highest position of any member of B, two things have to happen:

*

*Positions $1,2,\cdots,(k-1)$ must be taken by someone in either A or C.  If $k$ = 1, we can denote this event as having probability $= 1.$  For $k > 1$ then the probability of this happening is 
$\displaystyle \frac{33!}{[33 - (k-1)]!} \times \frac{[43 - (k-1)]!}{43!} = \frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!}.$

*Position $k$ must be taken by someone in B, after all of the previous positions have been taken by someone in either A or C.  The probability of this happening is 
$\displaystyle \frac{10}{44 - k}.$
Therefore, the expected value of the position of the highest ranking member of B is
$$\left\{1 \times \frac{10}{43}\right\} ~+~
\left\{ ~\sum_{k=2}^{34} \left[k \times 
\frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!} \times \frac{10}{44 - k}\right]
~\right\}.
$$

$\underline{\text{Problem 2:}}$
Since most of the groundwork has already been done in Problem 1, the easiest approach for Problem 2 is to consider it to be the mirror problem to Problem 1.  That is, the math is all exactly the same, but for each event that would be multiplied by the scalar $k$, you instead multiply it by the scalar $(44 - k)$.
Therefore, there is no need to repeat the analysis from Problem 1.  The expected value of the position of the lowest ranking member of B is
$$\left\{43 \times \frac{10}{43}\right\} ~+~
\left\{ ~\sum_{k=2}^{34} \left[(44 - k) \times 
\frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!} \times \frac{10}{44 - k}\right]
~\right\}.
$$
A: The position of the highest placed person from $B$ (call this $X$) can be anywhere from $1$ to $34$, but these are not equally likely. For $X=1$ you simply need person $1$ to be from club $B$. This has probability $18/43$. For $X=10$, say, you need person $10$ to be from club $B$ (probability $18/43$) and none of the previous $9$ people to be from $B$. So you need the remaining $17$ members of $B$ (who can be in any $17$ of the remaining $42$ positions) all to be within the last $33$. This has probability $\frac{\binom{33}{9}}{\binom{42}{9}}$, so overall you get
$$\Pr(X=10)=\frac{18}{43}\times\frac{\binom{33}{9}}{\binom{42}{9}}.$$
