, 2022

Solution:

Insert a player with skill level 0, who will be the first active player (and lose their first game).
If Alice plays after any of the players with skill level $12,13, \ldots, 21$, which happens with probability $\frac{10}{11}$, then she will play exactly 1 game.
If Alice is the first of the players with skill level $11,12, \ldots, 21$, which happens with probability $\frac{1}{11}$, then there are an expected $\frac{10}{12}$ players between her and someone better than her. Thus, she plays an expected $2+\frac{10}{12}=\frac{17}{6}$ games.
Alice will only play the player with skill 0 if she is the first of all other players, which happens with probability $\frac{1}{21}$.
The final answer is
$$\frac{10}{11} \cdot 1+\frac{1}{11} \cdot \frac{17}{6}-\frac{1}{21}=\frac{47}{42}$$

Solution:

Replace 21 by $n$ and 11 by $k$. The general formula is $\frac{n+1}{(n-k+1)(n-k+2)}+1-\frac{1}{n}-[k=n]$. The problem is roughly equivalent to picking a random permutation of $1, \ldots, n$ and asking the expected number of prefix maximums that are equal to $k$. For the first $m$ elements, the probability is equal to
$$\begin{aligned} P(\max \text{ of first } m=k) & =P(\max \text{ of first } m \leq k)-P(\max \text{ of first } m \leq k-1) \\ & =\frac{\binom{k}{m} \cdot m!\cdot(n-m)!}{n!}-\frac{\binom{k-1}{m} \cdot m!\cdot(n-m)!}{n!} \\ & =\frac{\binom{k}{m}}{\binom{n}{m}}-\frac{\binom{k-1}{m}}{\binom{n}{m}} \\ & =\frac{\binom{k-1}{m-1}}{\binom{n}{m}} \end{aligned}$$
$$\begin{aligned} E[\text{ prefix } \max =k] & =\sum_{m=1}^{k} \frac{\binom{k-1}{m-1}}{\binom{n}{m}} \\ & =\sum_{m=1}^{k} \frac{(k-1)!m!(n-m)!}{(k-m)!(m-1)!n!} \\ & =\frac{(k-1)!}{n!} \sum_{m=1}^{k} \frac{m(n-m)!}{(k-m)!} \\ & =\frac{(k-1)!(n-k)!}{n!} \sum_{m=1}^{k} m\binom{n-m}{n-k} \end{aligned}$$
Now a combinatorial interpretation of the sum is having $n$ balls in a row, choosing a divider between them, and choosing 1 ball on the left side of the divider and $n-k$ balls on the right side of the divider ( $m$ corresponds to the number of balls left of the divider). This is equal to choosing $n-k+2$ objects among $n+1$ objects and letting the second smallest one correspond to the divider, which is $\binom{n+1}{n-k+2}$. Therefore the answer is
$$\frac{(k-1)!(n-k)!}{n!} \cdot \frac{(n+1)!}{(n-k+2)!(k-1)!}=\frac{n+1}{(n-k+1)(n-k+2)}$$