Intuitive Understanding of exponential distrbution

Question

$\textbf{Problem}$: An athletic facility has $5$ tennis courts. Pairs of players arrive at the courts and use a court for an exponentially distributed time with mean $40$ minutes. Suppose a pair of players arrives and finds all courts busy and k other pairs waiting in queue. What is the expected waiting time to get a court?

$\textbf{Answer}$: As long as the pair of players are waiting, all five courts are occupied by other players. When all five courts are occupied, the time until a court is freed up is exponentially distributed with mean $40/5=8 $ minutes. For our pair of players to get a court, a court must be freed up $k+1$ times. Thus, the expected waiting time is $8(k + 1)$.

The issue that I'm having in understanding is why does "The time until a court is freed up is exponentially distributed with mean $40/5=8$ minutes" rather than simply $40$ minutes. Since every court is independent and using the memoryless property.

On the contrary, if I look at $$Z=min(X_{1}, X_{2}, X_{3}, X_{4}, X_{5})$$. where $X_{i}$ is time taken by a pair to play in court $i$ and is exponentially distributed time with mean $40$ minutes.
Then $Z$ itself becomes and exponentially distributed with mean $40*5 = 200$ minutes.

I am unable to understand it intuitively. Any help is appreciated.

Answer 1

Each $X_i$ has rate $\lambda=\frac1{40}$ (with minutes as the time unit), so $Z$ has $\lambda=\frac5{40}=\frac18$ for a mean of $8$ minutes. You have applied the formula incorrectly: the rates add, not the means.

Answer 2

Let $Z=\min(X_1,...,X_5)$.

We have $X_1=Z$ with probability ${\large{\frac{1}{5}}}$, and by the memoryless property we have $$ E[X_1-Z{\,\mid\,}X_1 > Z]=40 $$ hence \begin{align*} & E[X_1-Z] = \left({\small{\frac{1}{5}}}\right)\!(0) + \left({\small{\frac{4}{5}}}\right)\!(40) \\[4pt] \implies\;& E[X_1-Z]=32 \\[4pt] \implies\;& E[X_1]-E[Z]=32 \\[4pt] \implies\;& 40-E[Z]=32 \\[4pt] \implies\;& E[Z]=8 \\[4pt] \end{align*}