Expected number of games so that every player has been an Imposter in Among Us Imagine you have $n$ players. In each game, $k$ $(k\leq n)$ players are chosen randomly to do whatever.

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*By game $G$, what is the probability that every player has been chosen at least once?

*What is the expected number of games that have to be played so that every player was chosen at least once?

Of course, when $G < n/k$ the probability is zero.
This idea came from the game Among Us, where there are $n$ players ($n\leq10$) and in each game you have $k$ imposters (usually $1$, $2$ or $3$).
 A: This can be found using the principle of inclusion exclusion. You first find the probability that someone had not been picked yet by adding up, for each player, the probability they had not been picked. The result is $n\big(\binom{n-1}k/\binom{n}k\big)^G$. But then you must add back in the probabilities, for each pair of players, the probability that they were both not picked, which is $\binom{n}2\big(\binom{n-2}k/\binom{n}k\big)^G$. Continuing in this fashion, the probability that everyone was picked is
$$
\mathbb P(\text{all players were picked in $G$ rounds})=\sum_{i=0}^n(-1)^i\binom{n}i\left[\binom{n-i}k\big/\binom{n}k\right]^G
$$
Finally, let $X$ be the number of rounds. Note that the event $\{X> G\}$ occurs if and only if not everyone was picked in $G$ rounds. Using the "layer cake formula" for expected value,
\begin{align}
\mathbb E[X]
&=\sum_{G=0}^\infty P(X>G)
\\&=\sum_{G=0}^\infty\sum_{i=1}^n(-1)^{i+1}\binom{n}i\left[\binom{n-i}k\big/\binom{n}k\right]^G
\\&=\sum_{i=1}^n(-1)^{i+1}\binom{n}i\frac{1}{1-\binom{n-i}k\big/\binom{n}k}
\end{align}
Here is the formula in Wolfram Alpha so you can see specific values. As a special case, when $k=1$, this is just the Coupon Collector's problem, for which the expected number of trials is $n(1+1/2+\dots+1/n)$.
