Expected number of rolls until all dice are removed (followup)

Question

Suppose we roll $n$ dice, remove all the dice that come up 1, and roll the rest again. If we repeat this process, eventually all the dice will be eliminated. How many rolls, on average, will we make?

I believe this problem can be interpreted in 2 ways:

1. Calculating the total number of rolls (where rolling a single dice adds $+1$ to our count

2. Calculating the total rounds of rolls (where rolling all of the remaining dice adds $+1$ to our count

The second interpretation is discussed here and can be thought of as the maximum of n geometric random variables, which I understand.

For the first interpretation, could you use linearity of expectation? Given that each of the dice can be seen as a random variable $\mathrm{Geometric}\left(\frac16\right)$, the expected number of rolls until a $1$ is rolled is $6$. Then, there are $n$ RVs, so the sum of them is $6n$.

Answer 1

Let $X_i$ be the number of times you roll $i$-th die until you get $1$, then the number of rolls you need to make until you remove all dice is $$ Y=\max_{i=1,2,\dots,n} X_i. $$ Since $X_i$s are independent, for $k=1,2,\dots$ $$ \mathbb{P}(Y\le k)=\mathbb{P}(X_1\le k,X_2\le k,\dots,X_n\le k) =\mathbb{P}(X_i\le k)\mathbb{P}(X_2\le k)\dots\mathbb{P}(X_n\le k) =\mathbb{P}(X_i\le k)^n. $$ Since $X_i$ are Geometric, $\mathbb{P}(X_i\le k)=1-\mathbb{P}(X_i> k)=1-\left(\frac56\right)^k$. Hence $$ \mathbb{P}(Y\le k)=\left[1-\left(\frac56\right)^k\right]^n $$ and \begin{align*} \mathbb{E}(Y)&=\sum_{k=0}^\infty \mathbb{P}(Y> k) =\sum_{k=0}^\infty \left(1-\left[1-\left(\frac56\right)^k\right]^n\right)\\ &=\sum_{k=0}^\infty \sum_{i=1}^n (-1)^{i+1}\binom{n}{i} \left(\frac56\right)^{ki} =\sum_{i=1}^n (-1)^{i+1}\binom{n}{i}\sum_{k=0}^\infty \left(\frac56\right)^{ki} \\ & =\sum_{i=1}^n \frac{(-1)^{i+1}}{1-(5/6)^i}\binom{n}{i} \end{align*} which is $6$ for $n=1$, $8\frac8{11}$ for $n=2$, $10\frac{556}{1001}$ for $n=3$, etc.