Suppose I get in an elevator every morning and am in it for exactly one minute. A continuous loop of music plays in the elevator and is 10 minutes long.

If my daily arrival time is uniform in the time frame from 8:00 to 8:10, what is the expected number of days until I have heard the entire 10 minutes of the loop? Assume I take the stairs down at the end of the day and so am only in the elevator for one minute per day.

  • $\begingroup$ No - I have found a published solution, but have not had time to digest it and post. $\endgroup$
    – soakley
    Apr 5, 2018 at 13:23

1 Answer 1


This question reminds me of Wilfrid Kendall's dead leaves simulation, which he uses to explain the difference between forward and backward sampling.

Given that the problem can be formalised through uniform spacings, this highly detailed answer on CV is connected with this question.

Indeed, if $U_1,\ldots,U_T$ denote the mid-times of the elevator trips on days 1, 2, ..., T, assumed to be Uniform on $(0,L)$, and if $U_{(1)},U_{(2)},\ldots,U_{(T)}$ are the corresponding order statistics, the condition for covering the entire L mn musical programme is $$U_{(2)}-U_{(1)}<1,U_{(3)}-U_{(2)}<1,\ldots,U_{(T)}-U_{(T-1)}<1,$$ and $$U_{(1)}+L-U_{(T)}<1$$ Defining the Dirichlet vector $\Delta$ associated with the spacings $U_{(i)}-U_{(i-1)}$ $(2\le i\le T+1)$, with $U_{(T+1)}=L$, $$\Delta=\frac{1}{L}(U_{(1)},U_{(2)}-U_{(1)}, U_{(3)}-U_{(2)},\ldots,U_{(T)}-U_{(T-1)},1-U_{(T)})$$ which is Dirichlet $\mathcal{D}_{T+1}(1,\ldots,1)$, the question is thus almost equivalent to finding the law of the maximal component of $\Delta$, to determine$$\mathbb{P}\left(\max_{1\le i\le T+1}\Delta_i<1\big/L\right)$$for which the approximation provided in the above mentioned answer applies. Obviously, this is not the entire answer to the question but it provides an interesting entry.

The expectation of $T$ can then be deduced from $$\mathbb{P}(\max_i\Delta_i\le 1/L) = \sum_{j=0}^{L} { T+1 \choose j } (-1)^j (1-j/L)^T,$$ since \begin{align*} \mathbb{E}[T]&=\sum_{t=L+1}^\infty t \mathbb{P}(T=t)\\ &=\sum_{t=L+1}^\infty t \mathbb{P}(T\ge t)-\sum_{t=1}^\infty t \mathbb{P}(T\ge t+1)\\ &=(L+1)\mathbb{P}(T\ge L+1)+\sum_{t=L+2}^\infty \mathbb{P}(T\ge t)\\ &=(L+1)\{1-\mathbb{P}(T\le L)\}+\sum_{t=L+2}^\infty \{1-\mathbb{P}(T\le t-1)\}\\ &=L+1+\sum_{t=L+1}^\infty \{1-\mathbb{P}(T\le t)\}\\ &=L+1+\sum_{t=L+1}^\infty \left\{1-\mathbb{P}\left(\max_{1\le i\le T+1}\Delta_i\le1\big/L\right)\right\}\\ &=L+1+\sum_{t=L+1}^\infty \left\{1-\sum_{j=0}^{L} { t+1 \choose j } (-1)^j (1-j/L)^T\right\}\\ \end{align*} A quick simulation shows the accuracy of the expectation

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The approximation to the actual problem of "the continuous coupon collector" can also be evaluated by simulation and the regression of the simulated expected number $\hat T$ on the (approximate) expected number $\hat T_0$ shows a good fit of the formula $$\hat T=3/2+\hat T_0$$

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  • 1
    $\begingroup$ So, how many days? $\endgroup$
    – pajonk
    Mar 23, 2018 at 6:04

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