Expected Value of Elevator Buttons

For each $1\leq i\leq 10$ , let $A_i$ denote the event that button $i$ is pressed, let $A'_i$ denote the complementary event that button $i$ is not pressed, and let $\mathbb{I}_{A_i} = \begin{cases}1 \text{ if button i is pressed} \\ 0 \text{ otherwise}\end{cases}$ The first feature of these random variables (called "indicator random variables") that we are interested in right now is $\mathbb{E}\left[\mathbb{I}_{A_i}\right] = \mathbb{P}\left(A_i\right) = 1 - \mathbb{P}\left(A'_i\right) = 1 - \left(\frac{9}{10}\right)^7 \approx 0.5217.$

The second feature is that $N = \text{number of distinct buttons pushed} = \sum_{i=1}^{10} \mathbb{I}_{A_i}.$ and therefore we may compute the expected number of distinct buttons pushed by $\mathbb{E}\left[N\right] = \mathbb{E}\left[\sum_{i=1}^{10} \mathbb{I}_{A_i}\right] = \sum_{i=1}^{10}\mathbb{E}\left[\mathbb{I}_{A_i}\right] = \sum_{i=1}^{10} \left(1 - \left(\frac{9}{10}\right)^7\right) = 10\cdot\left(1 - \left(\frac{9}{10}\right)^7\right) \approx \boxed{5.217}$

Suppose that $r$ passengers get into a lift with $n \ge r$ possible destination floors. If $N$ is the number of distinct floor buttons pressed, then $\mathbb{P}[N=m] \; = \; \frac{1}{n^r} \binom{n}{m} m! \left\{ {r \atop m} \right\} \hspace{2cm} 1 \le m \le r$ where $\left\{ {r \atop m} \right\}$ is a Stirling number of the second kind. This is because there are $\left\{ {r \atop m} \right\}$ ways of sorting $r$ passengers into $m$ groups, and then $\binom{n}{m} m!$ ways of determing which floor button is associated to each group, while there are a total of $n^r$ different combinations of floor choices that $r$ passengers can make.

Using the standard combinatorial identity $\left(x\frac{d}{dx}\right)^r f(x) \; = \; \sum_{m=1}^r \left\{ {r \atop m}\right\} f^{(m)}(x) x^m$ for any $r \ge 1$ and any function $f$ , we deduce, applying this result to $f(x) = (1 + x)^n$ , that $\sum_{k=0}^n \binom{n}{k}k^r x^k \; = \; \sum_{m=1}^r \left\{ {r \atop m} \right\} \binom{n}{m} m! (1+x)^{n-m}x^m$ and hence, after a change of variable, $\sum_{m=1}^r \binom{n}{m} m! \left\{{r \atop m}\right\} t^m \; = \; \sum_{k=0}^n \binom{n}{k} k^r t^k(1-t)^{n-k}$ we see that the probability generating function of $N$ is $G_N(t) \; = \; \frac{1}{n^r} \sum_{m=1}^r \binom{n}{m} m! \left\{{r \atop m}\right\} t^m \; = \; \frac{1}{n^r}\sum_{k=0}^n \binom{n}{k} k^r t^k(1-t)^{n-k}$ Thus $G_N'(t) \; = \; \frac{1}{n^r} \sum_{k=0}^n \binom{n}{k} k^r \left\{ kt^{k-1}(1-t)^{n-k} - (n-k)t^k(1-t)^{n-k-1}\right\}$ and nearly all terms in this expression contain a factor of $1-t$ , so that $\mathbb{E}[N] \; = \; G_N'(1) \; = \; \frac{1}{n^r}\left\{ n \times (n-1)^r \times (-1) + 1 \times n^r \times n\right\} \; = \; \frac{n^r - (n-1)^r}{n^{r-1}}$ In the case that $r=7$ and $n=10$ we have $\mathbb{E}[N] \; = \; \frac{10^7 - 9^7}{10^6} \; = \; \frac{5217031}{10^6} \; =\; \boxed{5.217031}$