Elevators arrival time Question: A ten-floor building has two elevators. Suppose the positions of the elevators are independent and follow uniform distributions. On average it takes 1 minute for you to wait, on the bottom floor. Today only one elevator is available. On average how long do you need to wait?
Any hints on how to approach this problem would be appreciated.
 A: I assume both elevators are randomly distributed and that the elevator closest to the ground floor goes immediately to the ground floor. I work with a continuous distribution, but perhaps a discrete distribution is more what you want.
We have to find first the distrtibution of $X$, the distance of the elevator that is closest to the ground floor.
For $x$ between 0 and 10, we can calculate the CDF as
$$P(X < x) = 1 - \left( \frac{10-x}{10} \right)^2.$$
Now you can calculate the PDF by deriving the distribution as
$$f_X(x) = \frac{d}{dx} P(X < x) =  \frac{10-x}{50}.$$
The average distance for the closest elevator is now
$$\int_0^{10} x\cdot\frac{10-x}{50} \mathrm{d}x = 10/3.$$
When you have only one elevator the average distance is of course 5. This means that the average waiting time in case of 1 elevator will be
$$\frac{5}{10/3}\cdot 1\text{ min} = 90\;\text{s}.$$
A: When there are two lifts, the waiting time is the minimum of the waiting time for either lift.
We know that the $\text{cdf}_{\min}$ of the minimum of two IID random variables is $1-(1-\text{cdf})^2$*. Considering a uniform distribution in a unit interval,
$$\text{cdf}(t)=t\to\text{cdf}_{min}(t)=2t-t^2,$$
so that
$$\text{pdf}_{min}(t)=2-2t,$$
and the corresponding expectations are $\dfrac12$ and $\dfrac13$.
Hence the ratio of the two waiting times is $\dfrac32$.

*Justification:$$\mathbb P(\max(X_1,X_2)\le t)=\mathbb P(X_1\le t\land X_2\le t)=\mathbb P(X_1\le t)\mathbb P(X_2\le t)$$
and by symmetry
$$1-\mathbb P(\min(X_1,X_2)\le t)
=(1-P(X_1\le t))(1-\mathbb P(X_2\le t)).$$
A: When you wait for a single lift, all waiting times are equiprobable, in $[0,T]$. When there is a second, the probability of waiting for a time $t$ for the first lift is weighted by the probability that the second lift arrives before, and this probability is proportional to $t$.
Hence by averaging $t$ over the distributions ($\dfrac1T$ and $\dfrac t{T^2}$), the expectations of the waiting times are $\dfrac T2$ and $\dfrac T3$.
